Circles Class 10 Extra Questions Maths Chapter 10 with Solutions Answers

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Extra Questions for Class 10 Maths Circles with Answers Solutions

Extra Questions for Class 10 Maths Chapter 10 Circles with Solutions Answers

Circles Class 10 Extra Questions Very Short Answer Type

Question 1.
If a point P is 17 cm from the centre of a circle of radius 8 cm, then find the length of the tangent drawn to the circle from point P.
Solution:
Circles Class 10 Extra Questions Maths Chapter 10 with Solutions Answers 1
OA ⊥ PA (∵ radius is ⊥ to tangent at point of contact)
∴ In ∆OAP, we have
PO2 = PA2 + AO2
⇒ (17)2 = (PA)2 + (8)2
(PA)2 = 289 – 64 = 225
⇒ PA = √225 = 15
Hence, the length of the tangent from point P is 15 cm.

Question 2.
The length of the tangent to a circle from a point P, which is 25 cm away from the centre, is 24 cm. What is the radius of the circle?
Solution:
Circles Class 10 Extra Questions Maths Chapter 10 with Solutions Answers 2
∵ OQ ⊥ PQ
∴ PQ2 + QO2 = OP2
⇒ 252 = OQ2 + 242
or OQ = √625 – √576
= √49 = 7 cm

Question 3.
In Fig. 8.6, ABCD is a cyclic quadrilateral. If ∠BAC = 50° and ∠DBC = 60° then find ∠BCD.
Solution:
Circles Class 10 Extra Questions Maths Chapter 10 with Solutions Answers 3
Here ∠BDC = ∠BAC = 50° (angles in same segment are equal)
In ABCD, we have
∠BCD = 180° – (∠BDC + ∠DBC)
= 180° – (50° + 60°)= 70°

Question 4.
In Fig. 8.7, the quadrilateral ABCD circumscribes a circle with centre O. If ∠AOB = 115°, then find ∠COD.
Solution:
Circles Class 10 Extra Questions Maths Chapter 10 with Solutions Answers 4
∵ ∠AOB = ∠COD (vertically opposite angles)
∴ ∠COD = 115°

Question 5.
In Fig. 8.8, AABC is circumscribing a circle. Find the length of BC.
Circles Class 10 Extra Questions Maths Chapter 10 with Solutions Answers 5
Solution:
AN = AM = 3 cm [Tangents drawn from an external point]
BN = BL = 4 cm [Tangents drawn from an external point]
CL = CM = AC – AM = 9 – 3 = 6 cm
⇒ BC = BL + CL = 4 + 6 = 10 cm.

Question 6.
In Fig. 8.9, O is the centre of a circle, PQ is a chord and the tangent PR at P makes an angle of 50° with PQ. Find ∠POQ.
Solution:
Circles Class 10 Extra Questions Maths Chapter 10 with Solutions Answers 6
∠OPQ = 90° – 50° = 40°
OP = OQ [Radii of a circle]
∠OPQ = ∠OQP = 40°
(Equal opposite sides have equal opposite angles)
∠POQ = 180° – ∠OPQ – ∠OQP
= 180° – 40° – 40° = 100°

Question 7.
If two tangents inclined at an angle 60° are drawn to a circle of radius 3 cm, then find the length of each tangent.
Solution:
Circles Class 10 Extra Questions Maths Chapter 10 with Solutions Answers 7
In Fig. 8.10
∆AOP ≅ ∆BOP (By SSS congruence criterion)
∠APO = ∠BPO = \(\frac{60°}{2}\) = 30°
In ∆AOP, OA ⊥ AP
∴ tan 30° = \(\frac{OA}{AP}\)
⇒ \(\frac{1}{\sqrt{3}}\) = \(\frac{3}{AP}\)
⇒ AP = 3√3 cm

Question 8.
If radii of two concentric circles are 4 cm and 5 cm, then find the length of each chord of one circle which is tangent to the other circle.
Solution:
Circles Class 10 Extra Questions Maths Chapter 10 with Solutions Answers 8
OA = 4 cm, OB = 5 cm
Also, OA ⊥ BC
∴ OB2 = OA2 + AB2
⇒ 52 = 42 + AB2
⇒ AB = √25 – √16 = 3 cm
⇒ BC = 2 AB = 2 × 3 = 6 cm

Question 9.
PQ is a tangent drawn from a point P to a circle with centre O and QOR is a diameter of the circle such that ∠POR = 120° then find ∠OPQ.
Solution:
Circles Class 10 Extra Questions Maths Chapter 10 with Solutions Answers 9
∠OQP = 90°
∠QOP = 180° – 120° = 60°
∠OPQ = 180° – ∠OQP – ∠QOP
= 180° – 90° – 60°
= 30°

Question 10.
From an external point P, tangents PA and PB are drawn to a circle with centre O. If ∠PAB = 50°, then find ∠AOB.
Circles Class 10 Extra Questions Maths Chapter 10 with Solutions Answers 10
Solution:
∵ PA = PB ⇒ ∠BAP = ∠ABP = 50°
∴ ∠APB = 180° – 50° – 50° = 80°
∴ ∠AOB = 180° – 80° = 100°

Question 11.
In PQ is a tangent at a point C to a circle with centre O. If AB is a diameter and ∠CAB = 30°, find ∠PCA.
Solution:
Circles Class 10 Extra Questions Maths Chapter 10 with Solutions Answers 11
∠ACB = 90° (Angle in the semicircle)
∠CAB = 30° (given)
In ∆ABC,
90° + 30° + ∠ABC = 180°
⇒ ∠ABC = 60°
Now, ∠PCA = ∠ABC (Angles in the alternate segment)
∴ ∠PCA = 60°
or
Construction: Join O to C.
∠PCO = 90° [∵ Line joining centre to point of contact is perpendicular to PQ]
In ∆AOC, OA = OC [Radii of circle]
∴ ∠OAC = ∠OCA = 30° [Equal sides have equal opp. angles]
Now, ∠PCA = ∠PCO – ∠ACO
= 90° – 30° = 60°

Question 12.
In fig. 8.16, there are two concentric circles with centre O. PRT and PQS are tangents to the inner circle from a point P lying on the outer circle. If PR = 5 cm, find the length of PS.
Solution:
Circles Class 10 Extra Questions Maths Chapter 10 with Solutions Answers 12
PQ = PR = 5 cm [∵ Tangents drawn from external point are equal] PK
∴ PS = 2PQ = 10 cm [∵ Perpendicular drawn from centre to the chord bisects the chord]

Circles Class 10 Extra Questions Short Answer Type 1

State true or false for each of the following and justify your answer (Q. 1 to 3)

Question 1.
AB is a diameter of a circle and AC is its chord such that ∠BAC = 30°. If the tangent at C intersects AB extended at D, then BC = BD.
Solution:
Circles Class 10 Extra Questions Maths Chapter 10 with Solutions Answers 13
True, Join OC,
∠ACB = 90° (Angle in semi-circle)
∴ ∠OBC = 180o – (90° + 30°) = 60°
Since, OB = OC = radii of same circle [Fig. 8.16]
∴ ∠OBC = ∠OCB = 60°
Also, ∠OCD = 90°
⇒ ∠BCD = 90° – 60° = 30°
Now, ∠OBC = ∠BCD + ∠BDC (Exterior angle property)
⇒ 60° = 30° + ∠BDC
⇒ ∠BDC = 30°
∵ ∠BCD = ∠BDC = 30°
∴ BC = BD

Question 2.
The length of tangent from an external point P on a circle with centre O is always less than OP.
Circles Class 10 Extra Questions Maths Chapter 10 with Solutions Answers 14
Solution:
True, let PQ be the tangent from the external point P.
Then ∆PQO is always a right angled triangle with OP as the hypotenuse. So, PQ is always less than OP.

Question 3.
If angle between two tangents drawn from a point P to a circle of radius ‘a’ and centre 0 is 90°, then OP = a√2.
Circles Class 10 Extra Questions Maths Chapter 10 with Solutions Answers 15
Solution:
True, let PQ and PR be the tangents
Since ∠P = 90°, so ∠QOR = 90°
Also, OR = OQ = a
∴ PQOR is a square
Circles Class 10 Extra Questions Maths Chapter 10 with Solutions Answers 16

Question 4.
In Fig. 8.20, PA and PB are tangents to the circle drawn from an external point P. CD is the third tangent touching the circle at Q. If PA = 15 cm, find the perimeter of ∆PCD.
Circles Class 10 Extra Questions Maths Chapter 10 with Solutions Answers 17
Solution:
∵ PA and PB are tangent from same external point
∴ PA = PB = 15 cm
Now, Perimeter of ∆PCD = PC + CD + DP = PC + CQ + QD + DP
= PC + CA + DB + DP
= PA + PB = 15 + 15 = 30 cm

Question 5.
In Fig. 8.21, PA and PB are tangents to the circle from an external point P. CD is another tangent touching the circle at Q. If PA = 12 cm, QC = QD = 3 cm, then find PC + PD.
Circles Class 10 Extra Questions Maths Chapter 10 with Solutions Answers 18
Solution:
PA = PC + CA = PC + CQ [∵ CA = CQ (tangents drawn An from external point are equal)]
⇒ 12 = PC + 3 = PC = 9 cm
∵ PA = PB = PA – AC = PB – BD
⇒ PC = PD
∴ PD = 9 cm
Hence, PC + PD = 18 cm

Question 6.
Prove that the line segment joining the points of contact of two parallel tangents of a circle, passes through its centre.
Circles Class 10 Extra Questions Maths Chapter 10 with Solutions Answers 19
Solution:
Let the tangents to a circle with centre O be ABC and XYZ.
Construction : Join OB and OY.
Draw OP||AC
Since AB||PO
∠ABO + ∠POB = 180° (Adjacent interior angles)
∠ABO = 90° (A tangent to a circle is perpendicular to the radius through the point of contact)
90° + ∠POB = 180° = ∠POB = 90°
Similarly ∠POY = 90°
∠POB + ∠POY = 90° + 90° = 180°
Hence, BOY is a straight line passing through the centre of the circle.

Question 7.
If from an external point P of a circle with centre 0, two tangents PQ and PR are drawn such that QPR = 120°, prove that 2PQ = PO.
Circles Class 10 Extra Questions Maths Chapter 10 with Solutions Answers 20
Solution:
Given, ∠QPR = 120°
Radius is perpendicular to the tangent at the point of contact.
∠OQP = 90°
⇒ ∠QPO = 60°
(Tangents drawn to a circle from an external point are equally inclined to the segment, joining the centre to that point)
Circles Class 10 Extra Questions Maths Chapter 10 with Solutions Answers 21
2PQ = PO

Question 8.
In Fig. 8.24, common tangents AB and CD to two circles with centres , and 0, intersect at E. Prove that AB = CD.
Circles Class 10 Extra Questions Maths Chapter 10 with Solutions Answers 22
Solution:
AE = CE and BE = ED [Tangents drawn from an external point are equal]
On addition, we get
AE + BE = CE + ED
∠QPO = 60°
⇒ AB = CD

Question 9.
The incircle of an isosceles triangle ABC, in which AB = AC, touches the sides BC, CA and AB at D, E and F respectively. Prove that BD = DC.
OR
In Fig. 8.25, if AB = AC, prove that BE = EC.
[Note: D, E, F replace by F, D, E]
Circles Class 10 Extra Questions Maths Chapter 10 with Solutions Answers 23
Solution:
Circles Class 10 Extra Questions Maths Chapter 10 with Solutions Answers 24
Given, AB = AC
We have, BF + AF = AE + CE ….(i)
AB, BC and CA are tangents to the circle at F, D and E respectively.
∴ BF = BD, AE = AF and CE = CD ….(ii)
From (i) and (ii)
BD + AE = AE + CD (∵ AF = AE)
⇒ BD = CD

Question 10.
In Fig. 8.27, XP and XQ are two tangents to the circle with centre O, drawn from an external point X. ARB is another tangent, touching the circle at R. Prove that XA + AR = XB + BR.
Circles Class 10 Extra Questions Maths Chapter 10 with Solutions Answers 25
Solution:
In the given figure,
AP = AR
BR = BQ
XP = XQ [Tangent to a circle from an external point are equal]
XA + AP = XB + BQ
XA + AR = XB + BR [AP = AR, BQ = BR]

Question 11.
In Fig. 8.28, a circle is inscribed in a AABC, such that it touches the sides AB, BC and CA at points D, E and F respectively. If the lengths of sides AB, BC and CA are 12 cm, 8 cm and 10 cm respectively, find the lengths of AD, BE and CF.
Circles Class 10 Extra Questions Maths Chapter 10 with Solutions Answers 26
Solution:
Let AD = AF = x
∴ DB = BE = 12 – x
and CF = CE = 10 – x
BC = BE + EC
⇒ 8 = 12 – x + 10 – x
⇒ x = 7
∴ AD = 7 cm, BE = 12 – 7 = 5 cm, CF = 10 – 7 = 3 cm

Question 12.
In Fig. 8.29, AP and BP are tangents to a circle with centre 0, such that AP = 5 cm and ∠APB = 60°. Find the length of chord AB.
Circles Class 10 Extra Questions Maths Chapter 10 with Solutions Answers 27
Solution:
PA = PB (Tangents from an external point are equal)
and ∠APB = 60°
⇒ ∠PAB = ∠PBA = 60°
∴ ∆PAB is an equilateral triangle.
Hence AB = PA = 5 cm.

Question 13.
In Fig. 8.30 from an external point P, two tangents PT and PS are drawn to a circle with centre 0 and radius r. If OP = 2r, show that ∠OTS = ∠OST = 30°.
Circles Class 10 Extra Questions Maths Chapter 10 with Solutions Answers 28
Solution:
Let ∠TOP = θ
∴ cos θ = \(\frac{OT}{OP}\) = \(\frac{r}{2r}\) =\(\frac{1}{2}\)
⇒ cos θ = cos 60°
⇒ θ = 60°
Hence, ∠TOS = 120°
In ∆OTS, OT = OS [Radii of circle]
⇒ ∠OTS = ∠OST = \(\frac{60^{\circ}}{2}\) = 30°

Question 14.
In Fig. 8.31, are two concentric circles of radii 6 cm and 4 cm with centre O. If AP is a tangent to the larger circle and BP to the smaller circle and length of AP is 8 cm, find the length of BP.
Circles Class 10 Extra Questions Maths Chapter 10 with Solutions Answers 29
Solution:
OA = 6 cm, OB = 4 cm, AP = 8 cm
OP2 = OA2 + AP2 = 36 + 64 = 100
⇒ OP = 10 cm
BP2 = OP2 – OB2 = 100 – 16 = 84
⇒ BP = 2√21 cm

Question 15.
In fig. 8.32, PQ is a tangent from an external point P to a circle with centre ( and OP cuts the circle at T and QOR is a diameter. If ∠POR = 130° and S is a point on the circle, find R 21 + 22.
Circles Class 10 Extra Questions Maths Chapter 10 with Solutions Answers 30
Solution:
∵ ∠POR = 130o = ∠ROT [∵ Angle subtended by an arc at the centre is double than the angle subtended by it at any part of circumference]
∠2 = \(\frac{1}{2}\) ∠ROT = \(\frac{1}{2}\) × 130° = 65°
∠POQ = 180° – 130° = 50° [Linear pair]
and ∠Q = 90°
∴ ∠1 = 40°[∵ Line drawn from centre to the point of contact is perpendicular to the tangent]
Hence ∠2 + ∠1 = 65° + 40° = 105° .

Circles Class 10 Extra Questions Short Answer Type 2

Question 1.
A tangent PQ at a point P of a circle of radius 5 cm meets a line through the centre 0 at a point so that OQ = 12 cm. Find the length of PQ.
Circles Class 10 Extra Questions Maths Chapter 10 with Solutions Answers 31
Solution:
We have, ∠OPQ = 90°
OQ = 12 cm and OP = 5 cm
∴ By Pythagoras Theorem
OQ2 = OP2 + QP2
⇒ 122 = 52 + QP2
⇒ QP2 = 144 – 25 = 119
= QP = √119 cm

Question 2.
From a point l, the length of the tangent to a circle is 24 cm and the distance of Q from the centre is 25 cm. Find the radius of the circle.
Solution:
Circles Class 10 Extra Questions Maths Chapter 10 with Solutions Answers 32
Let QT be the tangent and OT be the radius of circle. Therefore
OT ⊥ QT i.e., ∠OTQ = 90°
and OQ = 25 cm and QT = 24 cm
Now, by Pythagoras Theorem, we have
OQ2 = QT2 + OT2
⇒ 252 = 242 + OT2
⇒ OT2 = 252 – 242
⇒ 625 – 576
OT2 – 49
∴ OT = 7 cm

Question 3.
In Fig. 8.35, if TP and TQ are the two tangents to a circle with centre 0 so that ∠POQ = 110°, then find ∠PTQ.
Solution:
Circles Class 10 Extra Questions Maths Chapter 10 with Solutions Answers 33
Since TP and TQ are the tangents to the circle with centre O
So, OPIPT and OQ ⊥ QT
⇒ ∠OPT = 90°, ∠OQT = 90° and ∠POQ = 110°
So, in quadrilateral OPTQ, we have
∠POQ + ∠OPT + ∠PTQ + ∠TQO = 360°
⇒ 110° + 90° + ∠PTQ + 90° = 360°
⇒ ∠PTQ + 290° = 360°
∴ ∠PTQ = 360° – 290°
= ∠PTQ = 70°

Question 4.
Prove that the tangents drawn at the ends of a diameter of a circle are parallel.
Solution:
Circles Class 10 Extra Questions Maths Chapter 10 with Solutions Answers 34
Let AB be the diameter of the given circle with centre O, and two tangents PQ and LM are drawn at the end of diameter AB respectively.
Р. Now, since the tangent at a point to a circle is perpendicular to the radius through the point of contact.
Therefore, OA ⊥ PQ and OB ⊥ LM
i.e., AB ⊥ PQ and also AB ⊥ LM
⇒ ∠BAQ = ∠ABL (each 90°)
∴ PQ||LM (∵ ∠BAQ and ∠ABL are alternate angles)

Question 5.
If tangents PA and PB from a point P to a circle with centre O are inclined to each other at angle of 80°, then find ∠POA.
Solution:
Circles Class 10 Extra Questions Maths Chapter 10 with Solutions Answers 35
∵ PA and PB are tangents to a circle with centre O,
∴ OA ⊥ AP and OB ⊥ PB
i.e., ∠APB = 80°, ∠OAP = 90°, and ∠OBP = 90°
Now, in quadrilateral OAPB, we have
∠APB + ∠PBO + ∠BOA + ∠OAP = 360°
⇒ 80° + 90° + ∠BOA + 90o = 360°
⇒ 260° + ∠BOA = 360°
∴ ∠BOA = 360° – 260°
⇒ ∠BOA = 100°
Now, in ∆POA and APOB we have
OP = OP (Common)
ОА = ОВ (Radii of the same circle)
∠OAP = ∠OBP = 90°
∴ ∆POA ≅ APOB (RHS congruence condition)
⇒ ∠POA = ∠POB (CPCT)
Now, ∠POA = \(\frac{1}{2}\) = ∠BOA = \(\frac{1}{2}\) × 100 = 50°

Question 6.
The length of a tangent from a point A at distance 5 cm from the centre of the circle is 4 cm. Find the radius of the circle.
Circles Class 10 Extra Questions Maths Chapter 10 with Solutions Answers 36
Solution:
Let O be the centre and P be the point of contact.
Since tangent to a circle is perpendicular to the radius through the point of contact,
∴ ∠OPA = 90° Now, in right ∆OPA we have
OA2 = OP2 + PA2 [By Pythagoras Theorem]
52 = OP2 + 42
= 25 = OP2 + 16
⇒ OP2 = 25 – 16 = 9
∴ OP = 3cm
Hence, the radius of the circle is 3 cm.

Question 7.
Two concentric circles are of radii 5 cm and 3 cm. Find the length of the chord of the larger circle which touches the smaller circle.
Solution:
Circles Class 10 Extra Questions Maths Chapter 10 with Solutions Answers 37
Let O be the common centre of two concentric circles and let AB be a chord of larger circle
touching the smaller circle at P. Join OP.
Since OP is the radius of the smaller circle and AB is tangent to this circle at P,
∴ OP ⊥ AB
We know that the perpendicular drawn from the centre of a circle to any chord of the circle bisects the chord.
Therefore, AP = BP
In right ∆APO we have
⇒ OA2 = AP2 + OP2
⇒ 52 = AP2 + 32
⇒ 25 – 9 = AP2
⇒ AP2 = 16
⇒ AP = 4
Now, AB = 2.AP = 2 × 4 = 8 [∵ AP = PB]
Hence, the length of the chord of the larger circle which touches the smaller circle is 8 cm.

Question 8.
Prove that the tangents drawn at the ends of a chord of circle make equal angles with the chord.
Solution:
Circles Class 10 Extra Questions Maths Chapter 10 with Solutions Answers 38
Given: A circle with centre O, PA and PB are tangents drawn at the ends A and B on chord AB.
To prove: ∠PAB = ∠PBA
Construction: Join OA and OB.
Proof: In ∆OAB we have
ОА = ОВ … (i) [Radii of the same circle]
∠2 = ∠1 … (ii) [Angles opposite to equal sides of a A]
Also (∠2 + ∠3 = ∠1 + 24) …(iii) [Both 90° as Radius ⊥ Tangent]
Subtracting (ii) from (iii), we have
∴ ∠3 = ∠4 = ∠PAB = ∠PBA

Question 9.
Prove that the perpendicular at the point of contact to the tangent to a circle passes through the centre.
Circles Class 10 Extra Questions Maths Chapter 10 with Solutions Answers 39
Solution:
Let LM be tangent drawn at the point P on the circle with centre O. Join OP. If possible, let PQ be perpendicular to LM, not passing through O.
Now, since tangent at a point to a circle is perpendicular to the radius through the point.
∴ OP ⊥ LM ⇒ ∠OPM = 90°
Also, ∠QPM = 90° (as assumed above)
∴ ∠OPM = ∠QPM,
which is possible only when points O and I coincide
Hence, the perpendicular at the point of contact to tangent to a circle passes through the centre.

Question 10.
A quadrilateral ABCD is drawn to circumscribe a circle (Fig. 8.42). Prove that AB + CD = AD + BC.
OR
A circle touches all the four sides of a quadrilateral ABCD. Prove that AB + CD = BC + DA
Circles Class 10 Extra Questions Maths Chapter 10 with Solutions Answers 40
Solution:
Since lengths of two tangents drawn from an external point of circle are equal,
Therefore, AP = AS, BP = BQ and DR = DS
CR = CQ (Where P, Q, R and S are the points of contact]
Adding all these, we have
(AP + BP) + (CR + RD) = (BQ + CQ) + (DS + AS)
⇒ AB + CD = BC + DA

Question 11.
A circle is touching the side BC of AABC at P and touching AB and AC produced at Q and R respectively. Prove that AQ = \(\frac{1}{2}\) (perimeter of ∆ABC).
Solution:
Circles Class 10 Extra Questions Maths Chapter 10 with Solutions Answers 41
Since tangents from an exterior point to a circle are equal in length.
∴ BP = BQ [Tangents from B] …(i)
CP = CR [Tangents from C] … (ii)
and, AQ = AR [Tangents from A] …(iii)
From (iii), we have
AQ = AR
⇒ AB + BQ = AC + CR
AB + BP
⇒ AC + CP (Using (i) and (ii)] …(iv)
Now, perimeter of AABC = AB + BC + AC
= AB + (BP.+ PC) + AC
= (AB + BP) + (AC + PC)
= 2(AB + BP) [Uisng (iv)]
= 2(AB + BQ) = 2AQ [Using (i)]
AQ = \(\frac{1}{2}\) (Perimeter of ∆ABC)

Question 12.
The difference between the radii of the smaller circle and the larger circle is 7 cm and the difference between the areas of the two circles is 1078 sq. cm. Find the radius of the smaller circle.
Solution:
Given: r2 – r1 = 7 (r2 > r1) …(i)
and π(r22 – r12) = 1078
π (r2 – r1) (r2 + r1) = 1078
π (r2 + r1) = 1078 [(From equation (i)]
⇒ r2 + r1 = \(\frac{1078 \times 7}{22 \times 7}\) = 49…(ii)
Adding (i) and (ii), we get
2r2 = 56
⇒ r2 = 28 cm
r1 = 21 cm [From equation (ii)]
∴ Radius of smaller circle = 21 cm.

Circles Class 10 Extra Questions Long Answer Type 1

Question 1.
Prove that the tangent to a circle is perpendicular to the radius through the point of contact.
Solution:
Given: A circle C(O,r) and a tangent AB at a point P.
To Prove: OP ⊥ AB.
Construction: Take any point l, other than P, on the tangent AB. Join OQ. Suppose OQ meets the circle at R.
Proof: We know that among all line segments joining the point to a point on AB, the shortest one is perpendicular to AB. So, to prove that OP ⊥ AB it is sufficient to prove that OP is shorter than any other segment joining O to any point of AB.
Circles Class 10 Extra Questions Maths Chapter 10 with Solutions Answers 42
Clearly, OP = OR [Radii of the same circle]
Now, OQ = OR + RQ
⇒ OQ > OR
⇒ OQ > OP [∵OP = OR]
Thus, OP is shorter than any other segment joining O to any point on AB.
Hence, OP ⊥ AB.

Question 2.
Prove that the lengths of two tangents drawn from an external point to a circle are equal.
Solution:
Circles Class 10 Extra Questions Maths Chapter 10 with Solutions Answers 43
Given: AP and AQ are two tangents from a point A to a circle C (O, r).
To Prove: AP = AQ
Construction: Join OP, OQ and OA.
Proof: In order to prove that AP = AQ, we shall first prove that ∆OPA ≅ ∆OQA.
Since a tangent at any point of a circle is perpendicular to the radius through the point of contact.
∴ OP ⊥ AP and OQ ⊥ AQ
⇒ ∠OPA = ∠OQA = 90°
Now, in right triangles OPA and OQA, we have
OP = OQ [Radii of a circle]
∠OPA = ∠OQA [Each 90°]
and OA = OA [Common]
So, by RHS-criterion of congruence, we get
∆OPA ≅ OQA
⇒ AP = AQ [CPCT]
Hence, lengths of two tangents from an external point are equal.

Question 3.
Prove that the parallelogram circumscribing a circle is a rhombus.
Solution:
Circles Class 10 Extra Questions Maths Chapter 10 with Solutions Answers 44
Let ABCD be a parallelogram such that its sides touch a circle with centre O.
We know that the tangents to a circle from an exterior point are equal in length.
Therefore, we have
AP = AS [Tangents from A]
BP = BQ [Tangents from B] …. (ii)
CR = CQ [Tangents from C] …. (iii)
And DR = DS [Tangents from D] …. (iv)
Adding (i), (ii), (iii) and (iv), we have
(AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ)
AB + CD = AD + BC
AB + AB = BC + BC [∵ ABCD is a parallelogram ∴ AB = CD, BC = DA]
2AB = 2BC ⇒ AB = BC
Thus, AB = BC = CD = AD
Hence, ABCD is a rhombus.

Question 4.
In Fig. 8.47, PQ is a chord of length 16 cm, of a circle of radius 10 cm. The tangents at P and Q intersect at a point T. Find the length of TP.
Solution:
Given: PQ = 16 cm
PO = 10 cm
Circles Class 10 Extra Questions Maths Chapter 10 with Solutions Answers 45

Question 5.
If PQ is a tangent drawn from an external point P to a circle with centre O and QOR is a diameter where length of QOR is 8 cm such that ∠POR = 120°, then find OP and PQ.
Solution:
Let O be the centre and QOR = 8 cm is diameter of a circle. PQ is tangent such that ∠POR = 120°
Circles Class 10 Extra Questions Maths Chapter 10 with Solutions Answers 46

Question 6.
In Fig. 8.49, two equal circles, with centres O and O’, touch each other at X.OO’ produced meets the circle with centre O’ at A. AC is tangent to the circle with centre 0, at the point C. O’D is perpendicular to AC. Find the value of \(\frac{DO’}{CO}\) .
Solution:
Circles Class 10 Extra Questions Maths Chapter 10 with Solutions Answers 47
AC is tangent to circle with centre O.
Thus ∠ACO = 90°
In ∆AO’D and ∆AOC
∠ADO’ = ∠ACO = 90°
∠A = ∠A (Common)
∴ ∆AO’D – ∠AOC (By AA similarity)
Circles Class 10 Extra Questions Maths Chapter 10 with Solutions Answers 48

Question 7.
In Fig. 8.50, O is the centre of a circle of radius 5 cm. T is a point such that OT = 13 cm and OT intersects circle at E. If AB is a tangent to the circle at E, find the length of AB, where TP and TQ are two tangents to the circle.
Circles Class 10 Extra Questions Maths Chapter 10 with Solutions Answers 49
Solution:
In right ∆POT
Circles Class 10 Extra Questions Maths Chapter 10 with Solutions Answers 50
TE = 8 cm
Let PA = AE = x
(Tangents from an external point to a circle are equal)
In right ∆AET
TA2 = TE2 + EA2
⇒ (12 – x)2 = 64 + x2
⇒ 144 + x2 – 24x = 64 + x2
⇒ x = \(\frac{80}{24}\)
⇒ x = 3.3 cm
Thus, AB = 6.6 cm

Circles Class 10 Extra Questions HOTS

Question 1.
Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.
Solution:
Circles Class 10 Extra Questions Maths Chapter 10 with Solutions Answers 51
Let a circle with centre O touches the sides AB, BC, CD and DA of a D quadrilateral ABCD at the points P, Q, R and S respectively. Then, we have to prove that
∠AOB + ∠COD = 180° and ∠AOD + ∠BOC = 180°
Now, Join OP, OQ, OR and OS.
Since the two tangents drawn from an external point to a circle subtend equal angles at the centre.
∴ ∠1 = ∠2, ∠3 = 24, 25 = 26 and 27 = 28 …(i)
Now, 21 + 22 +23 + 24 + 25 +26+ 27 + ∠8 = 360° … (ii)
[sum of all the angles subtended at a point is 360°]
⇒ 2(∠2 + ∠3 + ∠6 + ∠7) = 360° [using equation (i) and (ii)]
= (∠2 + ∠3) + (∠6 + ∠7) = 180°
∠AOB + ∠COD = 180°
again 2(∠1 + ∠8 +∠4 + ∠5) = 360° [from (i) and (ii)]
(∠1 + ∠8) + (∠4 + ∠5) = 180°
∠AOD + ∠BOC = 180°

Question 2.
A triangle ABC [Fig. 8.52] is drawn to circumscribe a circle of radius 4 cm such that the segments BD and DC into which BC is divided by the point of contact D are of lengths 8 cm and 6 cm respectively. Find the sides AB and AC.
Solution:
Circles Class 10 Extra Questions Maths Chapter 10 with Solutions Answers 52
Let ∆ABC be drawn to circumscribe a circle with centre O and radius 4 cm and circle touches the sides BC, CA and AB at D, E and 6 cm F respectively.
We have given that CD = 6 cm and BD = 8 cm
∴ BF = BD = 8 cm and CE = CD = 6 cm
{Length of two tangents drawn from an external point of circle are equal}
Now, let AF = AE = x cm
Then, AB = c = (x + 8) cm, BC = a = 14 cm, CA = b = (x + 6) cm
2s = (x + 8) + 14 + (x + 6) 25 = 2x + 28 or s = x + 14
s – a = (x + 14) – 14 = x
s – b = (x + 14) – (x + 6) = 8
s – c = (x + 14) – (x + 8) = 6
Circles Class 10 Extra Questions Maths Chapter 10 with Solutions Answers 53

Squaring both sides, we have
48x (x + 14) = 16(x + 14)2 = 48x (x + 14) – 16 (x + 14)2 = 0
16 (x + 14) (3x – (x + 14)] = 0
⇒ 16(x + 14)(2x – 14) = 0
either 16(x + 14) = 0 or 2x – 14 = 0
⇒ x = -14 or 2x = 14
⇒ x = -14 or x = 7
But x cannot be negative so x ≠ – 14 .
∴ x = 7 cm
Hence, the sides AB = x + 8 = 7 + 8 = 15 cm
AC = x + 6 = 7 + 6 = 13 cm.

Question 3.
In Fig. 8.53, XY and X’Y are two parallel tangents to a circle with centre O and another tangent AB with point of contact C intersecting XY at A and ∠X’Y at B. Prove that ∠AOB = 90°.
Solution:
Circles Class 10 Extra Questions Maths Chapter 10 with Solutions Answers 54
Join OC. In ∆APO and ∆ACO, we have
AP = AC (Tangents drawn from external point A)
AO = OA (Common)
PO = OC (Radii of the same circle)
∴ ∆APO ≅ ∆ACO (By SSS criterion of congruence)
∴∠PAO = ∠CAO (CPCT)
⇒ ∠PAC = 2∠CAO
Similarly, we can prove that
∆OQB ≅ ∆OCB
∴∠QBO = 2CBO
⇒ ∠CBQ = 22CBO
Now, ∠PAC + ∠CBQ = 180° [Sum of interior angles on the same side of transversal is 180°]
⇒ 2∠CAO + 2∠CBO = 180°
⇒ ∠CAO + ∠CBO = 90°
⇒ 180° – ∠AOB = 90°
[∵ ∠CAO + ∠CBO + ∠AOB = 180°]
⇒ 180° – 90° = ∠AOB
⇒ ∠AOB = 90°

Question 4.
Let A be one point of intersection of two intersecting circles with centres O and Q. The tangents at A to the two circles meet the circles again at B and C respectively. Let the point P be located so that AOPQ is a parallelogram. Prove that P is the circumcentre of the triangle ABC.
Solution:
Circles Class 10 Extra Questions Maths Chapter 10 with Solutions Answers 55
In order to prove that P is the circumcentre of ∆ABC it is sufficient to show that P is the point of intersection of perpendicular bisectors of the sides of AABC i.e., OP and PQ are perpendicular bisectors of sides AB and AC respectively. Now, AC is tangent at A to the circle with centre at 0 and OA is its radius.
∴ OA ⊥ AC
⇒ PQ ⊥ AC [∵ OAQP is a parallelogram so, OA ||PQ]
Also, Q is the centre of the circle
QP bisects AC [Perpendicular from the centre to the chord bisects the chord]
⇒ PQ is the perpendicular bisector of AC.
Similarly, BA is the tangent to the circle at A and AQ is its radius through A.
∴ BA ⊥ AQ [∵ AQPO is parallelogram]
BA ⊥ OP [∴ OP || AQ]
Also, OP bisects AB [∵ 0 is the centre of the circle]
⇒ OP is the perpendicular bisector of AB. Thus, P is the point of intersection of perpendicular bisectors PQ and PO of sides AC and AB respectively.
Hence, P is the circumcentre of ∆ABC.

Important Questions for Class 10 Science CBSE Chapter Wise PDF

NCERT CBSE Important Questions for Class 10 Science: Students who are struggling to find out what are the important question asked in the annual exams? Here is the list of CBSE Class 10 Science Chapter Wise Question Bank Important Questions which are prepared by subject experts as per the latest CBSE syllabus curriculum. All these questions are designed after analyzing the previous questions papers & model papers. So, make sure to include practicing these NCERT extra important science questions and attain good marks in CBSE Board Exams.

Class 10 Science Important Questions with Answers PDF Download

Access all CBSE NCERT Chapter Wise Important Questions of Class 10 Science with answers and solutions by clicking on the particular chapter link available over here.

  1. Chemical Reactions and Equations Class 10 Important Questions
  2. Acids Bases and Salts Class 10 Important Questions
  3. Metals and Non-metals Class 10 Important Questions
  4. Carbon and its Compounds Class 10 Important Questions
  5. Periodic Classification of Elements Class 10 Important Questions
  6. Life Processes Class 10 Important Questions
  7. Control and Coordination Class 10 Important Questions
  8. How do Organisms Reproduce Class 10 Important Questions
  9. Heredity and Evolution Class 10 Important Questions
  10. Light Reflection and Refraction Class 10 Important Questions
  11. Human Eye and Colourful World Class 10 Important Questions
  12. Electricity Class 10 Important Questions
  13. Magnetic Effects of Electric Current Class 10 Important Questions
  14. Sources of Energy Class 10 Important Questions
  15. Our Environment Class 10 Important Questions
  16. Management of Natural Resources Class 10 Important Questions

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Important Questions for Class 9 Science CBSE Chapter Wise Pdf

NCERT CBSE Important Questions for Class 9 Science: Students who are struggling to find out what are the important question asked in the annual exams? Here is the list of CBSE Important questions for class 9 science chapterwise which are prepared by subject experts as per the latest CBSE syllabus curriculum. All these questions are designed after analyzing the previous questions papers & model papers. So, make sure to include practicing these NCERT extra important science questions and attain good marks in Exams.

Class 9 Science Important Questions with Answers

Access all CBSE NCERT Chapter Wise Important Questions of Class 9 Science with answers and solutions by clicking on the particular chapter link available over here.

  1. Matter in Our Surroundings Class 9 Important Questions
  2. Is Matter Around Us Pure Class 9 Important Questions
  3. Atoms and Molecules Class 9 Important Questions
  4. Structure of the Atom Class 9 Important Questions
  5. The Fundamental Unit of Life Class 9 Important Questions
  6. Tissues Class 9 Important Questions
  7. Diversity in Living Organisms Class 9 Important Questions
  8. Motion Class 9 Important Questions
  9. Force and Laws of Motion Class 9 Important Questions
  10. Gravitation Class 9 Important Questions
  11. Work, Power and Energy Class 9 Important Questions
  12. Sound Class 9 Important Questions
  13. Why Do we Fall Ill Class 9 Important Questions
  14. Natural Resources Class 9 Important Questions
  15. Improvement in Food Resources Class 9 Important Questions
  16. Floatation Class 9 Important Questions

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Constructions Class 10 Extra Questions Maths Chapter 11 with Solutions Answers

Here we are providing Constructions Class 10 Extra Questions Maths Chapter 11 with Answers Solutions, Extra Questions for Class 10 Maths was designed by subject expert teachers.

Extra Questions for Class 10 Maths Constructions with Answers Solutions

Extra Questions for Class 10 Maths Chapter 11 Constructions with Solutions Answers

Constructions Class 10 Extra Questions Very Short Answer Type

Question 1.
Is construction of a triangle with sides 8 cm, 4 cm, 4 cm possible?
Solution:
No, we know that in a triangle sum of two sides of a triangle is greater than the third side. So the condition is not satisfied.

Question 2.
To divide the line segment AB in the ratio 5 : 6, draw a ray AX such that ∠BAX is an acute angle, then draw a ray BY parallel to AX and the point A1, A2, A3… and B1, B2, B3… are located at equal distances on ray AX and BY respectively. Then which points should be joined?
Solution:
A5 and B6.

Question 3.
To draw a pair of tangents to a circle which are inclined to each other at an angle of 60°, it is required to draw tangents at end points of those two radii of the circle. What should be the angle between them?
Solution:
120°

Question 4.
In Fig. 9.1 by what ratio does P divide AB internally.
Solution:
From Fig. 9.1, it is clear that there are 3 points at equal distances on AX and 4 points at equal distances on BY. Here P divides AB on joining A3 B4. So P divides internally by 3 : 4.
Constructions Class 10 Extra Questions Maths Chapter 11 with Solutions Answers 1

Question 5.
Given a triangle with side AB = 8 cm. To get a line segment AB’ = 2 of AB, in what ratio will line segment AB be divided?
Solution:

Constructions Class 10 Extra Questions Maths Chapter 11 with Solutions Answers 2
Given AB = 8 cm
AB’ = \(\frac{3}{4}\) of AB
= \(\frac{3}{4}\) × 8 = 6 cm
BB’ = AB – AB’ = 8 – 6 = 2 cm.
⇒ AB’: BB’ = 6 : 2 = 3 : 1
Hence the required ratio is 3 : 1.

Constructions Class 10 Extra Questions Short Answer Type I and II

Question 1.
Construct a triangle of sides 4 cm, 5 cm and 6 cm and then a triangle similar to it whose sides are \(\frac{2}{3}\) of the corresponding sides of the first triangle.
OR
Draw a triangle with sides 4 cm, 5 cm and 6 cm. Then construct another triangle whose sides are \(\frac{2}{3}\) of the corresponding sides of first triangle.
Solution:
Steps of Construction:
Step I: Draw a line segment BC = 6 cm
Step II: Draw an arc with B as centre and radius equal to 5 cm.
Step III: Draw an arc, with C as centre and radius equal to 4 cm intersecting the previous drawn arc at A.
Step IV: Join AB and AC, then ∆ABC is the required triangle.
Step V: Below BC make an acute angle CBX
Step VI: Along BX mark off three points at equal distance: B1, B2, B3, such that BB1 = B1B2, = B2B3.
Step VII: Join BC3.
Step VIII: From B2, draw B2, D || B3,C, meeting BC at D.
Step IX: From D draw ED || AC meeting BA at E. Then we have ∆EDB which is the required triangle.
Constructions Class 10 Extra Questions Maths Chapter 11 with Solutions Answers 3

Justification:
Since DE || CA
Constructions Class 10 Extra Questions Maths Chapter 11 with Solutions Answers 4
Hence, we have the new AEBD similar to the given ∆ABC, whose sides are equal to \(\frac{2}{3}\) of the corresponding sides of ∆ABC.

Question 2.
Draw a line segment of length 7.6 cm and divide it in the ratio 5 : 8. Measure the two parts.
Solution:
Steps of Construction:
Step I: Draw a line segment AB = 7.6 cm
Step II: Draw any ray AX making an acute angle ∠BAX with AB.
Step III: On ray AX starting from A, mark 5 + 8 = 13 equal arcs. AA1, A1A2, A2A3, A3A4, … A11A12, and A12A13.
Step IV: Join A13B.
Step V: From A5, draw A5P || A13B, meeting AB at P. Thus, P divides AB in the ratio 5 : 8. On measuring the two parts. We find AP = 2.9 cm and PB = 4.7 cm (approx).
Constructions Class 10 Extra Questions Maths Chapter 11 with Solutions Answers 5
Justification:
In ∆ABA13, PA5 || BA13 .
∴ By Basic Proportionality Theorem
Constructions Class 10 Extra Questions Maths Chapter 11 with Solutions Answers 6

Question 3.
Construct an isosceles triangle whose base is 8 cm and altitude 4 cm and then draw another triangle whose sides are 1\(\frac{1}{2}\) times the corresponding sides of the isosceles triangle.
Solution:
Steps of Construction:
Step 1: Draw BC = 8 cm.
Step II: Construct XY, the perpendicular bisector of line segment BC, meeting BC at M.
Step III: Along MP, cut-off MA = 4 cm.
Step IV: Join BA and CA. Then ∆ABC so obtained is the required ∆ABC.
Step V: Extend BC to D, such that BD = 12 cm
Step VI: Draw DE || CA meeting BA produced at E. Then AEBD is the required triangle.
Constructions Class 10 Extra Questions Maths Chapter 11 with Solutions Answers 7
Justification:
Since, DE || CA .
Constructions Class 10 Extra Questions Maths Chapter 11 with Solutions Answers 8
Hence, we have the new triangle similar to the given triangle whose sides are 1 \(\frac{1}{2}\) i.e, \(\frac{3}{2}\) times the corresponding sides of the isosceles ABC.

Question 4.
Draw a triangle ABC with side BC = 6 cm, AB = 5 cm and ∠ABC = 60°. Then construct a triangle whose sides of the corresponding sides of the triangle ABC.
Solution:
Steps of Construction:
Step 1: Construct a ∆ABC in which BC = 6 cm and, AB = 5 cm and ∠ABC = 60°.
Step II: Below BC make an acute ∠CBX.
Step III: Along BX mark off four arcs: B1, B2, B3 such that BB1 = B1B2 = B2B3 = B3B4.
Step IV: Join B4C.
Step V: From B3, draw B3D || B4C, meeting BC at D.
Step VI: From D, draw ED || AC, meeting BA at E.
Now, we have AEBD which is the required triangle whose sides are \(\frac{3}{4}\)th of the corresponding sides of ∆ABC.
Constructions Class 10 Extra Questions Maths Chapter 11 with Solutions Answers 9
Justification:
Here, DE || CA
∴ ∆ABC ~ ∆EBD.
Constructions Class 10 Extra Questions Maths Chapter 11 with Solutions Answers 10
Hence, we get the new triangle similar to the given triangle whose sides are equal to \(\frac{3}{4}\)th of the corresponding sides of ∆ABC.

Question 5.
Draw a circle of radius 6 cm. From a point 10 cm away from its centre, construct the pair of tangents to the circle and measure their lengths.
Solution:
Steps of Construction:
Step 1: Take a point O and draw a circle of radius 6 cm.
Step II: Take a point P at a distance of 10 cm from the centre 0.
Step III: Join OP and bisect it. Let M be the mid-point.
Step IV: With M as centre and MP as radius, draw a circle to intersect the circle at Q and R.
Step V: Join PQ and PR. Then, PQ and PR are the required tangents. On measuring, we find, PQ = PR = 8cm.
Constructions Class 10 Extra Questions Maths Chapter 11 with Solutions Answers 11
Justification:
On joining OQ, we find that ∠PQO = 90°, as ∠PQO is the angle in the Semicircle.
∴ PQ ⊥ OQ
Since OQ is the radius of the given circle, PQ has to be a tangent to the circle. Similarly, PR is
also a tangent to the circle.

Question 6.
Construct a tangent to a circle of radius 4 cm from a point on the concentric circle of radius 6 cm and measure its length. Also, verify the measurement by actual calculation.
Solution:
Steps of Construction:
Step 1: Take a point O and draw a circle of radius OA = 4 cm. Also, draw a concentric circle of radius OB = 6 cm
Step II: Find the mid-point C of OB and draw a circle of radius OC = BC. Suppose this circle intersects the circle of radius 4 cm at P and Q.
Step III: Join BP and BQ to get the desired tangents from a point B on the circle of radius 6 cm. By actual measurement, we find BP = BQ = 4.5 cm.
Constructions Class 10 Extra Questions Maths Chapter 11 with Solutions Answers 12
Justification:
In ∆BPO, we have
∠BPO = 90°, OB = 6 cm and OP = 4 cm
∴ OB2 = BP2 + OP2 [Using Pythagoras theorem]
Constructions Class 10 Extra Questions Maths Chapter 11 with Solutions Answers 13
Similarly, BQ = 4.47 cm

Question 7.
Draw a line segment AB of length 8 cm. Taking A as centre, draw a circle of radius 4 cm and
taking B as centre, draw another circle of radius 3 cm. Construct tangents to each circle from the centre of the other circle.
Solution:
Steps of Construction:
Step I: Draw a line segment AB = 8 cm.
Step II: With A as centre, draw a circle of radius 4 cm and let it intersect the line segment AB in M.
Step III: With B as centre, draw a circle of radius 3 cm.
Step IV: With M as centre, draw a circle of radius AM and let it intersect the given two circles in P, e and R, S.
Step V: Join AP, AQ, BR and BS.
These are the required tangents.
Constructions Class 10 Extra Questions Maths Chapter 11 with Solutions Answers 14
Justification:
On joining BP, we have ∠BPA = 90°, as ∠BPA is the angle in the semicircle.
∴ AP ⊥ PB
Since BP is the radius of given circle, so AP has to be a tangent to the circle. Similarly, AQ, BR and BS are the tangents.

Constructions Class 10 Extra Questions Long Answer Type

Question 1.
Construct a triangle similar to a given triangle ABC with its sides equal to \(\frac{5}{3}\) of the corresponding sides of the triangle ABC (i.e., of scale factor ).
Solution:
Steps of Construction:
Step I: Draw any ray BX making an acute angle with BC on the side opposite to the vertex A.
Step II: From B cut off 5 arcs
B1, B2, B3, B4 and B5 on BX so that
BB1 = B1B2 = B2B3 = B3B4 = B4B5.
Step III: Join B3 to C and draw a line through B5, parallel to B3C intersecting the extended line segment BC at C’.
Step IV: Draw a line through C’ parallel to CA intersecting the
extended line segment BA at A’ (see figure). Then, A’ BC’ is the required triangle.
Constructions Class 10 Extra Questions Maths Chapter 11 with Solutions Answers 15
Justification:
Note that ∆ABC ~ ∆A’BC” (Since AC || A’C’)
Constructions Class 10 Extra Questions Maths Chapter 11 with Solutions Answers 16

Question 2.
Draw a circle of radius of 3 cm. Take two points P and Q on one of its diameters extended on both sides, each at a distance of 7 cm on opposite sides of its centre. Draw tangents to the circle from these two points P and Q.
Solution:
Steps of Construction:
Step 1: Taking a point ( as centre, draw a circle of radius 3 cm.
Step II: Take two points P and Q on one of its extended diameter such that OP = OQ = 7 cm.
Step III: Bisect OP and OQ and let M1 and M2 be the mid-points of OP and OQ respectively.
Step IV: Draw a circle with M1 as centre and M1 P as radius to intersect the circle at T1, and T2.
Step V: Join PT1 and PT2.
Then, PT1 and PT2 are the required tangents. Similarly, the tangents QT3 and QT4 can be obtained
Constructions Class 10 Extra Questions Maths Chapter 11 with Solutions Answers 17
Justification:
On joining OT1, we find ∠PT1O = 90°, as it is an angle in the semicircle.
PT1 ⊥ OT1
Since OT1 is a radius of the given circle, so PT1 has to be a tangent to the circle.
Similarly, PT2, QT3 and QT4 are also tangents to the circle.

Question 3.
Let ABC be a right triangle in which AB = 6 cm, BC = 8 cm and ∠B = 90°. BD is the perpendicular from B on AC. The circle through B, C, D is drawn. Construct the tangents from A to this circle.
Solution:
Constructions Class 10 Extra Questions Maths Chapter 11 with Solutions Answers 18
Steps of Construction:
Step I: Draw ∆ABC and perpendicular BD from B on AC.
Step II: Draw a circle with BC as a diameter. This circle will pass through D.
Step III: Let O be the mid-point of BC. Join A0.
Step IV: Draw a circle with AO as diameter. This circle cuts the circle drawn in step II at B and E.
Step V: Join AE. AE and AB are desired tangents drawn from A to the circle passing through B, C and D.

Question 4.
Draw a right triangle in which the sides (other than hypotenuse) are of lengths 4 cm and 3 cm. Then construct another triangle whose sides are \(\frac{5}{3}\) times the corresponding sides of the given triangle.
Solution:
Steps of Construction:
Step I: Construct a SABC in which BC = 4 cm, ∠B = 90° and BA = 3 cm.
Step II: Below BC, make an acute ∠CBX.
Step III: Along BX mark off five arcs: B1, B2, B3, B4 and B5 such that
BB1 = B1B2 = B2B3 = B3B4 = B4B5.
Step IV: Join B3C.
Step V: From B5, draw B5D || B3C, meeting BC produced at D.
Step VI: From D, draw ED || AC, meeting BA produced at E. Then EBD is the required triangle whose sides are \(\frac{5}{3}\) times the corresponding sides of ∆ABC.
Constructions Class 10 Extra Questions Maths Chapter 11 with Solutions Answers 19
Justification:
Since, DE || CA
Constructions Class 10 Extra Questions Maths Chapter 11 with Solutions Answers 20
Hence, we have the new triangle similar to the given triangle whose sides are equal to \(\frac{5}{3}\) times the corresponding sides of ∆ABC.

Question 5.
Construct a triangle similar to a given triangle ABC with its sides equal to \(\frac{3}{4}\) of the corresponding sides of the triangle ABC ( i.e., of scale factor \(\frac{3}{4}\)).
Solution:
Steps of Construction:
Step I: Draw any ray BX making an acute angle with BC on the side opposite to the vertex A.
Step II: Locate 4 arcs B1, B2, B3, and B4 on BX so that
BB1 = B1B2 = B2B3 = B3B4.
Step III: Join B4C and draw a line through B3 parallel to B4C to intersect BC at C’.
Step VI: Draw a line through C’ parallel to the line CA to intersect BA at A’ (Fig. 9.14).
Then, ∆A’ BC’ is the required triangle.
Let us now see how this construction gives the required triangle.
Constructions Class 10 Extra Questions Maths Chapter 11 with Solutions Answers 21
Justification:
Constructions Class 10 Extra Questions Maths Chapter 11 with Solutions Answers 22

Question 6.
Draw a triangle ABC in which AB = 4 cm, BC = 6 cm and AC = 9 cm. Construct a triangle similar to ∆ABC with scale factor \(\frac{3}{2}\). Justify the construction. Are the two triangles congruent?
Note that all the three angles and two sides of the two triangles are equal.
Solution:
Steps of Construction:
Step I: Draw a line segment BC = 6 cm.
Step II: With centre B and radius 4 cm draw an arc.
Step III: With centre C and radius 9 cm draw another arc which intersects the previous arc at A.
Step IV: Join BA and CA. ABC is the required triangle.
Step V: Through B, draw an acute angle CBX on the side opposite to vertex A.
Step VI: Locate three arcs B1, B2, and B3 on BX such that BB1 = B1B2 = B2B3.
Step VII: Join B2C.
Step VIII: Draw B3C’ || B2C intersecting the extended line segment BC at ∠C’.
Step IX: Draw C’A’ || CA intersecting the extended line segment BA to A’.
Thus, ∆A’BC’ is the required triangle (∆A’BC’ ~ ∆ABC).
Constructions Class 10 Extra Questions Maths Chapter 11 with Solutions Answers 23
Justification:
∵ B2C || B3C’
Constructions Class 10 Extra Questions Maths Chapter 11 with Solutions Answers 24

Constructions Class 10 Extra Questions HOTS

Question 1.
Draw a circle with the help of a bangle. Take a point outside the circle. Construct the pair of tangents from this point to the circle.
Solution:
Constructions Class 10 Extra Questions Maths Chapter 11 with Solutions Answers 25
Steps of Construction:
Step I: Draw a circle with the help of a bangle.
Step II: Let P be the external point from where the tangents are to be drawn to the given circle. Through P, draw a secant PAB to intersect the circle at A and B (say).
Step III: Produce AP to a point C, such that AP = PC, i.e., P, is the mid-point of AC.
Step IV: Draw a semicircle with BC as diameter.
Step V: Draw PD ⊥ CB, intersecting the semicircle at D.
Step VI: With P as centre and PD as radius, draw arcs to intersect the given circle at T and T1.
Step VII: Join PT and PT1. Then, PT and PT1 are the required tangents.

Question 2.
Draw a ∆ABC with side BC = 7 cm, ∠B = 45°, ∠A = 105°. Then, construct a triangle whose sides are \(\frac{3}{4}\) times the corresponding sides of ∆ABC.
Solution:
Steps of Construction:
Step 1: Construct a ∆ABC in which BC = 7 cm,
∠B = 45°, ∠C = 180° – (∠A + ∠B)
= 180° – (105° + 45°) = 180o – 150° = 30°.
Step II: Below BC, make an acute angle ∠CBX.
Step III: Along BX, mark off four arcs: B1, B2, B3, and B4 such that BB1 = B1B2 = B2B3 =B3B4.
Step IV: Join B4C
Step V: From B3, draw B3D || B4C meeting BC at D.
Step VI: From D, draw ED || AC, meeting BA at E. Then EBD is the required triangle whose sides are \(\frac{3}{4}\) times the corresponding sides of ∆ABC.
Constructions Class 10 Extra Questions Maths Chapter 11 with Solutions Answers 26
Justification:
Constructions Class 10 Extra Questions Maths Chapter 11 with Solutions Answers 27
Hence, we have the new triangle similar to the given triangle whose sides are equal to \(\frac{3}{4}\) times the corresponding sides of ∆ABC.

Question 3.
Draw a pair of tangents to a circle of radius 4 cm which are inclined to each other at an angle of 60°
OR
Draw a circle of radius 4 cm. Construct a pair of tangents to it, the angle between which is 60°. Also justify the construction. Measure the distance between the centre of the circle and the point of intersection of tangents.
Solution:
Steps of Construction:
Step I: Draw a circle with centre 0 and radius 4 cm.
Step II: Draw any diameter AOB.
Step III: Draw a radius OC such that ∠BOC = 60°.
Step IV: At C, we draw CM ⊥ OC and at A, we draw AN ⊥ OA.
Step V: Let the two perpendiculars intersect each other at P. Then, PA and PC are required tangents.
Constructions Class 10 Extra Questions Maths Chapter 11 with Solutions Answers 28
Justification:
Since OA is the radius, so PA has to be a tangent to the circle. Similarly, PC is also tangent to the circle.
∠APC = 360° – (∠OAP + ∠OCP + ∠AOC)
= 360° – (90° + 90° + 120°) = 360° – 300° = 60°
Hence, tangents PA and PC are inclined to each other at an angle of 60°

Unseen Passage for Class 9 Discursive CBSE With Questions And Answers

This grammar section explains English Grammar in a clear and simple way. There are example sentences to show how the language is used. Unseen Passage for Class 9 Discursive CBSE With Questions And Answers are Explained in this article

Unseen Passage for Class 9 Discursive CBSE With Questions And Answers PDF

Discursive writing expresses an opinion on issues. It can be argumentative. It may give reasons, explanations or explores cause and effect relationship. You can also visit the most accurate and elaborate NCERT Solutions for Class 9 English. Every question of the textbook has been answered here.

Reading Comprehension for Grade 9 With Questions And Answers PDF (Solved)

1. Read the passage given below and answer the questions that follow:

1. Experts say that what customers buy in the name of herbal cosmetics is often the same synthetic cosmetic with a herb or two added, which works to lure customers. A product that says it is 100% natural may be misleading in the sense that it’s difficult to stabilize a fully natural base and is always quite expensive, requiring advanced technology. The desired shelf life of a product is expected to be at least two years, but that of a purely herbal product would be restricted to about six months. Also, the product won’t be aesthetically appealing as a synthetic product. Hence manufacturers of herbal products have to include synthetic base ingredients to balance the formula.

2. Herbal ingredients might also cause allergies as they may contain a large number of constituents which may be allergic to that user. Thus the concept of using ‘total extracts’ may be harmful, because there are several other ingredients going into the product. Also, things like pH need to be considered before using it.

3. According to the law experts, customers suing someone is considered extreme. Given the legal system in India and the long-drawn litigation process, consumers tend to pursue legal action against manufacturers/distributors in the event of any problem. However, with the dawn of consumer forums and more effective laws protecting the interests of the consumers, the trend is slowly moving towards this direction.

4. At times it is not just manipulative manufacturers and passive consumers but sometimes weak and absent rules regarding the process of manufacturing can also get fake production licences. Also one cannot make a generalisation that synthetics are safe while naturals are harmful. I would say that the difference is that if one is a known devil (synthetics) the other is unknown. Unlike the case of synthetic cosmetic, herbal cosmetics are still in a nascent stage. You don’t have well-defined standards for the use of raw materials in the production of herbal cosmetics.

5. Finally, customers can also check out their cosmetics by using the ‘Cosmetics ingredient dictionary, a database of 19,000 ingredients. It claims to help users check out what chemical compounds they use daily. This App lets you feed the names of the ingredients written on the container into the app and then get the reading about it.

1.1 Read the given questions and write the answer in about 30–40 words:

(a) Why is it misleading when a product is labelled as 100% natural?
Answer:
It is misleading because it is difficult to stabilise a fully herbal base for the cosmetic. Also, natural cosmetics are expensive and require advanced technology.

(b) Mention two drawbacks of natural products.
Answer:
Natural cosmetics have a much shorter shelf life of just six months unlike synthetic products that last two years. Also, they are not as appealing aesthetically as synthetic products.

(c) Why is taking recourse to legal action gaining momentum?
Answer:
Legal action is gaining momentum because of the emergence of consumer forums and . more effective laws to protect the interest of consumers. This trend is making a difference.

(d) What is the cosmetics ingredient dictionary?
Answer:
The Cosmetics Ingredients Dictionary is an App device with a data base of 19,000 names of ingredients. It helps check out claims of what chemical compounds they use daily, by means of its readings.

1.2 On the basis of your reading of the passage, answer the following:

(a) As per the passage, customers in India don’t easily sue someone because:
(i) legal systems are pathetic
(ii) legal systems are long-drawn
(iii) clients lose the time and money
(iv) all of the above
Answer:
(iv) all of the above

(b) The desired shelf life of a purely herbal product is about:
(i) eight months
(ii) six months
(iii) nine months
(iv) one year
Answer:
(ii) six months

(c) The Cosmetics Ingredients Dictionary is a dictionary about cosmetics. (True/False)
Answer:
False

(d) Find the word from the passage that mean the same as ‘skilful at influencing in an unfair way. (para 4)
Answer:
“Manipulative’

2. Read the passage given below and answer the questions that follow:

1. One of the great values of punctuality is that it gives discipline to life. We have to get up in time.

We have to do things at the appointed time. All these entail certain amount of sacrifice. It dispels laziness and removes our ‘take-it-easy attitude’. A disciplined person always gets recognition and social acceptance. He is wanted and appreciated. Therefore, punctuality can make us socially acceptable people.

2. Another significant merit of punctuality is that it provides ample time to do our work correctly and properly. Doing things hurriedly or haphazardly can have disastrous consequences. When we do things in time there is every chance that they end up as fine works.

3. The virtue of punctuality is said to be the key to success. Look at the great world leaders who have achieved fame and success. Punctuality was their hallmark. They kept their promises. Punctuality is a virtue that is appreciated by all. Washington once took his secretary to task for being late. The secretary laid the blame upon his watch. Washington reported: “Then, Sir either you must get a new watch or I must get a new secretary.” People like them are ideals whom we should follow in earnest.

4. When individuals are not punctual they cause a lot of inconvenience to others. People have to wait for them and waste their valuable time. Want of punctuality reveals want of culture and is discourteous to the person we fail. Unpunctuality invites trouble and worry. History is full of cases which show that lack of punctuality has caused defeat, loss of kingdom and golden opportunities. It is said that Napoleon lost the battle of Waterloo in 1815 because one of his generals came late. Many people lose good. opportunities of job or promotion when they reach late for appointment.

5. All of us are not born with the virtue of punctuality. We have to cultivate it painstakingly. Only constant vigil and practice can implant this virtue. It calls for great deal of sacrifice. It calls for courage to root out laziness and the ‘take-it easy attitude’. It demands a disciplined life. That is why very few individuals have the virtue of punctuality. But, know it for certain that it is the surest way to success.

2.1 Read the given questions and write the answer in about 30-40 words:

(a) What is the writer’s concern in this passage?
Answer:
In this passage, the writer intends to say that no one is born with the virtue of punctuality but one has to cultivate it through constant vigil and practice.

(b) What would be the consequences of not maintaining punctuality in your work?
Answer:
If punctuality is not maintained in our work, it invites trouble and worry. We cannot do our work successfully. As a result, it has disastrous consequences—we lose opportunity and have to suffer for unpunctuality in the work.

(c) Give some examples to show that lack of punctuality has caused trouble and worry.
Answer:
Lack of punctuality causes trouble and worry. Look at the great world leader like Washington. His secretary was taken to task for being late. He lost his job. Again, Napoleon lost the battle of Waterloo in 1815 because one of his generals came late.

(d) How did the great world leaders achieve fame and success in their life?
Answer:
The world leader achieved fame and success in their life by dint of punctuality and strict discipline.

2.2 On the basis of your reading of the passage, answer the following:

(a) A disciplined person always gets:
(i) recognition
(ii) social acceptance
(iii) work done correctly and property
(iv) all of the above
Answer:
(iv) all of the above

(b) If you are not disciplined, you will:
(i) Work hurriedly or haphazardly
(ii) You will be punctual
(iii) be appreciated by people
(iv) none of the above
Answer:
(i) Work hurriedly or haphazardly

(c) We are not born with the virtue of punctuality.
Answer:
True

(d) Give one word for ‘to keep a strict watch’. (para 5)
Answer:
vigil

3. Read the passage given below and answer the questions that follow:

1. The apologists of terror tell us that the root cause of terrorism is the deprivation of national and civic rights, and that the way to stop terror is to redress the supposed grievances that arise from this deprivation.

2. But the root cause of terrorism, the deliberate targeting of civilians, is not the deprivation of rights. If it were, then in the thousands of conflicts and struggles for national and civil rights in modern times we would see countless instances of terrorism. But we do not.

3. Mahatma Gandhi fought for the Independence of India without resorting to terrorism. So too did the people of Eastern Europe in their struggle to bring down the Berlin Wall and Martin Luther. King’s campaign for equal rights for all Americans eschewed all violence, much less terrorism.

4. If the deprivation of rights is indeed the root cause of terrorism, why did all these people pursue their cause without resorting to terror? Put simply, because they were democrats, not terrorists. They believed in the sanctity of each human life, were committed to the ideals of liberty, and championed the values of democracy.

5. But those who practise terrorism, do not believe in these things. In fact, they believe in the very opposite. For them, the cause they espouse, is so all encompassing, so total, that it justifies anything. It allows them to break any law, discard any moral code and trample all human rights in the dust. In their eyes, it permits them to indiscriminately murder and maim innocent men and women, and lets them blow up a bus full of children.

6. There is a name for the doctrine that produces this evil. It is called totalitarianism. Only a totalitarian regime, by systemically brainwashing its subjects, can indoctrinate hordes of killers to suspend all moral constraints for the sake of a twisted cause. That is why from its inception totalitarianism has always been wedded to terrorism–from Lenin to Stalin to Hitler to the Ayatollahs to Saddam Hussein, right down to Osama Bin Laden and Yasser Arafat

7. It is merely that the goals of terrorists do not justify the means they choose, it is that the means they choose tell us what true goals are. Those who fight as terrorists, rule as terrorists. People who deliberately target the innocent, never become leaders who protect freedom and human rights. When terrorists seize power, they invariably set up the darkest of dictatorships—whether in Iraq, Iran, Afghanistan or Arafatistan.

3.1 Read the given questions and write the answer in about 30–40 words:

(a) What according to some, is the root cause of terrorism? How can it be stopped?
Answer:
The root cause of terrorism is the deprivation of national and civic rights. Terrorism, according to apologists, can be stopped redressing the grievances of those deprived of their rights.

(b) Prove that the root cause of terrorism is not the deprivation of rights.
Answer:
Deprivation is not the root cause of terrorism. This is proved by the fact that conflicts and struggles for national and civil rights have been fought without resorting to terrorism.

(c) Mention two international personalities who fought for rights without resorting to terrorism.
Answer:
Gandhiji fought for the independence of India without resorting to terrorism. In the same manner when Martin Luther King fought for equal rights for all Americans, he too did not resort to any form of terrorism and eschewed all forms of violence.

(d) What are the beliefs of terrorists?
Answer:
Terrorists believe in their cause and for fulfilling that, they are willing to break any law, discard moral codes, trample human rights and indiscriminately take the lives of innocent women and children and blow up buses and bridges.

3.2 On the basis of your reading of the passage, answer the following:

(a) People who fought for equal rights without resorting to terrorism were ……..
(i) terrorists
(ii) autocrats
(iii) democrats
(iv) socialists
Answer:
(iii) democrats

(b) Those who practice terrorism justify ………….
(i) murdering people indiscriminately
(ii) maim innocent women and children
(iii) blow up buses filled with children
(iv) all of the above
Answer:
(iv) all of the above

(c) The name for the doctrine that produces this evil is called ………
Answer:
totalitarianism

(d) Find the word from the passage that is similar to ‘aiming’. (para 2)
Answer:
‘targeting

4. Read the passage given below and answer the questions that follow:

Role of Music in Life Today, we are rushing through life and everyone seems to be pressurised and stressed about one or the other thing. Stress has become an inescapable part of our lives. Many of our illnesses are a result of stress, and to deal with it, we need certain ways to relax our body and mind. Listening to music, particularly gentle, calming and melodious music, relaxes people both physically and mentally. Music releases endorphins, the natural painkillers in our brain, which relieve us from the sense of pain. Music can normalise heart rate and blood pressure. The common occurrences of fatigue and boredom decrease when we listen to music.

It diverts our attention from everyday anxieties, thereby reducing the stress caused by such concerns. While meditating, listening to music helps us to connect with our soul. Students can improve their concentration and boost their memory by listening to soft music. Soothing music, when played at night, helps us to rest and relax for a good night’s sleep. Listening to good music helps in overcoming negative traits such as anger or worry, thereby improving our personality.

It creates positive energy and happiness. Every cell in the body becomes energetic with increased peace of mind. Music is a therapy for everyone plants, birds, animals and human beings — to flourish and rejuvenate. It certainly plays an important part in making the world better for all living beings.

Complete the following statements by selecting the most appropriate options given below.

(a) helps in relaxation of our body and mind.
(i) Pop music
(ii) Soothing music
(iii) Jazz music
(iv) Rock music
Answer:
(ii) Soothing music

(b) The natural painkillers released in our brain are
(i) Insulin
(ii) Endorphins
(iii) Endocrine
(iv) Glycerin
Answer:
(ii) Endorphins

(c) Music is helpful in improving the _________ of students.
(i) concentration
(ii) emotion
(iii) relaxation
(iv) anxiety
Answer:
(i) concentration

(d) Which word from the passage means the same as ‘to renew”?
(i) rejuvenate
(ii) enhance
(iii) release
(iv) relieve
Answer:
(i) rejuvenate

(e) List the benefits of music mentioned in the passage.
Answer:
Music relaxes people both physically and mentally. Music releases endorphins, the natural painkillers in our brain, which relieve us from the sense of pain. Music can normalize heart rate and blood pressure. The common occurrence of fatigue and boredom decrease when we listen to music. Music diverts our attention from everyday anxieties thereby reducing stress.

5. Read the passage given below and answer the questions that follow:

1. With the next round of the Commonwealth Games coming up this year, sports fans are already speculating about the likely winners from India. While chances appear bright in some sporting activities, in others the picture appears dismal. Chances of India throwing up a few surprises is always discussed among sports lovers. Most game watchers predict that while India can add more to its medals tally in the shooting events, the chances of improving her status in the swimming category appear slim, despite the potential of creating good swimmers in this country.

2. One sport in the country with tremendous potential but pathetic performance is swimming. The country is flooded with talented swimmers and coaches whose potential is not fully utilised due to red-tapism and bureaucratic hurdles. In spite of being a country full of rivers and canals, swimming has failed to capture the imagination of Indians at large. The Government, on its part, has done very little to boost the sport.

3. All those persons, who are interested in swimming, realise that bad quality of water is one rampant problem with almost 90% of swimming pools. As for schools, it requires exorbitant sums of money and the schools cannot afford it. A handful of the privileged few, who enjoy this luxury, fall in the category of the elite. These are out of reach of the common man.

4. Where do the Indians stand today in comparison with international swimmers? The history of swimming in India has not been too bright. International winners have excellent facilities in terms of coaching, nutrition, tactics, positive attitude and hard work. Paucity of high calibre international coaches is one setback Indians have suffered in all the arenas of sport. Thanks to the petty gains and trivial politics, the good ones are dropped like a hot potato and the blue-eyed ones taken over for participating in international matches. Except for the metros, the country is deprived of good Olympic-size swimming pools.

5. The government and sports organisations will have to make serious efforts to transform the future of Indian sports. Till then, it is a long, long wait.

5.1 Read the given questions and write the answer in about 30-40 words:

(a) What do sports fans predict about India’s chances in the Commonwealth Games this year?
Answer:
India’s chances of throwing up a few surprises are predicted by sports fans in India. Also, her chances in shooting hold promise but her chances of improving her status in swimming are bleak, despite the potential of creating good Indian swimmers.

(b) What factors make for India’s potential in swimming? What are the drawbacks?
Answer:
Despite several rivers and canals, swimming as a sport is not popular. The country has several talented swimmers and coaches, but red-tapism and bureaucratic hurdles make it difficult to utilise their potential and thus swimming has failed to inspire youngsters.

(c) What problem do interested swimmers face in India?
Answer:
Swimmers find that the quality of water in 90% of swimming pools is bad. Schools are unable to afford the exorbitant funds needed for the sport. Only the select elite can enjoy swimming as it remains out of reach of the common man.

(d) What is India’s tally in the international swimming arena?
Answer:
Indian swimmers lack facilities in terms of coaching, nutrition, tactics, positive attitude and hard work and Olympic size swimming pools as against International standards. Also, there is a paucity of high calibre trainers and good coaches get sidelined while favourites represent India in international matches.

5.2 On the basis of your reading of the passage, answer the following:

(a) One sport of our country that has tremendous potential is per the passage is:
(i) Kho-Kho
(ii) swimming
(iii) wrestling
(iv) javelin
Answer:
(ii) swimming

(b) International winners have excellent facilities. They have: trained coaches
(ii) nutritional supplements
(iii) the right tactics and hard work
(iv) all of the above
Answer:
(iv) all of the above

(c) India has the potential of creating good …………….. in the country.
Answer:
swimmers.

(d) Give the synonym of ‘inundated’. (para 2)
Answer:
‘flooded

6. Read the passage given below and answer the questions that follow:

1. Photographer Marie-Caroline Senlis came to India for the first time 15 years ago to photograph a wedding in Udaipur, and succumbed to the charms of this experience. She found herself irrevocably fascinated by the country and returned again in 2009. So when she had to go back to France, once again, she took on a project. She chose a few expats who were about to leave India after a long stay and decided to tell the Indian slice of their life through stories and a portrait. As a single portrait cannot give the idea of the life of her subjects, she decided to tell their stories, through a small interview with each portrait and the series culminated into a book Before Leaving, Indian Snapshots.

2. Talking of the kind of preparation that went into each of her portrait sessions with her subjects, Marie Caroline says: “Before I met these people for the shoot I asked them to think of something that would represent their stay in India because I wanted to include that in the portrait. Some chose books, some chose pictures, some chose a place in their house. It was symbolic for them in some way.

3. Senlis chose a day close to the time of their departure for the portrait and did everything in a single session. This is because she thinks that just before leaving some place you realise a lot of things. Earlier you are just living in the city but just as you are about to leave, the emotions are stronger, deeper.

4. She does not like to pick a favourite from among the portraits of the series. But she often re-reads the entire script of the interview with her subjects during the session of the portrait. Though a lot of her subjects have had a lot to say about this country, there were some people who refused to be part of her project.

5. Senlis has also included herself in this project and chose for her photographs, some books and her music book, for the portrait. As she had started singing in India, she included the music book. As she read a lot of Indian literature during her stay here, those too find a place in the portrait. But what she misses most according to her own statement is: ‘the noise, the smells and the food.’

6.1 Read the given questions and write the answer in about 30–40 words:

(a) What do we know about Senlis’ coming to India?
Answer:
Marie-Caroline Senlis first came to India 15 years ago as a professional photographer to shoot an Indian wedding in Udaipur. She next came to India in 2009, and when leaving India once again, she took on a project.

(b) What is unique about the portraits that Senlis makes?
Answer:
The portraits are of expats in India on the verge of their leaving India. The portraits give viewers an idea of the slice of their Indian stay through stories and a portrait. The portraits also include an interview with Senlis.

(c) How are the portraits made?
Answer:
Her subjects think of something that would represent their stay in India, which are then included in the portrait. As people make varied choices such as books, pictures or favourite corners of their home in India, the portraits become unique.

(d) How does Senlis sum up her overall stay in India?
Answer:
Senlis says that she had started singing in India, so she included the music book in the portrait. She says that she will miss the foods, the smells and the noise that she had experienced in India. Her self-portrait depicts her books of Indian literature and her music book.

6.2 On the basis of your reading of the passage, answer the following:

(a) When Senlis had to go back to France:
(i) she refused to go
(ii) she took on a new project
(iii) she preferred to go to London instead
(iv) she started photography classes in India
Answer:
(ii) she took on a new project

(b) Senlis feels that when one is about to leave a country the emotions are:
(i) stronger
(ii) deeper
(iii) more intense (with a lot of realisation)
(iv) all of the above
Answer:
(iv) all of the above

(c) Maria-Caroline Senlis had come to India to shoot a wedding in Udaipur. (True/False)
Answer:
True

(d) Find the antonym of the word “whole’. (para 4)
Answer:
part

7. Read the passage given below and answer the questions that follow:

When we think of tourism, we think primarily of people who are visiting a particular place for sightseeing, visiting friends and relatives, taking a vacation, and having a good time. They may spend their leisure time engaging in various sports, sunbathing, talking, singing, taking rides, touring, reading or simply enjoying further. We may include in our definition of tourism people who are participating in a convention, a business conference, or some other kind of business or professional activity. Those who are taking a study tour under an expert guide or doing some kind of scientific research or study are also doing tourism.

These visitors use all forms of transportation from hiking in a wilderness park to flying in a jet to an exciting city. Transportation can include taking a chairlift up a Colorado mountainside or standing at the rail of a cruise ship looking across the blue Caribbean. Whether people travel by one of these means or by car, motor coach camper, train, motorbike or bicycle, they are taking a trip and thus are engaging in tourism.

(Adapted from Tourism: Principles, Practices, Philosophies by Charles R. Goeldner and J.R. Brent Ritchie, p. 4)

(a) Tourism is about taking a tour
(i) to a place under an expert guide
(ii) for hiking in remote places
(iii) for visiting a place for sightseeing
(iv) All of the above
Answer:
(iv) All of the above

(b) By ‘hiking in wilderness’ the author means
(i) taking a long distance walk in abandoned areas
(ii) walk in the forest areas
(iii) marathon in uninhabited areas
(iv) sprinting in abandoned, uninhabited areas
Answer:
(ii) walk in the forest areas

(c) A cruise ship is
(i) a large ship that carries people on voyages of pleasure.
(ii) a big ship that carries people and goods on special mission.
(iii) a large watercraft for carrying passengers from one point to another.
(iv) a large ship that carries mail, goods and first aid facilities.
Answer:
(i) a large ship that carries people on voyages of pleasure.

(d) What all activities do tourists engage in?
Answer:
Tourists engage in all kinds of activities like sports, sunbathing, business, hiking, etc.

(e) What are the forms of transportation tourists use for visiting places?
Answer:
Cruise ship, car, aeroplane, train, motorbike, bicycle, etc.

8. Read the passage given below and answer the questions that follow:

1. In 1923, a New York based agriculturalist wrote about the Ansault pear. U. P. Hetrick, the agriculturalist, praised the pear and called it better than any other kind of pear. He commented that it had a rich sweet flavour and a distinct but delicate perfume. But where can you get an Ansault pear these days? The answer is: nowhere. Thus is because this variety of pear is extinct. A similar situation has arisen in the case of apples. Once apple growers had more than 3000 varieties of apples to choose from. Today, they have barely a thousand.

2. The story of dietary shrinking is not restricted to fruits alone. In North America there were once hundreds of different breeds of cattle. Today, a single breed, the Holstein Freesian, accounts for 90% of dairy cattle raised in the US, and another 4% are Jersey cattle. All other dairy breeds occupy the remaining 6%.

3. Another interesting feature about the disappearance of breeds and varieties is that it is those varieties that are unique to a single local region which tend to disappear fast. This disappearance is due to the fact that in these regions, the community have never expanded their agricultural varieties beyond local confines. They have never propagated their varieties of agricultural produce beyond their own requirements. Thus when small farms or backyard operations close up, or decide to switch over to the conventional breeds, the local varieties disappear. As a result, compared to pre-1900 days, about 75% of global farmed plant diversity is gone.

4. The real culprit behind this mass destruction is the agribusiness industry. Giant agricultural operations develop and grow fruit and vegetables specifically for giant farms. Such giant production centres concentrate on a single variety of fruit or breed, chosen for its high-yielding potential. These are then hybridised for higher yield. Many of them cannot even produce offspring and thus have put to an end the age-old tradition of gathering seeds for the next year’s crop. Besides being non-productive, such plants also require intensive fertilizers, pesticides, and insecticides. They are grown only if they can withstand mechanical harvesting and the rigours of shipping to distant markets. The sweet taste of local produce, grown in one’s own orchard is now all but forgotten.

8.1 Read the given questions and write the answer in about 30-40 words:

(a) What is the Ansault pear?
Answer:
The Ansault pear is a variety of fruit that was grown extensively and had a sweet flavour and distinct but delicate perfume. Today, this variety of fruit has become extinct.

(b) Why have certain varieties of agricultural produce become extinct?
Answer:
Certain varieties of agricultural produce have shrunk because they were confined to a single region and a particular community that did not propagate it further. Thus, when the community decided to switch over to conventional breeds, the local varieties disappeared.

(c) How has agribusiness contributed to the disappearance of varieties?
Answer:
Agribusiness depends on giant production centres, growing fruit and vegetables specifically for giant farms. Such giant production centres concentrate on a single variety of fruit or breed, chosen for its high-yielding potential, which are further hybridised for giving higher yield.

(d) How are high yielding varieties produced?
Answer:
High yielding varieties are grown in giant farms using hybrid seeds, intensive fertilising, and the use of pesticides and insecticides. They are grown to withstand mechanical harvesting and the rigours of shipping to distant markets.

8.2 On the basis of your reading of the passage, answer the following:

(a) U. P. Hetrick, the New York based agriculturalist wrote in 1923 about:
(i) Red apple
(ii) The Durian
(iii) Ansault Pear
(iv) blueborries
Answer:
(iii) Ansault Pear

(b) The local varieties disappeared because:
(i) of pesticides and insecticides
(ii) the small farms close up or decide to switch over to the conventional breeds
(iii) there is tough competition in the market
(iv) none of the above Answer:
(ii) the small farms close up or decide to switch over to the conventional breeds

(c) The Ansault Pear variety is now extinct.
Answer: True

(d) Find the antonym of ‘construction’. (para 4)
Answer:
destruction

9. Read the passage given below and answer the questions that follow:

1. Certain foods can rejuvenate and activate the body, inducing even stable mental health and the advisory positions about the remarkable healing power of food. To recognise, isolate and increase the intake of foods that have large amounts of disease fighting antioxidants, to identify the two kinds of fat; the beneficial Omega-3 and the Omega-6, in which foods are commonly cooked; to alienate allergies caused by foods that work against the human metabolism.

2. Even oxygen has certain toxic forms called oxides, which spark off lethal reactions that have been linked to sixty odd chronic diseases, one of which is ageing. Antioxidants minimise the effects of the oxidants. Plant foods, thankfully are packed with antioxidant agents. Scientists are now researching into an antioxidant “Status report” based on individual blood tests; if the antioxidants are funnying low, specific food should be prescribed to boost the levels.

3. Fat comes in two types – Omega-3 which is found in marine life and Omega-6 which is . concentrated in vegetable oils. The first is good, the other is plain rotten.

4. The best source of Omega-3 is preferably sea fish. But frying it in Omega-6 rich vegetable oil kills all its goodness. The third imperative in codifying food health is through identifying irritants.

5. While some foods cause obvious and easily identified allergies like rashes, others cause either delayed reactions or minor irritants which could, nonetheless, be a serious deterrent to general well-being. Obstinate amoebiosis, nagging depression and persistent headaches are the most obvious symptoms. Food plays a dramatic role in alerting and fine-tuning of brain cells to give them sharper concentration. An innocuous combination of red wine and cheese can trigger off migraine.

6. Ageing brains have low levels of thiamin, which is concentrated in wheat-germ and bran, nuts, meat and cereals. More good brain-food comes from liver, milk and almonds, which are rich in riboflavin and extremely good for memory. Carotene, available in deep green leafy vegetables and fruits, is also good for geriatric brains. So is a high iron diet: it can make old brains gallop hyperactively like young ones. Iron comes from greens, liver, shell-fish, red meat and soyabeans Sea-food, very high in iron, is an excellent diet supplement.

7. The New England Journal of Medicine reported in its May 1985 issue that 30 grams of fish a day could result in a dramatic drop in the chances of acquiring a cardiovascular disease.

9.1 Read the given questions and write the answer in about 30–40 words:

(a) What are oxides? What effect do they have on the human body?
Answer:
Certain toxic forms of oxygen are known as oxides. Oxides spark off lethal reaction in the body and have been linked to around 60 chronic diseases, including the process of ageing.

(b) Why are antioxidants necessary? Which foods are rich in antioxidants?
Answer:
Antioxidants are useful in fighting diseases and minimising the effects of oxidants. Plant foods and other specific foods that are rich in antioxidant agents are prescribed for boosting antioxidant levels in humAnswer: These are prescribed following individual blood tests.

(c) Where is Omega-3 found? How can the good effect of Omega-3 fats be killed by Omega-6 fats?
Answer:
Omega-3 fats are found in marine life, particularly in sea fish. The positive properties of
Omega-3 fat get nullified when the fish is fried in vegetable oils containing Omega-6.

(d) What foods are necessary for geriatric brains? Answer: Foods with concentrated levels of thiamine, such as wheat-germ, and bran, nuts, meat and cereals, are good for geriatric brains. Also liver, milk, almonds, carotene-rich foods, fruits and an overall high iron diet are good for geriatric brains.

9.2 On the basis of your reading of the passage, answer the following:

(a) A harmless combination of red wine and cheese can trigger off:
(i) bodyache
(ii) nausea
(iii) cough and sneezing
(iv) headache
Answer:
(iv) headache

(b) Iron comes from:
(i) liver and shell-fish
(ii) greens
(iii) sea food and red meat
(iv) all of the above
Answer:
(iv) all of the above

(c) Fat comes in two types Omega-3 and …..
Answer:
Omega-6

(d) Write the antonym of the word “marked’. (para 5)
Answer:
‘innocuous’

10. Read the passage given below and answer the questions that follow:

1. Vegetarianism promotes a natural way of life. But despite its implicit message of universal love and non-violence it has not spread, as it should have. This may be because it usually is an inward looking habit and is best cultivated in the mind.

2. Leading a vegetarian way of life helps the animal kingdom to co-exist with man. The animals supply milk, manure and energy. This has been centred to the Indian culture for thousands of years. A vegetarian lifestyle is natural, multifaceted and helps self-preservation in a healthy way. Food and health are closely related.

3. In India, a vegetarian is usually a lacto-vegetarian. In the Western world vegetarians are sub divided as “vegans” (pure vegetarians who do not take any food coming from animal kingdom), lacto-vegetarians, who use dairy products of the vegetable kingdom and lacto-ova-vegetarians. The last category includes eggs, in addition to dairy products.

4. The Western science of food considers food as something to sustain only the human body, whereas the Indian science considers food as something which sustains not only the body, but also maintains the purity of heart, mind and the soul. Thus an item of food which is injurious to the mind, is not considered to be fit for consumption, even if it is otherwise beneficial to the body or satisfies the taste. Indian food science does not give so much importance to protein or even to the balanced diet but it gives importance to food that increases the strength of the body and its virility.

5. Vegetarian foods provide an infinite variety of flavours, whereas non-vegetarian foods have hardly any taste of their own. In fact, non-vegetarian foods have to be seasoned with ingredients from the vegetable kingdom to make them palatable.

6. In most sports disciplines, vegetarians lead in endurance tests. “You are what you eat” is an old saying and it is a fact that it is the food that makes the man. The food we eat, its quality, quantity, its timing and combination – is of utmost importance to healthy life.

7. It is significant to note that in the USA nearly 30 to 40 million people have adopted vegetarianism in the last decade. More and more intellectuals in the UK, Germany, France, Switzerland, Italy, Israel, Mexico, Russia and its former allies, are gradually turning to vegetarian diet not only on ethical and humanitarian grounds but also for health and ecological reasons.

10.1 Read the given questions and write the answer in about 30-40 words:

(a) Why has vegetarianism not spread?
Answer:
Vegetarianism has an inner message of universal love and non-violence. This message has. not spread because it is usually an inward-looking habit that promotes a natural way of life and is best cultivated in the mind.

(b) How are vegetarians classified in the western world?
Answer:
In the western world, vegetarians are classified into vegans, who are pure vegetarians and lacto-vegetarians who use dairy products of the vegetable kingdom and also lacto-ova vegetarians who take eggs.

(c) Besides sustaining the body, what else does Indian science consider necessary for food intake?
Answer:
According to Indian science, food is something which sustains not just the body but also maintains the purity of the mind, heart and soul. Thus, a food that is just beneficial to the body, satisfies taste and increases bodily strength and virility, is not given much importance.

(d) Why are more and more intellectuals in some European countries and Israel adopting a vegetarian diet?
Answer:
More and more intellectuals in some countries in Europe, including Russia and its former allies, as well as Israel, are adopting a vegetarian diet not only on ethical and humanitarian grounds, but also for health and ecological reasons.

10.2 On the basis of your reading of the passage, answer the following:

(a) What is of utmost importance about food to lead a healthy life?
(i) its quality
(ii) its quantity
(iii) its timing and combination
(iv) all of the above
Answer:
(iv) all of the above

(b) More and more people are turning vegetarian on :
(i) ethical grounds
(ii) health and ecological reasons
(iii) humanitarian grounds
(iv) all of the above
Answer:
(iv) all of the above

(c) Food and health are closely related to each other. (True/False)
Answer:
True

(d) Write the synonym of ‘compassionate’. (para 7)
Answer:
‘humanitarian?

11. Read the passage given below and answer the questions that follow:

1. The three major effects of viewing TV violence are : children may become less sensitive to the pain and suffering of others, they may be more fearful of the world around them, and they may be more likely to behave in aggressive or harmful ways towards others. Parents can limit the effects of violence by viewing TV together and discussing the violence with the child. Talk about why the violence happened and how painful it is. They may ask the child how conflicts can be solved without violence and explain to the child how violence on an entertainment programme is actually “fake”. Parents must restrict violent videos. Encourage your child to watch programmes with characters that cooperate, help and care for one another. These programmes have been shown to have a positive influence on children.

2. Television programming is all about showing reality without censorship, meaning that child watchers are inundated with shows depicting physical and sexual violence, as well as the use of illegal substances and harsh language. According to the American Academy of Pediatrics, “Extensive research evidence indicates that media violence can contribute to aggressive behaviour, desensitization to violence, nightmares, and fear of being harmed.” Statistics read that some cartoons average twenty acts of violence in one hour, and that by the age of eighteen children will have seen 16,000 simulated murders and 200,000 acts of violence on television. Young people pardy of the negative effects of television violence because “many younger children cannot discriminate between what they see and what is real.

3. Read to your child rather than watch television. Children’s stories, written by thoughtful authors, portray conflict in a more realistic and constructive manner. The best TV teachers are entertaining as well as informative, and they are also child-centred that expand the horizons of the young minds.

4. Watching these programmes provides children with the right stimulus to acquire knowledge through an indirect play way methodology. Absorption of such knowledge and the love for learning are stirred and the child begins to understand the explorative thrills of the knowledge track, where he becomes the racing car driver speeding the knowledge vehicle through his own speedometer of knowledge acquisition.

11.1 Read the given questions and write the answer in about 30–40 words:

(a) What traits can children imbibe by watching violence on the TV?
Answer:
Children may become insensitive to others’ sufferings, after viewing scenes of people in distress. They may become fearful of the world around them or may begin to behave aggressively and harmfully towards others.

(b) How can parents limit the effects of viewing violence on TV?
Answer:
If parents sit along with their children and watch TV, then they can limit violence viewing of their children on TV. Alternately, they can talk about why the violence happened, its painful effects and how conflicts can be solved without resorting to violence.

(c) What is the statistical finding about cartoon content?
Answer:
Statistics has shown that some cartoons have average 20 acts of violence, in an hour, and by age 18, children are exposed to 16,000 acts of simulated murders and 200,000 acts of violence on their TV screens.

(d) What alternative practices can parents follow and what will be its outcome?
Answer:
Parents could read to their children instead of viewing television. On TV they could view entertaining and informative programmes. The effect of such viewing is that children will be set on the path of self-exploration and knowledge gaining at their own pace.

11.2 On the basis of your reading of the passage, answer the following:

(a) Media violence, according to an American Paediatrics Academy says, can contribute to:
(i) nightmares
(ii) desensitization to violence
(iii) aggressive behavior
(iv) all of the above
Answer:
(iv) all of the above

(b) Watching child-centered programs:
(i) become uninteresting for children
(ii) provide the right stimulus to acquire knowledge
(iii) are not encouraged much by parents
(iv) are not entertaining and informative
Answer:
(ii) provide the right stimulus to acquire knowledge

(c) Programmes promoting values have been shown to have a positive influence on children. (True/False)
Answer:
True

(d) Find from the passage the word that is opposite to the word ‘friendly’. (para 1).
Answer:
‘aggressive’

12. Read the passage given below.

1. Being healthy does not necessarily mean only physical fitness. It also includes mental and emotional well-being. You need to follow some guidelines and maintain a time table of your daily activities in order to stay fit and fine throughout your life.

2. The first and the most important point to be taken care of in order to remain healthy is to follow healthy eating habits. One must avoid consumption of unhealthy junk foods. Try to include those foodstuffs that are rich in all the nutrients that are very essential for the proper growth of our body tissues. Make sure that these foods are rich in vitamins, minerals, proteins, good carbohydrates as well as fats. It is a misconception that fats, whether good or bad, are always harmful for your body. Not all fats are bad for health: There are some types of fats that are essential for the body as well. To make sure that your heart keeps healthy all the time, have a wholesome meal. By doing this you will also have a healthy brain and a good immune system.

3. Burning of calories is also very necessary for good health. For this, you will have to plan your schedule and give some time for exercises every day. This proves to be very beneficial in maintaining the wear and tear of all your body muscles. Another important point that you should always keep in mind is that you cannot stay healthy, by skipping meals. Have your food at proper time intervals and avoid starvation. Instead of eating a large quantity of food at one single time, try to have 6 to 7 small courses of meals at regular intervals. This will help you maintain proper functioning of your digestive system.

4. Apart from the solid food, you must also pay attention towards liquids. Include beverages such as coconut water, fruit juices, and especially water. You must drink at least 10 to 12 glasses of water in the entire day. This maintains a very good rate of metabolisim.

5. Sleep is another important factor with regard to health. Along with proper liquid and solid food you also need a sound and sufficient sleep to keep you fresh all day. If you do not have a sound sleep of 6 to 7 hours at night then you will feel tired and lazy all day and this will also affect your mood.

12.1 Read the given questions and write the answer in about 30–40 words:

(a) What do you understand by the term ‘being healthy??
Answer:
“Being healthy’ does not mean only physical fitness but it includes mental and emotional well-being. By following certain guidelines and maintaining a timetable daily in order one can stay fit and fine.

(b) What are the essential nutrients required for proper growth of our body tissues?
Answer:
The essential nutrients required for proper growth of our body tissues are vitamins, minerals, proteins, carbohydrate as well as good fats.

(c) What are the advantages of exercise?
Answer:
Exercise helps us in keeping good health. By doing exercise daily it helps burning of calories which is necessary to keep good health. Secondly, it compensates the damage to one’s body muscles. Thirdly, it helps to function one’s digestive system properly.

(d) Besides solid food and exercise, what other things are needed to maintain a good rate of etabolism?
Answer:
Besides solid food and exercise, beverages such as coconut water, fruit juices, and especially water are required to maintain a good rate of metabolism in the body.

12.2 On the basis of your reading of the passage, answer the following:

(a) As per the passage, which is right?
(i) all fats are not bad
(ii) some types of fats are essential for the body
(iii) it is a misconception that fats good or bad are harmful for your body
(iv) all of the above
Answer:
(iv) all of the above

(b) Along with sufficient liquid intake, proper food, this is also important:
(i) starve yourself sometimes
(ii) 6 to 7 hours sound sleep at night
(iii) skip your breakfast
(iv) eat in between the meals
Answer:
(ii) 6 to 7 hours sound sleep at night

(c) ………. help us in keeping good health in multiple ways.
Answer:
Exercises

(d) Find the word which means the same as “wrong notion’. (para 2)
Answer:
‘misconception’

13. Read the passage given below and answer the questions that follow:

1. If you thought that overweight was a problem among humans alone, then think again. A recent study carried out in UK on the state of obesity among pets revealed startling facts. According to this report there are almost 45% pets in the country that are overweight.

2. According to the findings, what causes this overweight problem is the feeding of leftovers. In fact this custom is regarded as the biggest factor causing overweight, according to 200 vets and 1000 pet owners. The report found that nearly 45% of the dogs were overweight while the obesity rate among cats was slightly lower, at 40%. Almost one in three rabbits (28%) and guinea pigs fell into the obese category, while 15% of caged birds were also found to be too fat. Of their owners, what the report revealed was that two out of every three owners believed their pets were the correct weight. Around three in four vets believe that pet obesity is on the rise.

3. When contacted by the researchers, vets came forward with the opinion that the root cause of pet weight gain was that the owners of these animals and birds were not following feeding guidelines. Along with feeding them with leftovers, these pets were not given adequate exercise. This problem was most marked among dog owners where also 78% of them were found to be indulged in by their owners.

4. Besides, lethargy and eating problems, these overweight pets suffer from a myriad of health issues according to Zara Boland, founder Vet Voice. These animals run the risk of osteoarthritis, cardiovascular disease, and diabetes. There is nothing ‘cuddly’ about an obese pet, she adds. 5. Obese animals have the same symptoms as obese humans Their obesity causes them discomfort, and illness and can cause result in both emotional distress and financial pressures for owners, and has been proven to reduce actual life length.

6. These vets say that they are committed to continuing pushing the pet health message until overweight pets are no longer an increasing and widespread concern. To keep their pets in the best of health these vets advised their owners to exercise their dog for thirty minutes twice daily for adults dogs, and forty minutes for cats.

13.1 Read the given questions and write the answer in about 30–40 words:

(a) What were the findings of the UK study on animals?
Answer:
The UK study on animals found that about 45% of pets were overweight. Of these 45% were dogs, around 40% were cats, one in three rabbits (28%) were overweight while 15% of caged birds, too, were overweight.

(b) What according to the vets was the cause of pet weight gain?
Answer:
According to the 200 vets and 1000 pet owners, it was found that the feeding of leftovers “to pets caused overweight due to their negligence and disobeying feeding guidelines. This problem was more acute among dog owners as 78% of them indulged their pets.

(c) What are the health problems that overweight pets are prone to?
Answer:
According to Zara Boland’s Vet Voice, it was found that overweight pets are lethargic, and suffer from a myriad health issues ranging from osteoarthritis, cardiovascular diseases and diabetes.

(d) How are vets trying to overcome this problem?
Answer:
Committed vets are continuing to push the pet health message widely so that overweight among animals is no longer an increasing and widespread concern. They also advise pet owners to exercise their dogs (30 minutes twice daily) and their cats (40 minutes) daily.

13.2 On the basis of your reading of the passage, answer the following:

(a) The recent findings in UK reveal that:
(i) pet obesity is on the rise
(ii) pets are not well-maintained by the owners.
(iii) pets are very often abandoned.
(iv) more. cats are overweight than dogs.
Answer:
(i) pet obesity is on the rise.

(b) Do obese animals have the same symptoms as obese humans?
(i) obese animals too suffer discomfort
(ii) they frequently fall sick
(iii) their obesity causes emotional and financial distress to the owners
(iv) all of the above
Answer:
(iv) all of the above

(c) The researchers were of the opinion that pet owners were not following the guidelines. (True/False)
Answer:
True

(d) Find the word similar in meaning to ‘lazyness’. (para 4)
Answer:
‘lethargy’

14. Read the passage given below and answer the questions that follow:

1. Thirteen years ago, when the then American President Bill Clinton drove down to Rampur.
Maniharan village in western UP, he had come there to open a women’s polytechnic, funded by a prominent NRI. Today, the Bill Clinton School stands bright and shiny on the same campus, among low-slung buildings that house laboratories, libraries and classrooms. According to the school president Rajkamal Saxena, there are 565 students studying under CBSE, of whom 234 are girls. Besides the local folk, the affluent families across the social spectrum send their children to this institution.

2. The success story of this school in the area has given rise to a spate of educational institutions along the 66-km road from Shamli to Saharanpur. All of them promise to unlock exciting career options, especially for the landed class of the area. These schools sport trendy names and have sprung up in the years when the 42nd President of the US came calling. Sitting amidst sugar cane fields and mango orchards they present a picture of a society straining to change.

3. Despite such a positive social factor visible in the area, there is little evidence all round of infrastructure changes in the area. The roads are battered and dusty. The dream of an all-weather road remains a dream. Steady power supply is unthinkable and traditional industries that once thrived in the area, are now on the decline. Added to these woes are the problems of governance such as the law and order situation. The discourse at public gatherings and among social groups is about the growing tensions among communities.

4. According to the locals, people in the region are looking for a change. The driving factor behind s the youth. Most of the young men who have returned to the area have been educated in Delhi and have returned to work in their home town and for them the development of the region overrides all other concerns. Among the semi-agrarian middle classes, the call for a change is not through the improvement of the region’s facilities but through the ballot box. They are eager for a change in the very government of the state.

14.1 Read the given questions and write the answer in about 30–40 words:

(a) What is the origin of the Bill Clinton School? What is its condition today?
Answer:
The Bill Clinton School was founded by the 42nd US President Bill Clinton as a women’s polytechnic. Today, the school continues to be a prominent institution with its low-slung buildings, housing laboratories, libraries and classrooms, and an intake of 565 students.

(b) How has the presence of this school affected the area?
Answer:
Following the success of this school in the area, there are a number of schools that have sprung up along the 66 km-road from Shamli to Saharanpur. They offer to unlock exciting career options specially for the landed community of the area.

(c) What are the infrastructure drawbacks of the area?
Answer:
The infrastructure drawbacks of the area include lack of proper roads, with no all-weather road. Power supply is erratic and traditional industries are on the decline. Also the law and order situation is problematic and at public gatherings people discuss growing tensions between communities.

(d) What changes are the local youth and middle classes looking for?
Answer:
The youths of the region are keen to see development of the region, overriding all its concerns. The semi-agrarian middle classes in the area want a change of government to improve the region’s facilities.

14.2 On the basis of you reading of the passage, answer the following:

(a) The women’s polytechnic was founded by …
(i) a prominent leader
(ii) a prominent politician
(iii) a prominent NRI
(iv) a prominent industrialist
Answer:
(iii) a prominent NRI

(b) The driving factor behind the urge for change are ……….
(i) the locals
(ii) old people
(iii) the youth
(iv) none of the above
Answer:
(iii) the youth

(c) The Bill Clinton school houses ………… students of whom ..go.. girls.
Answer:
565, 234

(d) Pick up the word from the passage which has the same meaning as ‘prospered’. (para 3)
Answer:
‘thrived

15. Read the passage given below and answer the questions that follow:

This is an extract from Harper Lee’s To Kill a Mocking Bird. After supper, Atticus sat down with the paper and called, “Scout, ready to read?”. I ran crying, went to the front porch. Atticus followed me. “Something wrong, Scout?” I told Atticus I didn’t feel very well and didn’t think I’d go to school any more if it was alright with him. Atticus sat down in the swing and crossed his legs. His fingers wandered to his watch pocket; he said that was the only way he could think. He waited in amiable silence, and I sought to reinforce my position: “You never went to school and you do alright, so I’ll just stay home too.

You can teach me like Granddaddy taught you ‘n’ Uncle Jack.” “No I can’t”, said Atticus. “I have to make a living. Besides, they’d put me in jail if I kept you at home. Now what’s the matter? Bit by bit, I told him the day’s misfortunes. “And the teacher said you taught me all wrong, so we can’t ever read any more, ever. Please don’t send me back, please Sir.” Atticus stood up and walked to the end of the porch. When he completed his examination of the wisteria vine he strolled back to me.

“First of all”, he said, “if you can learn a simple trick, Scout, you’ll get along a lot better with all kinds of folks. You never really understand a person until you consider things from his point of view-” “Sir?” “-until you climb into his skin and walk around in it”. “But if I keep on goin’ to school, we can’t ever read any more…” “That’s really bothering you, isn’t it?” “Yes sir.” When Atticus looked down at me I saw the expression on his face that always made me expect something.

“Do you know what a compromise is?”, he asked. “Bending the law?” “No, an agreement reached by mutual concessions. It works this way”, he said. “If you’ll concede the necessity of going to school, we’ll go on reading every night just as we always have. Is it a bargain?” “Yes sir!”

“We’ll consider it sealed without the usual formality”, Atticus said, when he saw me preparing to. spit. As I opened the front screen door Atticus said, “By the way, Scout, you’d better not say anything at school about our agreement”. “Why not?” “I’m afraid our activities would be received with considerable disapprobation by the more learned authorities.” My brother and I were accustomed to our father’s diction, and we were at all times free to interrupt Atticus for a translation when it was beyond our understanding. “Huh, Sir?”

“I never went to school”, he said, “but I have a feeling that if you tell Miss Caroline we read every. night she’ll get after me, and I wouldn’t want her after me”.

(a) Why was Scout, the little girl upset?
(i) She didn’t like being told what to do
(ii) She didn’t want to read with her father
(iii) She didn’t want to read with her teacher
(iv) She didn’t want to stop reading with her father
Answer:
(iv) She didn’t want to stop reading with her father

(b) How did Atticus, her father react to her outburst?
(i) He was patient
(ii) He was annoyed
(iii) He was sad
(iv) He was angry
Answer:
(ii) He was annoyed

(c) What little advice did Atticus give to his little girl to cope up with situations that might upset her?
(i) to know and understand that life isn’t fair.
(ii) to stay calm and then run away from the problem.
(iii) to try and see from the other person’s point of view.
(iv) to face her fears by expressing how she felt.
Answer:
(iii) to try and see from the other person’s point of view.

(d) How was the matter eventually resolved?
(i) Atticus agreed to allow her to learn from home.
(ii) Atticus agreed to continue reading as before.
(iii) Atticus agreed to speak to her teacher.
(iv) Atticus agreed to allow only the teacher to teach her.
Answer:
(ii) Atticus agreed to continue reading as before.

(e) “I’m afraid our activities would be received with considerable disapprobation by the more learned authorities.” In this context, the word ‘disapprobation’ might mean
(i) disapproval
(ii) disturbance
(iii) disgrace
(iv) disorientation
Answer:
(i) disapproval

(f) Atticus decided to come to a compromise and to read to Scout but asked her not to mention it to her teacher because
(i) he respected the wishes of the teacher yet didn’t want to disappoint his daughter.
(ii) he might get into trouble with the teacher who might come after him.
(iii) he only cared for his daughter and thought the teacher was incorrect.
(iv) he was afraid that the teacher may not like it and Scout might get into trouble.
Answer:
(i) he respected the wishes of the teacher yet didn’t want to disappoint his daughter.

16. Read the passage given below and answer the questions that follow:

1. A four-year-old sapling of a cherry stone that spent time aboard the International Space Station (ISS), burst into bloom a full six years ahead of Mother Nature’s normal schedule. Its early blooming has baffled Buddhist brothers of the ancient temple in central Japan where the tree is growing.

The wonder pip was among 265 harvested from the celebrated Chujo-hime-seigan-zakura tree, selected as part of a project to gather seeds from different kinds of cherry trees, at 14 locations across Japan. The stones sent to ISS in November 2008, came back to earth in July the following year, after circling the globe 4,100 times. While some were sent for lab tests, most were ferried back to their places of origin and a selection were planted in nurseries near the Ganjoli temple.

3. The Ganjoli temple sapling is not the only early-flowering sapling. Of the 14 locations in which they were planted, blossoms have been spotted in 4 places. Two years ago, a young tree bore 11 flowers in Hokuto, a mountain region 115 km west of Tokyo. The seeds had been sent to the ISS as part of an educational and cultural project to let children gather stones and learn how they grow into trees and live on, after returning from space. The project organiser had expected the Gangoji (tree) to blossom in ten years after planting, when the children would come of age.

4. Kaori Tomita-Yokotani, a researcher at the University of Tsukuba who took part in the project, said she was stumped by the extra-terrestial mystery. She said they could still not rule out the possibility that it was somewhat influenced by its exposure to the space environment. She said it was difficult to explain why the temple tree had grown so fast because there was no control group to compare its growth with that of other trees. She said cross-pollination with another species could not be ruled out. But the lack of data was hampering an explanation. 5. There is also a possibility that exposure to stronger cosmic ray accelerated the process of sprouting and overall growth. At present scientists would like to acknowledge that they still don’t know.

16.1 Read the given questions and write the answer in about 30-40 words:

(a) What is unique about the sapling that travelled in space?
Answer:
The cherry sapling that spent time aboard the International Space Station, has bloomed in four years instead of the normal ten year period. Its early blooming has baffled Buddhist monks of the Gangoli temple where the tree has been growing.

(b) How did the cherry stone reach the space station?
Answer:
The selected sapling is one of 265 cherry stones harvested from the celebrated Chujo-hime seigan-zakura tree, and sent into space in November 2008. After circling the globe 4100 times, it was among the select few planted in the nurseries of the Gangoli Temple.

(c) Why were the seeds sent to the ISS?
Answer:
The seeds were part of an educational and cultural project to let children learn cherry stones live on as trees later, after returning from space. The project organizer had expected the tree to flower in ten years when the children came of age.

(d) What is Tomati-Yokotami’s explanation of the early flowering?
Answer:
Baffled by the early flowering, researcher Tomati-Yokotami feels it was somewhat influenced by exposure to the extra-terrestial environment, but further explanation was difficult as no comparative group of other trees had been set up. Also, cross-pollination could not be ruled out.

16.2 On the basis of your reading of the passage, answer the following:

(a) The project organizers had expected the cherry-stone sapling to blossom in:
(i) eight years after planting
(ii) ten years after planting
(iii) six years after planting
(iv) seven years after planting
Answer:
(ii) ten years after planting

(b) The stones after coming back to earth were:
(i) sent for lab tests
(ii) ferried back to their places of origin
(iii) a selection were planted in nurseries
(iv) all of the above
Answer:
(iv) all of the above

(c) The four-year-old sapling spent time aboard the International Space Station (ISS). (True/False)
Answer:
True

(d) Find the antonym of the world ‘deny’ from the text. (para 5)
Answer:
acknowledge

17. Read the passage given below and answer the questions that follow:

1. The family of blue macaws is back! It is the second installation of Carlos Saldanha’s musical film, ‘Rio 2’, where Blu and Tulio and their three kids undertake an adventurous journey to the Amazon forest. As Blu rightly puts it, they are not the birdliest of birds as they gorge on. pancakes for breakfast while listening to their iPods and live under the notion that they are truly the last of the species. But then Lindo, who raised Blu and Tulio, sees a documentary relating how they have discovered an entire family of blue macaws in the Amazon forest. It is then that the family decide to travel to the Amazon to seek Jewel’s family. But then they are pursued by evil Nigel, who can’t fly anymore, poisonous frog Gabi, who is love-struck by Nigel, and ant-eating Charlie.

2. Alongside the rescue idea, is the theme of preservation of wildlife, emphasized as Linda and Tulio take on a gang of deforesters who are trying to encroach the macaw habitat. At the outset ‘Rio 2 is a family drama centred on the themes of love but it also addresses the dilemma of choosing between the city’s comforts and living in the wild. Portraying these two themes in the same film, there is never a dull moment in the film, which is colourful, fast-paced and entertaining. It is bursting with jokes and musical delights right from samba, to rap and lullabies.

3. What viewers describe as highlights of the film are Gabi’s Broadway-like imitation song where she expresses her unrequited and forbidden love for Nigel. There is also the Amazonian version of the Brazilian carnival, filled with music. An avian football match in the film, reminds one of Harry Potter’s Quiddich matches. And the vibrant colours of the Amazon forest look brilliant with the animals, particularly the funky turtles, who take forever to dance and even highfive.

4. Of the bad hats in the movie, one must mention the annoying man who hires illegal loggers to encroach upon the forest and distributes lollypops around. Linda and Tulio, who had played the chief roles in the earlier version, do not come into the limelight in this episode and the other characters too, are so few and far between, that it becomes confusing at times. Altogether, this movie will leave the samba notes ringing in your ears long after it is over.

17.1 Read the given questions and write the answer in about 30–40 words:

(a) What is ‘Rio 2′?
Answer:
‘Rio 2′, the second installation of Carlos Saldanha’s musical film, is about the last surviving family of macaws, who undertake an adventurous journey into the Amazon forest, learning that there is another family of macaws living there.

(b) What is the parallel theme being portrayed in the film?
Answer:
The parallel theme is a family drama, but it also portrays the dilemma of choosing between living in the city as opposed to living in the forest. The two divergent aspects are knitted together with entertaining highlights in the form of musicals.

(c) Describe some of the characters in the film.
Answer:
Gabi, the poisonous frog, sings about his unrequited and forbidden love for Nigel. There are also illegal loggers who encroach upon the forest and distribute lollypops around. The macaw couple in the film are Blu and Tulio and they have three kids.

(d) What are some of the outstanding scenes depicted in the film?
Answer:
There is a Brazilian-style carnival included as a happening in the Amazon and football match played by aviAnswer: The songs are based on Brazilian forms such as samba, rap and lullabies, and the frog even sings a Broadway imitation.

17.2 On the basis of your reading of the passage, answer the following:

(a) The family of the Macaws is pursued by:
(i) evil Nigel
(ii) poisonous frog Gabi
(iii) ant-eating Charlie
(iv) all of the above
Answer:
(iv) all of the above

(b) The name of the musical film of Carlos is:
(i) Happy feet
(ii) 100 Dalmatians
(iii) Rio 2
(iv) Baby’s Day out
Answer:
(iii) Rio 2

(c) There is a parallel theme running, that of preservation of ……….
Answer:
wildlife

(d) Find the phrase from the passage that means ‘devour something greedily’. (para 1)
Answer:
gorge on

18. Read the passage given below and answer the questions that follow:

1. Struck at their work stations, the majority of the Indian workforce leads a stressful and sedentary life. An event like running a marathon is an interesting diversion in such a situation. A corporate run is a celebratory occasion as it provides employees a chance for bonding together to encourage the spirit of community wellness and help showcase their endurance and their tenacity.

2. Seeing the advantages of running corporate marathons, several large organizations have realized that teams that play together stay together. Last year a giant firm like Microsoft concluded its month-long campaign with a five-km run for its employees. They saw participation by its employees from over 150 employees across Microsoft businesses in Hyderabad. Another US based company in India organized a 10-km marathon recently and had 3500 employees putting their efforts together and demonstrating a sense of social responsibility.

3. Wipro has started running a club at their Bengaluru office where more than 50 employees along with their families come to train for long-distance running with an expert coach. Even though training costs make up nearly two to three per cent of the total staffing budget, it is regarded as a worthwhile investment. The facility is available at no cost to the employees and their friends. Another company organized its 10-km run driven by the credo of reduce, reuse and recycle. A participant in this event said: “I often go cycling for fun on the weekends but this was the first time I had participated in an event like this. I was really happy to cycle all the way.’ In addition volunteers constantly combine the running with educating and creating awareness about greener living amongst the team members through various initiatives.

4. Marathons prove to be a spectacular way of combining social wellness programmes with and building a sense of social responsibility at the same time. Corporate runs and marathons help people relate to a larger sense of direction. They also bring a culture of working hard for long term goals. For a runner, at a personal level, it helps the individual gain immensely in terms of bringing discipline in life being able to engage within, staying focused and sustaining persistence over a longer period. Thus the marathon and corporate have become new milestones forged in the history of corporate relations.

18.1 Read the given questions and write the answer in about 30–40 words:

(a) Why are corporate runs organized?
Answer:
With employees stuck at workstations they become stressful and sedentary. A corporate run provides the chance to bond together and encourages the spirit of community wellness as also show off their tenacity and endurance.

(b) How did the Microsoft employees engage in a marathon?
Answer:
The 150 odd employees at Hyderabad concluded a month-long campaign by organizing a 10-km marathon in which 3500 employees put in their efforts and demonstrated a sense of social responsibility.

(c) How have Wipro been helped in their bid to run a marathon?
Answer:
Wipro employees have started a club at Bengaluru where 50 employees and their families and friends are coached in marathon running. They are also encouraged to follow the credo of reuse, reduce and recycle, thus combining running with educating employees.

(d) What are the advantages of running marathons?
Answer:
They help people relate to a sense of direction and develops a culture of working hard for long-term goals. Personally runners develop discipline, remain focused and persistent over a longer period.

18.2 On the basis of your reading of the passage, answer the following:

(a) Seeing the success of corporate marathons, several large organisation realized that:
(i) teams that play together dine together
(ii) teams that play together stay together
(iii) teams that play together, work-out together
(iv) none of the above
Answer:
(ii) teams that play together stay together

(b) Driven by what credo, did another company organize a 10-km run:
(i) reduce
(ii) reuse
(iii) recycle
(iv) all of the above
Answer:
(iv) all of the above

(c) Running a marathon is an interesting …………………. for Indian workforce leading a stressful and sedentary lifestyle.
Answer:
diversion

(d) Find a word for ‘continuing in spite of difficulties’ from the text. (para 4)
Answer:
‘persistance’