RS Aggarwal Class 10 Solutions Chapter 12 Circles Test Yourself

RS Aggarwal Class 10 Solutions Chapter 12 Circles Test Yourself

These Solutions are part of RS Aggarwal Solutions Class 10. Here we have given RS Aggarwal Solutions Class 10 Chapter 12 Circles Test Yourself.

Other Exercises

MCQ
Question 1.
Solution:
In the given figure,
PT is the tangent and PQ is the chord of the circle with centre O.
∠OPT = 50°
RS Aggarwal Class 10 Solutions Chapter 12 Circles Test Yourself 1
OP is radius and PT is the tangent.
OP ⊥ PT ⇒ ∠OPT = 90°
∠OPQ + ∠QPT = 90°
⇒ ∠OPQ + 50° = 90°
⇒ ∠OPQ = 90° – 50° = 40°
In ∆OPQ,
OP = OQ (radii of the same circle)
∠OQP = ∠OPQ = 40°
In ∆OPQ,
∠POQ = 180° – (∠OPQ + ∠OQP)
= 180° – (40° + 40°) = 180° – 80°
= 100° (b)

Question 2.
Solution:
Angle between two radii of a circle = 130°
RS Aggarwal Class 10 Solutions Chapter 12 Circles Test Yourself 2
Then ∠APB = 180° – ∠AOB
= 180°- 130° = 50° (c)

Question 3.
Solution:
In the given figure,
PA and PB are the tangents drawn from P to the circle with centre O
∠APB = 80°
RS Aggarwal Class 10 Solutions Chapter 12 Circles Test Yourself 3
OA is radius of the circle and AP is the tangent
OA ⊥ AP ⇒ ∠OAP = 90°
OP bisects ∠APB,
∠APO = \(\frac { 1 }{ 2 }\) x 80 = 40°
∠POA = 180° – (90° + 40°)
= 180° – 130° = 50° (b)

Question 4.
Solution:
In the given figure, AD and AE are tangents to the circle with centre O.
BC is the tangent at F which meets AD at C and AE at B
AE = 5 cm
RS Aggarwal Class 10 Solutions Chapter 12 Circles Test Yourself 4
AE and AD are the tangents to the circle
AE = AD = 5 cm
Tangents from an external point drawn to the circle are equal
CD = CF and BE = BF
Now, perimeter of ∆ABC = AB + AC + BC
= AB + AC + BF + CF (BE = BF and CF = CD)
= AB + AC + BE + CD
= AB + BE + AC + CD
= AE + AD
= 5 + 5 = 10 cm (b)

Short-Answer Questions
Question 5.
Solution:
In the given figure, a quadrilateral ABCD is circumscribed a circle touching its sides at P, Q, R and S respectively.
AB = x cm, BC = 7 cm, CR = 3 cm and AS = 5 cm
RS Aggarwal Class 10 Solutions Chapter 12 Circles Test Yourself 5
A circle touches the sides of a quadrilateral ABCD.
AB + CD = BC + AD …(i)
Now, AP and AS are tangents to the circle
AP = AS = 5 cm …(ii)
Similarly, CQ = CR = 3 cm
BP = BQ = x – 5 = 4
BQ = BC – CQ = 7 – 3 = 4 cm
x – 5 = 4
⇒ x = 4 + 5 = 9cm

Question 6.
Solution:
In the given figure, PA and PB are the tangents drawn from P to the circle.
OA and OB are the radii of the circle and AP and BP are the tangents.
OA ⊥ AP and OB ⊥ BP
∠OAP = ∠OBP = 90°
In quad. AOBP
∠A + ∠B = 90° + 90° = 180°
But these are opposite angles of a quadrilateral
AOBP is a cyclic quadrilateral
A, O, B, P are concyclic

Question 7.
Solution:
In the given figure, PA and PB are two tangents to the circle with centre O from an external point P.
∠PBA = 65°,
To find : ∠OAB and ∠APB
In ∆APB
AP = BP (Tangents from P to the circle)
∠PAB = ∠PBA = 65°
∠APB = 180° – (∠PAB + ∠PBA)
= 180° – (65° + 65°) = 180° – 130° = 50°
OA is radius and AP is tangent
OA ⊥ AP
∠OAP = 90°
∠OAB = ∠OAP – ∠PAB = 90° – 65° = 25°
Hence, ∠OAB = 25° and ∠APB = 50°

Question 8.
Solution:
Given : In the figure,
BC and BD are the tangents drawn from B
to the circle with centre O.
∠CBD = 120°
To prove : OB = 2BC
Construction : Join OB.
Proof: OB bisects ∠CBD (OC = OD and BC = BD)
RS Aggarwal Class 10 Solutions Chapter 12 Circles Test Yourself 6

Question 9.
Solution:
(i) A line intersecting a circle in two distinct points is called a secant.
(ii) A circle can have two parallel tangents at the most.
(iii) The common point of a tangent to a circle and the circle is called the point of contact.
(iv) A circle can have infinitely many tangents.

Question 10.
Solution:
Given : In a circle, from an external point P, PA and PB are the tangents drawn to the circle with centre O.
To prove : PA = PB
Construction : Join OA, OB and OP.
Proof : OA and OB are the radii of the circle and AP and BP are tangents.
RS Aggarwal Class 10 Solutions Chapter 12 Circles Test Yourself 7
OA ⊥ AP and OB ⊥ BP
⇒ ∠OAP = ∠OBP = 90°
Now, in right ∆OAP and ∆OBP,
Hyp. OP = OP (common)
Side OA = OB (radii of the same circle)
∆OAP = ∆OBP (RHS axiom)
PA = PB (c.p.c.t.)
Hence proved.

Short-Answer Questions
Question 11.
Solution:
Given : In a circle with centre O and AB is its diameter.
From A and B, PQ and RS are the tangents drawn to the circle
RS Aggarwal Class 10 Solutions Chapter 12 Circles Test Yourself 8
To prove : PQ || RS
Proof : OA is radius and PAQ is the tangent
OA ⊥ PQ
∠PAO = 90° …(i)
Similarly, OB is the radius and RBS is tangent
∠OBS = 90° …(ii)
From (i) and (ii)
∠PAO = ∠OBS
But there are alternate angles
PQ || RS

Question 12.
Solution:
Given : In the given figure,
In ∆ABC,
AB = AC.
A circle is inscribed the triangle which touches it at D, E and F
To prove : BE = CE
Proof: AD and AF are the tangents drawn from A to the circle
AD = AF
But, AB = AC
AB – AD = AC -AF
⇒ BD = CF
But BD = BE and CF = CE (tangent drawn to the circle)
But BD = CF
BE = CE
Hence proved.

Question 13.
Solution:
Given : In a circle from an external point P, PA and PB are the tangents to the circle
OP, OA and OB are joined.
RS Aggarwal Class 10 Solutions Chapter 12 Circles Test Yourself 9
To prove: ∠POA = ∠POB
Proof: OA and OB are the radii of the circle and PA and PB are the tangents to the circle
OA ⊥ AP and OB ⊥ BP
∠OAP = ∠OBP = 90°
Now, in right ∆OAP and ∆OBP,
Hyp. OP = OP (common)
Side OA = OB (radii of the same circle)
∆OAP = ∆OBP (RHS axiom)
∠POA = ∠POB (c.p.c.t.)
Hence proved.

Question 14.
Solution:
Given : A circle with centre O, PA and PB are the tangents drawn from A and B which meets at P.
AB is chord of the circle
To prove : ∠PAB = ∠PBA
Construction : Join OA, OB and OP
RS Aggarwal Class 10 Solutions Chapter 12 Circles Test Yourself 10
Proof: OA is radius and AP is tangent
OA ⊥ AP ⇒ ∠OAP = 90°
Similarly, OB ⊥ BP ⇒ ∠OBP = 90°
In ∆OAB, OA = OB (radii of the circle)
∠OAB = ∠OBA
⇒ ∠OAP – ∠OAB = ∠OBP – ∠OBA
⇒ ∠PAB = ∠PBA
Hence proved.

Question 15.
Solution:
Given : A parallelogram ABCD is circumscribed a circle.
To prove : ABCD is a rhombus.
Proof: In a parallelogram ABCD.
Opposite sides are equal and parallel.
RS Aggarwal Class 10 Solutions Chapter 12 Circles Test Yourself 11
AB = CD and AD = BC
Tangents drawn from an external point of a circle to the circle are equal.
AP = AS BP = BQ
CQ = CR and DR = DS
Adding, we get
AP + BP + CR + DR = AS + BQ + CQ + DS
⇒ (AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ)
⇒ AB + CD = AD + BC [AB = CD and AD = BC]
⇒ AB + AB = BC + BC
⇒ 2AB = 2BC
⇒ AB = BC
But AB = CD and BC = AD
AB = BC = CD = AD
Hence || gm ABCD is a rhombus.

Question 16.
Solution:
Given : O is the centre of two concentric circles with radii 5 cm and 3 cm respectively.
AB is the chord of the larger circle which touches the smaller circle at P.
RS Aggarwal Class 10 Solutions Chapter 12 Circles Test Yourself 12
OP and OA are joined.
To find : Length of AB
Proof: OP is the radius of the smaller circle and touches the smaller circle at P
OP ⊥ AB and also bisects AB at P
AP = PB = \(\frac { 1 }{ 2 }\) AB
Now, in right ∆OAP,
OA² = OP² + AP² (Pythagoras Theorem)
⇒ (5)² = (3)² + AP²
⇒ 25 = 9 + AP²
⇒ AP² = 25 – 9 = 16 = (4)²
AP = 4 cm
Hence AB = 2 x AP = 2 x 4 = 8 cm

Long-Answer Questions
Question 17.
Solution:
In the figure, quad. ABCD is circumscribed about a circle which touches its sides at P, Q, R and S respectively
RS Aggarwal Class 10 Solutions Chapter 12 Circles Test Yourself 13
To prove : AB + CD = AD + BC
Proof: Tangents drawn from an external point to a circle are equal
AP = AS
BP = BQ
CR = CQ
DR = DS
Adding, we get,
AP + BP + CR + DR = AS + BQ + CQ + DS
⇒ (AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ)
⇒ AB + CD = AD + BC
Hence AB + CD = AD + BC

Question 18.
Solution:
Given : A quad. ABCD circumscribe a circle with centre O and touches at P, Q, R and S respectively
OA, OB, OC and OD are joined forming angles AOB, BOC, COD and DOA
To prove : ∠AOB + ∠COD = 180°
and ∠BOC + ∠AOD = 180°
Construction : Join OP, OQ, OR and OS
RS Aggarwal Class 10 Solutions Chapter 12 Circles Test Yourself 14
Proof: In right ∆AOP and ∆AOS,
Side OP = OS (radii of the same circle)
Hyp. OA = OA (common)
∆AOP = ∆AOS (RHS axiom)
∠1 = ∠2 (c.p.c.t.)
Similarly, we can prove that
∠4 = ∠3
∠5 = ∠6
∠8 = ∠7
Adding, we get
∠1 + ∠4 + ∠5 + ∠8 = ∠2 + ∠3 + ∠6 + ∠7
⇒ (∠1 + ∠8) + (∠4 + ∠5) = (∠2 + ∠3) + (∠6 + ∠7)
⇒ ∠AOB + ∠COD = ∠AOD + ∠BOC
But ∠AOB + ∠BOC + ∠COD + ∠DOA = 360° (angles at a point)
∠AOB + ∠COD = ∠AOD + ∠BOC = 180°
Hence proved

Question 19.
Solution:
Given : From an external point P, PA and PB are the tangents drawn to the circle,
OA and OB are joined.
To prove : ∠APB + ∠AOB = 180°
Construction : Join OP.
RS Aggarwal Class 10 Solutions Chapter 12 Circles Test Yourself 15
Proof : Now, in ∆POA and ∆PBO,
OP = OP (common)
PA = PB (Tangents from P to the circle)
OA = OB (Radii of the same circle)
∆POA = ∆PBO (SSS axiom)
∠APO = ∠BPO (c.p.c.t.)
and ∠AOP = ∠BOP (c.p.c.t.)
OA and OB are the radii and PA and PB are the tangents
OA ⊥ AP and OB ⊥ BP
⇒ ∠OAP = 90° and ∠OBP = 90°
In ∆POA,
∠OAP = 90°
∠APO + ∠AOP = 90°
Similarly, ∠BPO + ∠BOP = 90°
Adding, we get
(∠APO + ∠BPO) + (∠AOP + ∠BOP) = 90° + 90°
⇒ ∠APB + ∠AOB = 180°.
Hence proved.

Question 20.
Solution:
Given : PQ is chord of a circle with centre O.
TP and TQ are tangents to the circle
Radius of the circle = 10 cm
i.e. OP = OQ = 10 cm and PQ = 16 cm
RS Aggarwal Class 10 Solutions Chapter 12 Circles Test Yourself 16
To find : The length of TP.
OT bisects the chord PQ at M at right angle.
PM = MQ = \(\frac { 16 }{ 2 }\) = 8 cm
In right ∆PMO,
OP² = PM² + MO² (Pythagoras Theorem)
⇒ (10)² = (8)² + MO²
⇒ 100 = 64 + MO²
⇒ MO² = 100 – 64 = 36 = (6)²
MO = 6 cm
Let TP = x and TM = y
In right ∆TPM,
TP² = TM² + PM²
⇒ x² = y² + 8²
⇒ x² = y² + 64 …(i)
and in right ∆TPM
OT² = TP² + OP²
⇒ (y + 6)² = x² + 10²
⇒ y² + 12y + 36 = x² + 100
⇒ y² + 12y + 36 = y2 + 64 + 100 {From (i)}
⇒ 12y = 64 + 100 – 36 = 128
RS Aggarwal Class 10 Solutions Chapter 12 Circles Test Yourself 17

Hope given RS Aggarwal Solutions Class 10 Chapter 12 Circles Test Yourself are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RS Aggarwal Class 10 Solutions Chapter 12 Circles MCQS

RS Aggarwal Class 10 Solutions Chapter 12 Circles MCQS

These Solutions are part of RS Aggarwal Solutions Class 10. Here we have given RS Aggarwal Solutions Class 10 Chapter 12 Circles MCQS.

Other Exercises

Choose the correct answer in each of the following questions.
Question 1.
Solution:
Number of tangents drawn from an external point to a circle is 2. (b)

Question 2.
Solution:
In the given figure, RQ is tangent to the circle with centre O.
SQ = 6 cm, QR = 4 cm
OR = √(OQ² + QR²) (In right ∆OQR)
RS Aggarwal Class 10 Solutions Chapter 12 Circles MCQS 1

Question 3.
Solution:
In the given figure, PT is tangent to the circle with centre O and radius OT = 7 cm, PT = 24 cm
OT is the radius and PT is the tangent OT ⊥ PT
Now, in right ∆OTP,
OP² = OT² + PT²
OP² = (7)² + (24)²
OP² = 49 + 576 = 625 = (25)²
OP = 25 cm (c)

Question 4.
Solution:
Two diameters cannot be parallel. (d)

Question 5.
Solution:
A chord subtends a right angle at its centre
Radius of the circle = 10 cm
RS Aggarwal Class 10 Solutions Chapter 12 Circles MCQS 2

Question 6.
Solution:
In the given figure, PT is tangent to the circle with centre O and radius
OT = 6 cm OP = 10 cm
OT is the radius and PT is the tangent
OT ⊥ TB
Now, in right ∆OPT,
OP² = OT² + PT² (Pythagoras Theorem)
⇒ (10)² = (6)² + PT²
⇒ 100 = 36 + PT²
⇒ PT² = 100 – 36 = 64 = (8)².
PT = 8 cm (a)

Question 7.
Solution:
In the given figure, point P is 26 cm away from the centre O of the circle.
RS Aggarwal Class 10 Solutions Chapter 12 Circles MCQS 3
Length of tangent PT = 24 cm
Let radius = r
In right ∆OPT,
OP² = PT² + OT²
⇒ 26² = 24² + r²
⇒ r² = 26² – 24² = 676 – 576 = 100 = (10)²
r = 10
Radius = 10 cm (a)

Question 8.
Solution:
PQ is tangent to the circle with centre O at P. ∆OPQ is an isosceles triangle.
RS Aggarwal Class 10 Solutions Chapter 12 Circles MCQS 4
∠OQP = ?
∆OPQ is an isosceles triangle
OP = PQ
∠POQ = ∠OQP
But OP is radius and PQ is tangent
OP ⊥ PQ ⇒ ∠OPQ = 90°
∠POQ + ∠OQP = 90°
⇒ ∠POQ = ∠OQP = \(\frac { 90 }{ 2 }\) = 45°
Hence, ∠OQP = 45° (b)

Question 9.
Solution:
In the given figure, AB and AC are tangents to the circle with centre O such that ∠BAC = 40°, ∠BOC = ?
AB and AC are tangents and OB and OC are radii.
OB ⊥ AB and OC ⊥ AC
⇒ ∠OBA = 90° and ∠OCA = 90°
In quadrilateral ∆BOC,
∠BAC + ∠BOC = 180°
⇒ 40° + ∠BOC = 180°
⇒ ∠BOC = 180° – 40° = 140° (d)

Question 10.
Solution:
A chord AB subtends an angle of 60° at the centre of a circle with centre O.
RS Aggarwal Class 10 Solutions Chapter 12 Circles MCQS 5
TA and TB are tangents drawn to the circle.
Then, ∠ATB = 180° – ∠AOB = 180° – 60° = 120° (d)

Question 11.
Solution:
In the given figure, O is the centre of the two concentric circles of radii 6 cm and 10 cm.
AB is a chord of the outer circle and touches the inner circle at P.
OP = 6 cm, OA = 10 cm
OP is radius and APB is tangent to the inner circle.
OP ⊥ AB and P is the mid point of AB.
In right ∆OPA,
OA² = OP² + AP²
⇒ 10² = 6² + AP²
⇒ 100 = 36 + AP²
⇒ AP²= 100 – 36 = 64 = (8)²
AP = 8 cm
and AB = 2 x AP = 2 x 8 = 16 cm (c)

Question 12.
Solution:
In the given figure, AB and AC are tangents to a circle with centre O and radius 8 cm.
OA = 17 cm.
RS Aggarwal Class 10 Solutions Chapter 12 Circles MCQS 6

Question 13.
Solution:
In the given figure, O is the centre of the circle, AT is tangent, AOC is the diameter and ∠ACB = 50°
RS Aggarwal Class 10 Solutions Chapter 12 Circles MCQS 7
We have to find the measure of ∠BAT
AB is chord and AT is the tangent
∠ACB = ∠BAT (Angles in the alternate segment)
= 50° (b)

Question 14.
Solution:
O is the centre of circle, PQ is a chord, PT is tangent.
RS Aggarwal Class 10 Solutions Chapter 12 Circles MCQS 8
∠POQ = 70°, then ∠TPQ = ?
Take a point R on the major segment and join PR and QR
arc PQ subtends ∠POQ at the centre and ∠PRQ at the remaining part of the circle
∠PRQ = \(\frac { 1 }{ 2 }\) ∠POQ = \(\frac { 1 }{ 2 }\) x 70° = 35°
But ∠TPQ = ∠PRQ (Angles in the alternate segment)
∠TPQ = 35° (a)

Question 15.
Solution:
In the given figure, AT is the tangent to the circle with centre O and OA is its radius OT = 4 cm, ∠OTA = 30°
RS Aggarwal Class 10 Solutions Chapter 12 Circles MCQS 9
Now, we have to find the length of AT
RS Aggarwal Class 10 Solutions Chapter 12 Circles MCQS 10

Question 16.
Solution:
In the given figure, PA and PB are the two tangents to the circle with centre O, which subtends ∠AOB = 110°
RS Aggarwal Class 10 Solutions Chapter 12 Circles MCQS 11
Now, we have to find the measure ∠APB
OA and OB are the radii of the circle and AP and BP are the tangents
OA ⊥ AP and OB ⊥ BP
∠A = ∠B = 90°
In quadrilateral OAPB,
∠A + ∠B = 90° + 90°= 180°
∠AOB + ∠APB = 180°
⇒ 110° + ∠APB = 180°
⇒ ∠APB = 180° – 110° = 70° (c)

Question 17.
Solution:
In the given figure, in ∆ABC,
BC = ?
AF and AE are the tangents to the circle from A.
AE = AF = 4 cm CE = AC – AE = 11 – 4 = 7 cm
Similarly, CD and CE are tangents
CD = CE = 7 cm
and BF and BD are tangents BD = BF = 3 cm
BC = BD + CD = 3 + 7 = 10 cm (b)

Question 18.
Solution:
In the given figure, ∠AOD = 135°
We know that if a circle is inscribed in a quadrilateral, the opposite sides subtends supplementary angles.
∠AOD + ∠BOC = 180°
135° + ∠BOC = 180°
⇒ ∠BOC = 180° – 135° = 45° (b)

Question 19.
Solution:
In the given figure, PQ is a chord of a circle with centre O and PT is a tangent at P to the circle such that ∠QPT = 50°.
RS Aggarwal Class 10 Solutions Chapter 12 Circles MCQS 12
Then, we have to find ∠POQ.
PT is the tangent and OP is the radius
OP ⊥ PT ⇒ ∠OPT = 90°
∠OPQ = ∠OPT – ∠QPT = 90° – 50° = 40°
In ∆OPQ,
OP = OQ (radii of the same circle)
∠OPQ = ∠OQP = 40°
and ∠POQ = 180° – (∠OPQ + ∠OQP)
= 180° – (40° + 40°)
= 180°- 80° = 100° (a)

Question 20.
Solution:
In the given figure, PA and PB are two tangents to the circle with centre O.
∠APB = 60° then ∠OAB
Join OB.
RS Aggarwal Class 10 Solutions Chapter 12 Circles MCQS 13
PAOB is a cyclic quadrilateral.
∠APB + ∠AOB = 180°
OA is radius and PA is tangent
OA ⊥ AP ⇒ ∠OAP = 90°
PA = PB (Tangents to the circle)
∠PAB = ∠PBA
But, ∠PAB + ∠PBA = 180° – 60° = 120°
∠PAB = ∠PBA = \(\frac { 120 }{ 2 }\) = 60°
∠OAB = 90° – 60° = 30° (b)

Question 21.
Solution:
Two tangents inclined at an angle of 60° are drawn to a circle of radius 3 cm.
Join OA, OB and OP.
RS Aggarwal Class 10 Solutions Chapter 12 Circles MCQS 14
RS Aggarwal Class 10 Solutions Chapter 12 Circles MCQS 15

Question 22.
Solution:
In the given figure, PQ and PR are tangents drawn from an external point P to a circle with centre A.
RS Aggarwal Class 10 Solutions Chapter 12 Circles MCQS 16
∠QPA = 27°, ∠QAR = ?
AP bisects ∠QPR and ∠QPA = 27°
∠QPR = 2 x 27° = 54°
But ∠QPR + ∠QAR = 180° (QARP is a cyclic quadrilateral)
⇒ 54° + ∠QAR = 180°
⇒ ∠QAR = 180° – 54° = 126° (c)

Question 23.
Solution:
In the given figure, PA and PB are two tangents drawn from an external point P to a circle with centre C and radius 4 cm.
PA ⊥ PB, then length of tangent is = ?
Join GA and CB.
RS Aggarwal Class 10 Solutions Chapter 12 Circles MCQS 17
CAand CB are radii and PA, PB are tangents to the circle.
CA ⊥ PA and CB ⊥ PB But, ∠APB = 90°
∠ACB = 180° – 90° = 90°
PA = PB tangents of a circle
CAPB is a square
PA = PB = radius = 4 cm (b)

Question 24.
Solution:
In the given figure, PA and PB are tangents to the circle with centre O such that ∠APB = 80°.
Join OP
RS Aggarwal Class 10 Solutions Chapter 12 Circles MCQS 18
Now, in right ∆OAP, ∠A = 90°
∠AOP = 90° – 40° = 50° (b)

Question 25.
Solution:
In the given figure, O is the centre of a circle. AB is the tangent to the circle at point P.
RS Aggarwal Class 10 Solutions Chapter 12 Circles MCQS 19
∠APQ = 58°, ∠PQB = ?
∠QPR = 90° (Angle in a semi circle)
But, ∠RPB + ∠QPR + ∠APQ = 180° (Angles on one side of a line)
⇒ ∠RPB + 90° + 58° = 180°
⇒ ∠RPB + 148° = 180°
⇒ ∠RPB = 180° – 148° = 32°
∠PQR or ∠PQB = ∠RPB (Angles in the alternate segment)
⇒ ∠PQB = 32° (a)

Question 26.
Solution:
In the given figure, O is the centre of the circle. AB is tangent to the circle at P.
RS Aggarwal Class 10 Solutions Chapter 12 Circles MCQS 20
∠PAO = 30°
∠CPB + ∠ACP = ?
∠CPD = 90° (Angle in a semi circle)
∠DPA + ∠CPB = 90°
But, ∠DPA = ∠ACP (Angles in alternate segment)
∠CPB + ∠ACP = 90° (b)

Question 27.
Solution:
In the given figure, PQ is the tangent to the circle at A.
∠PAB = 67°, ∠AQB = ?
Join BC.
∠BAC = 90° (Angle in a semi circle)
But, ∠PAB + ∠BAC + ∠CAQ = 180°
⇒ 67° + 90° + ∠CAQ = 180°
⇒ 157° + ∠CAQ = 180°
∠CAQ = 182° – 157° = 23°
∠ACB = ∠PAB (Angles in the alternate segment)
∠ACB = 67°
In ∆ACQ,
Ext. ∠ACB = ∠CAQ + ∠AQC
⇒ 67° = 23° + ∠AQC
⇒ ∠AQC = 67° – 23° = 44°
⇒ ∠AQB = 44° (d)

Question 28.
Solution:
In the given figure, two circles touch each other at C. AB is the common tangent.
∠ACB = ?
Draw a tangent from C which meets AB at P.
PA and PC are tangents to the first circle.
RS Aggarwal Class 10 Solutions Chapter 12 Circles MCQS 21
PA = PC
∠PAC = ∠PCA …(i)
Similarly, PB = PC
∠PCB = ∠PBC …(ii)
Adding, ∠PAC + ∠PBC = ∠PCA + ∠PCB
⇒ ∠PAC + ∠PBC = ∠ACB
But, ∠PAC + ∠PBC + ∠ACB = 180° (Angles of a triangle)
∠ACB = 90° (c)

Question 29.
Solution:
In the given figure O is the centre of the circle with radius 5 cm P is a point out side the circle and OP = 13 cm
PQ and PR are the tangents to the circle drawn from P
RS Aggarwal Class 10 Solutions Chapter 12 Circles MCQS 22
We have to find the area of quad. PQOR
OQ is radius and PQ is the tangent
OQ ⊥ QP
In ∆OPQ,
OP² = OQ² + PQ² (Pythagoras Theorem)
⇒ (13)² = (5)² + PQ²
⇒ 169 = 25 + PQ²
⇒ PQ² = 169 – 25 = 144 = (12)²
PQ = 12 cm
PQ = PR = 12 cm
Now, diagonal OP bisects the quad. PQOR into two triangles equal in areas.
Now, area of ∆PQO = \(\frac { 1 }{ 2 }\) x PQ x OQ
= \(\frac { 1 }{ 2 }\) x 12 x 5 = 30 cm²
Area of quad. PQOR = 2 x area ∆PQO = 2 x 30 = 60 cm² (a)

Question 30.
Solution:
In the given figure,
PQR is a tangent drawn at Q to the circle with centre O.
AB is a chord parallel to PR such that ∠BQR = 70°
Then, we have to find ∠AQB
Join QO and produce it to AB meeting it at L.
OQ ⊥ PR ⇒ LQ ⊥ PR
QL bisects AB at L
QA = QB
∆QAB is an isosceles triangle
∠LQA = ∠LQB
∠LQA = ∠LQR – ∠BQR = 90° – 70° = 20°
∠AQB = 2 x 20° = 40° (c)

Question 31.
Solution:
Length of a tangent to the circle from an external point = 10 cm
Radius (r) = 5 cm OP = ?
RS Aggarwal Class 10 Solutions Chapter 12 Circles MCQS 23
OQ is radius and QP is tangent
OQ ⊥ QP
In right ∆OPQ,
OP² = OQ² + QP² (Pythagoras Theorem) = (5)² + (10)² = 25 + 100 = 125
OP = √125 cm (d)

Question 32.
Solution:
In the figure, O is the centre of the circle BOA is its diameter and PT is tangent at P which meets BA produced at T. ∠PBO = 30°.
RS Aggarwal Class 10 Solutions Chapter 12 Circles MCQS 24
We have to find ∠PTA
In ∆BOP,
OB = OP (radii of the same circle)
∠APB = ∠PBO = 30°
But, OP is radius and PT is the tangent
OP ⊥ PT ⇒ ∠OPT = 90°
∠BPT = ∠BPO + ∠OPT = 30° + 90° = 120°
Now, in ∆PBT,
∠BPT + ∠PBA + ∠PTA = 180° (sum of angles of a triangle)
⇒ 120° + 30° + ∠PTA = 180°
⇒ 150° + ∠PTA = 180°
⇒ ∠PTA = 180° – 150° = 30° (b)

Question 33.
Solution:
In the given figure, a circle touches the side DF of a AEDF at H and touches ED and EF on producing at K and M respectively.
EK = 9 cm.
RS Aggarwal Class 10 Solutions Chapter 12 Circles MCQS 25
Perimeter of ∆EDF.
DH and DK are tangents to the circle.
DH = DK
Similarly, ∠FH = ∠FM and EK = EH = 9 cm
EK = ED + DK ⇒ ED + DH = 9 cm…(i)
Similarly, EH = EF = FH = EF + FM = 9 cm …(ii)
Adding (i) and (ii)
ED + DH + EF + FH = 9 + 9 cm (DH + HF = DF)
ED + DF + FE = 18 cm
Perimeter of ∆EDF = 18 cm (d)

Question 34.
Solution:
In the given figure, PA and PB are two tangents drawn from an external point P which inclined at an angle of 45°.
OA and OB are radii of the circle.
RS Aggarwal Class 10 Solutions Chapter 12 Circles MCQS 26
To find ∠AOB
AOBP is a cyclic quadrilateral
∠AOB + ∠APB = 180°
⇒ ∠AOB + 45° = 180°
⇒ ∠AOB = 180° – 45° = 135° (b)

Question 35.
Solution:
In the given figure, O is the centre of the circle PQL and PRM are the tangents from P drawn to the circle meeting it at Q and R respectively ∠SQL = 50°, and ∠SRM = 60°
RS Aggarwal Class 10 Solutions Chapter 12 Circles MCQS 27
Now, we have to find ∠QSR,
Join OQ, OR and OS OQ is radius and QP is tangent
OQ ⊥ QP
Similarly, OR ⊥ RP
∠1 = 90° – 50° = 40° and ∠2 = 90° – 60° = 30°
OS = OQ (radii of the same circle)
∠3 = ∠1 = 40°
Similarly OS = OR
∠2 = ∠4 = 30°
∠QSR = ∠3 + ∠4 = 40° + 30° = 70° (d)

Question 36.
Solution:
In the given figure, a ∆PQR is drawn to inscribe a circle with centre O and radius 6 cm.
OT is radius.
RS Aggarwal Class 10 Solutions Chapter 12 Circles MCQS 28
QT = 12 cm, TR = 9 cm
Area ∆PQR =189 cm²
PQ = ?
QK = QT = 12 cm
RS = RT = 9 cm
Let PK = PS = x cm
PQ = 12 + x
PR = 9 + x cm
Area of A = \(\frac { 1 }{ 2 }\) x r x Perimeter of ∆PQR
⇒ 189 = \(\frac { 1 }{ 2 }\) x 6 x (PQ + QR + RP)
⇒ 189 = 3 (12 + x + 21 + 9 + x)
⇒ 63 = 42 + 2x
⇒ 2x = 63 – 42 = 21
x = 10.5
AB = 10.5 + 12 = 22.5 cm (c)

Question 37.
Solution:
In the given figure, QR is a common tangent to two given circles touching each other externally at point T.
A tangent PT is drawn from T which intersects QR at P.
PT = 3.8 cm, QR = ?
PT and PQ are tangents to the first circle.
PQ = PT …(i)
Similarly, PT and PR tangents to the second circle.
PR = PT …(ii)
From (i) and (ii),
PQ = PR = PT = 3.8 cm
QR = 3.8 + 3.8 = 7.6 cm (d)

Question 38.
Solution:
In the figure, quadrilateral ABCD is circumscribed touches the circle at P, Q, R and S
RS Aggarwal Class 10 Solutions Chapter 12 Circles MCQS 29
AP = 5 cm, BC = 7 cm, CS = 3 cm AB = ?
Tangents drawn from the external point to the circle are equal
AQ = AP = 5 cm
CR = CS = 3 cm
BQ = BR
Now, BR = BC – CR = 7 – 3 = 4 cm
BQ = 4 cm
Now, AB = AQ + BQ = 5 + 4 = 9 cm (a)

Question 39.
Solution:
In the given figure, quad. ABCD is circumscribed touching the circle at P, Q, R and S
AP = 6 cm, BP = 5 cm, CQ = 3 cm and DR = 4 cm.
Now, we have to find the perimeter of the quad. ABCD.
We know that tangents from an external point to the circle are equal.
AP = AS = 6 cm
BP = BQ = 5 cm
CQ = CR = 3 cm
DR = DS = 4 cm
AB = AP + BP = 6 + 5 = 11 cm
BC = BQ + CQ = 5 + 3 = 8 cm
CD = CR + DR = 3 + 4 = 7 cm
and DA = AS + DS = 6 + 4 = 10 cm
Perimeter of the quad. ABCD
= AB + BC + CD + DA
= (11 + 8 + 7 + 10) cm
= 36 cm (c)

Question 40.
Solution:
In the given figure, O is the centre of the circle, AB is chord and AT is the tangent at A.
∠AOB = 100°, ∠BAT = ?
Take a point P on the major segment of the circle and join AP and BP.
RS Aggarwal Class 10 Solutions Chapter 12 Circles MCQS 30
Arc AB subtends ∠AOB at the centre and ∠APB at the remaining part of the circle.
∠APB = \(\frac { 1 }{ 2 }\) ∠AOB = \(\frac { 1 }{ 2 }\) x 100° = 50°
Now, ∠BAT = ∠APB (Angles in the alternate segment)
∠BAT = 50° (b)

Question 41.
Solution:
In a right ∆ABC, right angled at B
BC = 12 cm, AB = 5 cm
A circle is inscribed in it touching its sides at P, Q and R.
Join OP, OQ and OR.
AC² = AB² + BC² (Pythagoras Theorem)
= 5² + 12² = 25 + 144 = 169 = (13)²
RS Aggarwal Class 10 Solutions Chapter 12 Circles MCQS 31

Question 42.
Solution:
In the given figure, a circle is inscribed in a quadrilateral ABCD touching its sides, AB, BC, CD and DA at P, Q, R and S respectively
Radius OS = 10 cm
BC = 38 cm, PB = 27 cm
RS Aggarwal Class 10 Solutions Chapter 12 Circles MCQS 32
AD ⊥ DC
Length of CD = ?
Join OR and OS
BP and BQ are tangents to the circle
BQ = BP = 27 cm
BC = 38 cm
QC = 38 – 27 = 11 cm
CQ and CR are the tangents to the circle
CR = CQ = 11 cm
DR and DS are the tangents to the circle
DR = DS
AD ⊥ CD
OS is the radius and AD is the tangent
OS ⊥ AD
Similarly, OR ⊥ DC
OSDR is a square whose each side is equal to the radius = 10 cm
DR = DS = 10 cm
CD = CR + DR = 11 + 10 = 21 cm (d)

Question 43.
Solution:
In the given figure, ∆ABC is a right angled triangle, right angle at ∠B.
BC = 6 cm, AB = 8 cm
A circle with centre O is inscribed inside the triangle ABC
OP ⊥ AB and OQ ⊥ BC and OR ⊥ AC
OP = OQ = OR = x cm
OPBQ is a square
In right ∆ABC,
AC² = AB² + BC² (Pythagoras Theorem)
= (8)² + (6)² = 64 + 36 = 100 = (10)²
AC = 10 cm
Tangents drawn from the external point to the circle are equal
BP = BQ = x
CQ = CR = 6 – x
AP = AR = 8 – x
AR + CR = AC
⇒ 8 – x + 6 – x = 10
⇒ 14 – 2x = 10
⇒ 2x = 14 – 10 = 4
x = 2
Hence r = 2cm (a)

Question 44.
Solution:
A quadrilateral ABCD is circumscribed to a circle with centre O.
AB = 6 cm, BC = 7 cm, CD = 4 cm, AD = 7 cm
RS Aggarwal Class 10 Solutions Chapter 12 Circles MCQS 33
ABCD circumscribed to a circle.
AB + CD = BC + AD
⇒ 6 + 4 = 7 + AD
⇒ 10 = 7 + AD
AD = 10 – 7 = 3 cm (a)

Question 45.
Solution:
In the given figure, PA and PB are the tangents to the circle with centre O from P
RS Aggarwal Class 10 Solutions Chapter 12 Circles MCQS 34
PA = 5 cm, ∠APB = 60°
PA = PB = 5 cm
In ∆APB, ∠P = 60° and PA = PB
PAB is an equilateral triangle
AB = AP = BP = 5 cm (b)

Question 46.
Solution:
In the given figure, DE and DF are tangents to the circle from an external point D.
A is the centre of the circle.
DF = 5 cm and DE ⊥ DF, radius of the circle = 3
Join EA and FA.
RS Aggarwal Class 10 Solutions Chapter 12 Circles MCQS 35
AE and AF are the radius of the circle and DE and BF are the tangents.
AE ⊥ DE and AF ⊥ DF
∠EAF = 180° – ∠EDF = 180° – 90° = 90°
AEDF is a square.
AE = 5 cm
Radius of the circle = 5 cm (c)

Question 47.
Solution:
In the given figure, three circles with centre A, B and C are drawn touching each other externally
AB = 5 cm, BC = 7 cm and CA = 6 cm
Let r1, r2, r3 be the radii of three circles respectively
RS Aggarwal Class 10 Solutions Chapter 12 Circles MCQS 36

Question 48.
Solution:
In the given figure, AP, AQ and BC are tangents to the circle.
RS Aggarwal Class 10 Solutions Chapter 12 Circles MCQS 37
AB = 5 cm, AC = 6 cm, BC = 4 cm
Length of AP = ?
BP and BR are the tangents to the circle.
BP = BR
Similarly, CR and CQ are tangents
CR = CQ
S and AP and AQ are tangents
AP = AQ
AP = AB + BP = AB + BR
AQ = AC + CQ = AC + CR
AP + AQ = AB + BR + AC + CR = AB + BR + CR + AC
AP + AP = AB + BC + AC
2AP = 5 + 4 + 6 = 15 cm
AP = \(\frac { 15 }{ 2 }\) = 7.5 cm (d)

Question 49.
Solution:
In the given figure, O is the centre of two concentric circles of radii 5 cm and 3 cm respectively.
From external point P, PA and PB are tangents are drawn to the external circle and internal circle respectively
PA = 12 cm, PB = ?
RS Aggarwal Class 10 Solutions Chapter 12 Circles MCQS 38
OA and OB are the radii
OA ⊥ AP and OB ⊥ BP
Now, in right ∆OAP,
OP² = OA² + AP² (Pythagoras Theorem)
= (5)² + (12)²
= 25 + 144 = 169 = (13)²
OP = 13 cm
and in right ∆OBP,
OP² = OB² + BP²
(13)² = (3)² + BP²
⇒ 169 = 9 + BP²
⇒ PB² = 169 – 9 = 160
PB = √160 = √(16 x 10) = 4√10 cm (c)

True/False Type
Question 50.
Solution:
(a) It is true that no tangent can be drawn from a point inside the circle.
(b) It is true, that one and only one tangent can be drawn from a point on the circle.
(c) True. If a point P is outside the circle, two tangents can be drawn to the circle.
(d) No, only two parallel tangents can be drawn which are parallel to a given line. (d)

Question 51.
Solution:
(a) It is true as a tangent intersects (touches) the circle exactly at one point.
(b) It is true that common point to the circle where the tangent touches the circle is called point of contact.
(c) It is true that the radius through the point of contact of a tangent is perpendicular to it.
(d) False as a straight line can meet at the most two points. (d)

Question 52.
Solution:
(a) It is true, that a secant is a line which intersects the circle at two points.
(b) It is true, as a tangent intersects the circle at only one point.
(c) It is true that the point at which a tangent touches the circle is called a point of contact.
(d) It is false, as no tangent can be drawn from a point in side the circle.

Assertion and Reason Type
Each question consists of two statements, namely, Assertion (A) and Reason (R).
For selecting the correct answer, use the following code :
(a) Both Assertion (A) and Reason (R) are true and Reason (R) is a correct explanation of Assertion (A).
(b) Both Assertion (A) and Reason (R) are true but Reason (R) is not a correct explanation of Assertion (A).
(c) Assertion (A) is true and Reason (R) is false.
(d) Assertion (A) is false and Reason (R) is true.
Question 53.
Solution:
In Assertion (A):
In right ∆OPQ, OP ⊥ PQ
RS Aggarwal Class 10 Solutions Chapter 12 Circles MCQS 39
OQ² = OP² + PQ² = (12)² + (16)² = 144 + 256 = 400 = (20)²
OQ = 20 cm, which is true
In Reason (R):
It is also with respect to (A) (a)

Question 54.
Solution:
Assertion (A):
The statement is true
In Reason (R):
It is also true but not with respect to (A) (b)

Question 55.
Solution:
In Assertion (A):
In the figure, ABCD is a quad, which is circumscribed a given circle.
Sum of opposite sides are equal
AB + CD = BC + AD
It is not true that AB + BC = AD + DC
In Reason (R):
It is true but not with respect to Assertion (A) (d)

Hope given RS Aggarwal Solutions Class 10 Chapter 12 Circles MCQS are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RS Aggarwal Class 10 Solutions Chapter 12 Circles Ex 12B

RS Aggarwal Class 10 Solutions Chapter 12 Circles Ex 12B

These Solutions are part of RS Aggarwal Solutions Class 10. Here we have given RS Aggarwal Solutions Class 10 Chapter 12 Circles Ex 12B.

Other Exercises

Very-Short-Answer Questions
Question 1.
Solution:
In the given figure, a circle touches the sides AB, BC, CD and DA of a quadrilateral ABCD at P, Q, R and S respectively.
AB = 6 cm, BC = 9 cm, CD = 8 cm.
RS Aggarwal Class 10 Solutions Chapter 12 Circles Ex 12B 1
To find : The length of side AD.
A circle touches the sides of a quadrilateral ABCD.
AB + CD = BC + AD
=> 6 + 8 = 9 + AD
=> 14 = 9 + AD
=> AD = 14 – 9 = 5 cm

Question 2.
Solution:
In the given figure, PA and PB are two tangents to the circle with centre O.
∠APB = 50°
To find : Measure of ∠OAB.
Construction : Join OB.
RS Aggarwal Class 10 Solutions Chapter 12 Circles Ex 12B 2
In ∆APB,
PA = PB (tangents of the circle)
∠PAB = ∠PBA
But, ∠PAB + ∠PBA + ∠APB = 180° (Angles of a triangle)
=> ∠PAB + ∠PAB + 50° = 180°
=> 2∠PAB = 180° – 50° = 130°
∠PAB = 65°
But ∠OAP = 90° (OA is radius and PA is tangent)
∠OAB = 90° – 65° = 25°

Question 3.
Solution:
In the given figure, O is the centre of a circle.
PT and PQ are tangents to the circle from an external point P.
R is any point on the circle. RT and RQ are joined.
∠TPQ = 70°
To find : ∠TRQ
Construction : Join TO and QO.
RS Aggarwal Class 10 Solutions Chapter 12 Circles Ex 12B 3
∠TPQ = 70°
∠TOQ = 180° – 70° = 110° (OT and OQ are perpendicular on TP and QP)
Now, ∠TOQ is on the centre and ∠TRQ is on the remaining part of the circle.
∠TRQ = \(\frac { 1 }{ 2 }\) x ∠TOQ = \(\frac { 1 }{ 2 }\) x 110° = 55°

Question 4.
Solution:
In the given figure, common tangents AB and CD to the two circles with centres O1 and O2 intersect each other at E.
To prove : AB = CD.
Proof : EA and EC are tangents to the circle O1
EA = EC …(i)
Similarly, EB and ED are tangents to the circle O2.
EB = ED …(ii)
Adding (i) and (ii),
EA + EB = EC + ED
=> AB = CD
Hence, AB = CD

Question 5.
Solution:
In the given figure, PT is the tangent to the circle with centre O at P.
PQ is a chord of the circle and ∠TPQ = 70°.
To find : The measure of ∠POQ.
PT is tangent and OP is the radius.
∠OPT = 90°
But ∠QPT = 70°
∠OPQ = 90° – 70° = 20°
In ∆OPQ,
OP = OQ (radii of the same circle)
∠OQP = ∠OPQ = 20°
and ∠POQ = 180° – (∠OPQ + ∠OQP)
= 180° – (20° + 20°)
= 180° – 40° = 140°

Short-Answer Questions
Question 6.
Solution:
In the given figure, ∆ABC is circumscribed a circle with centre O and radius 2 cm.
Point D divides BC in such a way that
BD = 4 cm, DC = 3 cm, OD = 2 cm
Area of ∆ABC = 21 cm²
To find : AB and AC.
Construction : Join OA, OB, OC, OE and OF.
RS Aggarwal Class 10 Solutions Chapter 12 Circles Ex 12B 4
BD and BF are tangents to the circle.
BF = BD = 4 cm.
Similarly, CD and CE are tangents.
CE = CD = 3 cm
and AF and AE are tangents
AE = AF = x (suppose)
Now, area of ∆ABC = \(\frac { 1 }{ 2 }\) x Perimeter of ∆ABC x Radius
21 = \(\frac { 1 }{ 2 }\) (AB + BC + CA) x OD
=> 21 x 2 = [4 + 3 + 3+ x + x + 4) x 2
=> 42 = (14 + 2x) x 2
=> 14 + 2x = \(\frac { 42 }{ 2 }\) = 21
=> 2x = 21 – 14 = 7
x = \(\frac { 7 }{ 2 }\) = 3.5
AB = AF + FB = 3.5 + 4 = 7.5 cm
AC = AE + CE = 3.5 + 3 = 6.5 cm

Question 7.
Solution:
Given : Two concentric circles with centre O and radii 5 cm and 3 cm respectively.
RS Aggarwal Class 10 Solutions Chapter 12 Circles Ex 12B 5
AB is chord of larger circle which touches the smaller circle at C.
To find : The length of chord AB.
Construction : Join OA and OC.
AB is tangent and OC is radius of the smaller circle.
OC ⊥ AB and OC bisects AB at C. (AB is chord and OC ⊥ AB)
In right ∆OAC,
OA² = OC² + AC² (Pythagoras Theorem)
=> (5)² = (3)² + AC²
=> 25 = 9 + AC²
=> AC² = 25 – 9 = 16 = (4)²
=> AC = 4
and AB = 2 x AC = 2 x 4 = 8cm

Question 8.
Solution:
Given : AB is the tangent to the circle with centre O at point P.
PL ⊥ AB
RS Aggarwal Class 10 Solutions Chapter 12 Circles Ex 12B 6
To prove : PL passes through O.
Let PQ ⊥PT where Q lies on the circle.
∠QPT = 90°
Let PQ does not pass through the centre O.
Join PO and produce it to meet the circle at L.
PO being the radius of the circle drawn from the point of contact P.
OP ⊥ AB
=> ∠OPB = 90° => ∠LPB = 90°
But, PQ ⊥ AB
∠QPB = 90°
It is possible only if L and Q coincide each other.
Hence, PQ passes through the centre and is perpendicular from the point of contact.

Question 9.
Solution:
In the given figure, two tangents RQ and RP are drawn from the external point R to the circle with centre O.
∠PRQ = 120°
To prove : OR = PR + RQ
Construction : Join OP and OQ.
Also join OR.
RS Aggarwal Class 10 Solutions Chapter 12 Circles Ex 12B 7
RS Aggarwal Class 10 Solutions Chapter 12 Circles Ex 12B 8

Question 10.
Solution:
In the given figure, a circle is inscribed in a ∆ABC touches the sides AB, BC and CA at D, E and F respectively.
AB = 14 cm, BC = 8 cm and CA = 12 cm.
RS Aggarwal Class 10 Solutions Chapter 12 Circles Ex 12B 9
To find : The length of AD, BE and CF.
Let AD = x, BE = y and CF = z
AD and AF are the tangents to the circle from A.
AD = AF = x
Similarly,
BE and BD are tangents
BD = BE = y
and CF and CE are the tangents
CE = CF = z
Now, AB + BC + CA = 14 + 8 + 12 = 34
=> (x + y) + (y + z) + (z + x) = 34
=> 2 (x + y + z) = 34
=> x + y + z = 17 …(i)
But x + y = 14 cm …(ii)
y + z = 8 cm …(iii)
z + x = 12 cm …(iv)
Subtracting (iii), (iv) and (ii) from (i) term by term
x = 17 – 8 = 9 cm
y = 17 – 12 = 5 cm
z = 17 – 14 = 3 cm
Hence, AD = 9 cm, BE = 5 cm and CF = 3 cm.

Question 11.
Solution:
In the given figure, O is the centre of the circle.
PA and PB are the tangents.
To prove : AOBP is a cyclic quadrilateral.
Proof: OA is radius and PA is tangent
OA ⊥ PA
=> ∠OAP = 90° ….. (i)
Similarly, OB is radius and PB is tangent.
OB ⊥ PB
=> ∠OBP = 90° ….. (ii)
Adding (i) and (ii),
∠OAP + ∠OBP = 90° + 90° = 180°
But these are opposite angles of the quadrilateral AOBP.
Quadrilateral AOBP is a cyclic.

Question 12.
Solution:
In two concentric circles with centre O, a chord AB of the laiger circle touches the smaller circle at C.
AB = 8 cm and radius of larger circle = 5 cm
Join OA, OC
To find, the radius of smaller circle,
AB is the tangent and OC is the radius
OC ⊥ AB
RS Aggarwal Class 10 Solutions Chapter 12 Circles Ex 12B 10
AC = CB = \(\frac { 8 }{ 2 }\) = 4 cm
OA = 5 cm
In right ∆OCA,
OA² = OC² + AC² (Pythagoras Theorem)
(5)² = OC² + (4)²
OC² = (5)² – (4)² = 25 – 16 = 9 = (3)²
OC = 3
Radius of smaller circle = 3 cm

Question 13.
Solution:
In the given figure, PQ is a chord of a circle with centre O.
PT is the tangent ∠QPT = 60°.
To find : ∠PRQ.
Construction : Take a point M on the alternate segment.
Join MP and MQ.
RS Aggarwal Class 10 Solutions Chapter 12 Circles Ex 12B 11
∠MPQ = ∠QPT = 60° (Angles in the alternate segment)
∠PMQ + ∠PRQ = 180° (Opposite angles of a cyclic quadrilateral)
=> 60° + ∠PRQ = 180°
∠PRQ = 180° – 60° = 120°
Hence, ∠PRQ = 120°

Question 14.
Solution:
In the given figure,
PA and PB are the two tangents to the circle.
With centre O, OA and AB are joined
∠APB = 60°
To find : The measure of ∠OAB
PA and PB are tangents to the circle from P
PA = PB
∠PAB = ∠PBA
But ∠APB = 60°
∠PAB + ∠PBA = 180° – 60° = 120°
2 ∠PAB = 120°
∠PBA = 60°
OA is radius and PA is tangent.
OA ⊥ PA
∠OAP = 90°
=> ∠OAB + ∠PAB = 90°
=> ∠OAB + 60° = 90°
=> ∠OAB = 90° – 60° = 30°
Hence, ∠OAB = 30°

Question 15.
Solution:
Since, tangents drawn from an external point are equally inclined to the line joining centre to that point.
RS Aggarwal Class 10 Solutions Chapter 12 Circles Ex 12B 12

Hope given RS Aggarwal Solutions Class 10 Chapter 12 Circles Ex 12B are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RS Aggarwal Class 10 Solutions Chapter 12 Circles Ex 12A

RS Aggarwal Class 10 Solutions Chapter 12 Circles Ex 12A

These Solutions are part of RS Aggarwal Solutions Class 10. Here we have given RS Aggarwal Solutions Class 10 Chapter 12 Circles Ex 12A.

Other Exercises

Question 1.
Solution:
PT is the tangent to the circle with centre O and radius OT = 20 cm.
P is a point 29 cm away from O.
RS Aggarwal Class 10 Solutions Chapter 12 Circles Ex 12A 1
OP = 29 cm, OT = 20 cm
OT is radius and PT is the tangent
OT ⊥ PT
Now, in right ∆OPT,
OP² = OT² + PT² (Pythagoras Theorem)
⇒ (29)² = (20)² + PT²
⇒ 841 = 400 + PT²
⇒ PT² = 841 – 400
⇒ PT² = 441 = (21)²
⇒ PT = 21
Length of tangent PT = 21 cm

Question 2.
Solution:
P is a point outside the circle with centre O and OP = 25 cm
PT is the tangent to the circle and OT is the radius
RS Aggarwal Class 10 Solutions Chapter 12 Circles Ex 12A 2
OT ⊥ PT
PT = 24 cm
Now, in right ∆OPT,
OP² = OT² + PT² (Pythagoras Theorem)
⇒ 25² = OT² + (24)²
⇒ 625 = OT² + 576
⇒ OT² = 625 – 576 = 49 = (7)²
OT = 7 cm
or radius of the circle is 7 cm.

Question 3.
Solution:
Given : Two concentric circles with centre O and radii 6.5 cm and 2.5 cm respectively.
AB is a chord of the larger circle which touches the smaller circle at C.
RS Aggarwal Class 10 Solutions Chapter 12 Circles Ex 12A 3
To find : The length of AB.
Join OC, OA.
AB is tangent to the smaller circle and OC is the radius.
OC ⊥ AB and OC bisects AB at C.
AC = CB
OA = 6.5 cm, OC = 2.5 cm
Now, in right ∆OAC,
OA² = OC² + AC² (Pythagoras Theorem)
6.5² = 2.5² + AC²
⇒ 42.25 = 6.25 + AC²
⇒ AC² = 42.25 – 6.25 = 36 = (6)²
⇒ AC = 6
Length of chord AB = 2 x AC = 2 x 6 = 12 cm

Question 4.
Solution:
Given : In the given figure, a circle is inscribed in a triangle ABC which touches the sides AB, BC, CA at D, E and F respectively.
AB = 12 cm, BC = 8 cm and AC = 10 cm.
RS Aggarwal Class 10 Solutions Chapter 12 Circles Ex 12A 4
To find : Lengths of AD, BE and CF.
AD and AF are tangents to the circle from A.
AD = AF = x.
Similarly, BD and BE are tangents to the circle.
BD = BE = y
and CE and CF are tangents to the circle
CE = CF = z
x + y + 12 …(i)
y + z = 8 …(ii)
z + x = 10 …(iii)
Adding, 2(x + y + z) = 12 + 8 + 10 = 30
x + y + z = 15 …(iv)
Now, subtracting (ii), (iii) and (i) respectively from (iv)
x = 15 – 8 = 7
y = 15 – 10 = 5
z = 15 – 12 = 3
AD = 7 cm, BE = 5 cm and CF = 3 cm

Question 5.
Solution:
Given : In the given figure,
PA and PB are the tangents drawn from P to the circle with centre O.
OA and OB are joined.
To prove : A, O, B and P are concyclic.
Proof : PA is tangent and OA is the radius.
OA ⊥ PA
∠OAP = 90° …(i)
Similarly, OB is the radius and PB is the tangent
OB ⊥ PB
∠OBP = 90° …(ii)
Adding (i) and (ii)
∠OAP + ∠OBP = 90° + 90° = 180°
But these are the opposite angles of the quadrilateral AOBP
Quadrilateral AOBP is a cyclic
Hence A, O, B and P are concylic

Question 6.
Solution:
Given : In the given figure, chord AB of larger circle of the two concentric circles with centre O, touches the smaller circle at C.
To prove : AC = CB.
Construction : Join OC, OA and OB.
RS Aggarwal Class 10 Solutions Chapter 12 Circles Ex 12A 5
Proof: AB is tangent to the smaller circles and OC is the radius.
OC ⊥ AB.
In right ∆OAC and ∆OBC,
Hypotenuse OA = OB (radii of the same circle)
Side OC = OC (common)
∆OAC = ∆OBC (RHS axiom)
AC = CB (c.p.c.t.)

Question 7.
Solution:
Given : In the figure, from an external point P of the circle, PA and PB are tangents to the circle with centre O.
CD is a tangent at E.
PA = 14 cm.
To find : Perimeter of ∆PCD.
Proof: PA and PB are tangents from P to the circle.
PA = PB …(i)
CA and CE are tangents to the circle.
CA = CE …(ii)
Similarly,
DB = DE
Now, perimeter of ∆PCD
= PC + CD + PD
= PC + CE + ED + PD
= PC + CA + BD + PD
= PA + PB [From (i) and (ii)]
= 14 + 14 = 28 cm

Question 8.
Solution:
A circle with centre O, is inscribed in a ∆ABC touching it at P, Q, R respectively
AB = 10 cm, AR = 7 cm and CR = 5 cm
RS Aggarwal Class 10 Solutions Chapter 12 Circles Ex 12A 6
To find, the length of BC
AP and AR are the tangents to the circle
AP = AR = 7 cm
BP = AB – AP = 10 – 7 = 3 cm
BP and BQ are the tangents to the circle
BQ = BP = 3 cm
Similarly, CQ = CR = 5 cm
BC = BQ + CQ = 3 + 5 = 8 cm

Question 9.
Solution:
In the figure, a circle with centre O, touches the sides of a quadrilateral ABCD at P, Q, R and S respectively.
RS Aggarwal Class 10 Solutions Chapter 12 Circles Ex 12A 7
AB = 6 cm, BC = 7 cm and CD = 4 cm
To find, the length of AD.
We know that tangents from an external point to a circle are equal.
AP = AS, BP = BQ
CR = CQ and DR = DS
Now, AP + BP + CR + DR = AS + BQ + CQ + DC
⇒ (AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ)
⇒ AB + CD = AD + BC
⇒ AD = (AB + CD) – BC = (6 + 4 – 7) = 3 cm (AB = 6 cm, CD = 4 cm and BC = 7 cm)
Hence, AD = 3 cm.

Question 10.
Solution:
In the given figure, an isosceles ∆ABC in which AB = AC, is circumscribed a circle.
The circle touches its sides BC, CA and AB at P, Q and R respectively.
To prove : P bisects the base BC.
i.e. BP = PC
Proof : AR and AQ are tangents to the circle.
AR = AQ But AB = AC
AB – AR = AC – AQ
⇒ BR = CQ …(i)
BR and BP are tangents to the circle.
BR = BP …(ii)
Similarly,
CP and CQ are tangents
CP = CQ …(iii)
BR = CQ (proved)
From (ii) and (iii),
BP = CP or BP = PC
Hence, P is midpoint of BC.

Question 11.
Solution:
In the given figure, O is the centre of two concentric circles with radii 4 cm and 6 cm respectively.
PA and PB are the tangents drawn from P, to the outer circle and inner circle respectively.
PA = 10 cm
RS Aggarwal Class 10 Solutions Chapter 12 Circles Ex 12A 8
To find, the length of PB (upto one place of decimal)
OA and OB are the radii and PA and PB are two tangents to the circles respectively
OA ⊥ PA and OB ⊥ PB
In right ∆OAP,
OP² = OA² + PA² (Pythagoras Theorem)
= (6)² + (10)² = 36 + 100 = 136
Similarly, in right ∆OBP,
OP² = OB² + PB²
136 = (4)² + PB²
⇒ 136 = 16 + PB²
⇒ PB² = 136 – 16 = 120
PB = √120 cm = 2√30 cm = 2 x 5.47 = 10.94 = 10.9 cm

Question 12.
Solution:
In the given figure, ∆ABC circumscribed the circle with centre O.
Radius OD = 3 cm
BD = 6 cm, DC = 9 cm
Area of ∆ABC = 54 cm²
To find : Lengths of AB and AC.
Construction : Join OA, OB, OC, OE and OF.
RS Aggarwal Class 10 Solutions Chapter 12 Circles Ex 12A 9
Proof : OE = OF = OD = 3 cm, radii of the same circle.
BD and BF are tangents to the circle.
BD = BF = 6 cm
Similarly, CD = CE = 9 cm and AE = AF = x (Suppose)
RS Aggarwal Class 10 Solutions Chapter 12 Circles Ex 12A 10
RS Aggarwal Class 10 Solutions Chapter 12 Circles Ex 12A 11
RS Aggarwal Class 10 Solutions Chapter 12 Circles Ex 12A 12

Question 13.
Solution:
In the given figure, PQ is a chord of the circle with centre O.
TP and TQ are tangents, OP and OT are joined.
Radius of the circle is 3 cm and PQ = 4.8 cm.
RS Aggarwal Class 10 Solutions Chapter 12 Circles Ex 12A 13
RS Aggarwal Class 10 Solutions Chapter 12 Circles Ex 12A 14

Question 14.
Solution:
PQ and RS are two parallel tangents which
touches the circle at A and B. O is the centre of the circle.
OA and OB are joined.
To prove : AB passes through the centre O of the circle.
Construction : Draw OC || PQ or RS.
RS Aggarwal Class 10 Solutions Chapter 12 Circles Ex 12A 15
Proof : OA is radius and PQ is tangent.
OA ⊥ PQ ⇒ ∠OAP = 90°
Similarly, OB is radius and RS is the tangent
OB ⊥ RS ⇒ ∠OBR = 90°
PQ || OC
∠AOC + ∠OAP = 180° (Co-interior angles)
∠AOC + 90° = 180°
∠AOC = 180° – 90° = 90°
Similarly, ∠BOC = 90°
∠AOC + ∠BOC = 90° + 90° = 180°
AOB is a straight line.
Hence, AB passes through the centre of the circle.

Question 15.
Solution:
In the given figure, a circle with centre O, is inscribed in a quadrilateral ABCD.
The circle touches the siaes of quadrilateral at P, Q, R and S respectively.
AB = 29 cm, AD = 23 cm, ∠B = 90°
DS = 5 cm
To find : The radius of the circle.
Construction : Join OP and OQ.
Proof: Let OP = OQ = r
∠B = 90°
PBQO is a square.
DR and DS are the tangents to the circle.
DR = DS = 5 cm
AR = AD – DR = 23 – 5 = 18 cm
AR and AQ are tangents to the circle.
AQ = AR = 18 cm
QB = AB – AQ = 29 – 18 = 11 cm
PBQO is a square.
OP = OQ = BQ = 11 cm
Hence, radius of the circle (r) = 11 cm

Question 16.
Solution:
In the given figure, TP is the tangent from an external point T and ∠PBT = 30°.
To prove : BA : AT = 2 : 1
Proof: ∠APB = 90° (Angle in a semicircle)
∠PBT = 30° (given)
∠PAB = 90° – 30° = 60°
But, ∠PAT + ∠PAB = 180° (Linear pair)
⇒ ∠PAT + 60° = 180°
⇒ ∠PAT = 180° – 60° = 120°
and ∠APT = ∠PBA = 30° (Angles in the alternate segment)
In ∆PAT,
∠PTA = 180° – (120° + 30°) = 180° – 150° = 30°
PA = AT
In right ∆APB,
RS Aggarwal Class 10 Solutions Chapter 12 Circles Ex 12A 16

Hope given RS Aggarwal Solutions Class 10 Chapter 12 Circles Ex 12A are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RS Aggarwal Class 10 Solutions Chapter 11 Arithmetic Progressions MCQS

RS Aggarwal Class 10 Solutions Chapter 11 Arithmetic Progressions MCQS

These Solutions are part of RS Aggarwal Solutions Class 10. Here we have given RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions MCQS

Other Exercises

Choose the correct answer in each of the following questions.
Question 1.
Solution:
RS Aggarwal Class 10 Solutions Chapter 11 Arithmetic Progressions MCQS 1

Question 2.
Solution:
RS Aggarwal Class 10 Solutions Chapter 11 Arithmetic Progressions MCQS 1

Question 3.
Solution:
RS Aggarwal Class 10 Solutions Chapter 11 Arithmetic Progressions MCQS 3

Question 4.
Solution:
RS Aggarwal Class 10 Solutions Chapter 11 Arithmetic Progressions MCQS 4

Question 5.
Solution:
RS Aggarwal Class 10 Solutions Chapter 11 Arithmetic Progressions MCQS 5

Question 6.
Solution:
RS Aggarwal Class 10 Solutions Chapter 11 Arithmetic Progressions MCQS 6

Question 7.
Solution:
RS Aggarwal Class 10 Solutions Chapter 11 Arithmetic Progressions MCQS 7

Question 8.
Solution:
RS Aggarwal Class 10 Solutions Chapter 11 Arithmetic Progressions MCQS 8
RS Aggarwal Class 10 Solutions Chapter 11 Arithmetic Progressions MCQS 9

Question 9.
Solution:
RS Aggarwal Class 10 Solutions Chapter 11 Arithmetic Progressions MCQS 10

Question 10.
Solution:
RS Aggarwal Class 10 Solutions Chapter 11 Arithmetic Progressions MCQS 11

Question 11.
Solution:
RS Aggarwal Class 10 Solutions Chapter 11 Arithmetic Progressions MCQS 12

Question 12.
Solution:
RS Aggarwal Class 10 Solutions Chapter 11 Arithmetic Progressions MCQS 13
RS Aggarwal Class 10 Solutions Chapter 11 Arithmetic Progressions MCQS 14

Question 13.
Solution:
RS Aggarwal Class 10 Solutions Chapter 11 Arithmetic Progressions MCQS 15

Question 14.
Solution:
Sum of first n odd natural numbers = (n)²
Sum of first 20 odd natural numbers = (20)² = 400 (c)

Question 15.
Solution:
First 40 positive integers divisible by 6 are 6, 12, 18, 24, … to 40 terms
Here, a = 6, d = 6, n = 40
RS Aggarwal Class 10 Solutions Chapter 11 Arithmetic Progressions MCQS 16
RS Aggarwal Class 10 Solutions Chapter 11 Arithmetic Progressions MCQS 17

Question 16.
Solution:
RS Aggarwal Class 10 Solutions Chapter 11 Arithmetic Progressions MCQS 18

Question 17.
Solution:
RS Aggarwal Class 10 Solutions Chapter 11 Arithmetic Progressions MCQS 19

Question 18.
Solution:
In an AP,
a18 – a14 = 32
Let a be the first term and d be the common difference, then
RS Aggarwal Class 10 Solutions Chapter 11 Arithmetic Progressions MCQS 20

Question 19.
Solution:
RS Aggarwal Class 10 Solutions Chapter 11 Arithmetic Progressions MCQS 21

Question 20.
Solution:
RS Aggarwal Class 10 Solutions Chapter 11 Arithmetic Progressions MCQS 22

Question 21.
Solution:
RS Aggarwal Class 10 Solutions Chapter 11 Arithmetic Progressions MCQS 23
RS Aggarwal Class 10 Solutions Chapter 11 Arithmetic Progressions MCQS 24

Question 22.
Solution:
RS Aggarwal Class 10 Solutions Chapter 11 Arithmetic Progressions MCQS 25

Question 23.
Solution:
RS Aggarwal Class 10 Solutions Chapter 11 Arithmetic Progressions MCQS 26

Question 24.
Solution:
RS Aggarwal Class 10 Solutions Chapter 11 Arithmetic Progressions MCQS 27
RS Aggarwal Class 10 Solutions Chapter 11 Arithmetic Progressions MCQS 28

Question 25.
Solution:
Sum of first 16 terms of AP 10, 6, 2, …
Here, a = 10, d = 6 – 10 = -4, n = 16
RS Aggarwal Class 10 Solutions Chapter 11 Arithmetic Progressions MCQS 29

Question 26.
Solution:
RS Aggarwal Class 10 Solutions Chapter 11 Arithmetic Progressions MCQS 30
RS Aggarwal Class 10 Solutions Chapter 11 Arithmetic Progressions MCQS 31

Question 27.
Solution:
RS Aggarwal Class 10 Solutions Chapter 11 Arithmetic Progressions MCQS 32

Question 28.
Solution:
In an AP, T17 = T10 + 21
Let a be the first term and d be the common difference, then
RS Aggarwal Class 10 Solutions Chapter 11 Arithmetic Progressions MCQS 33

Question 29.
Solution:
RS Aggarwal Class 10 Solutions Chapter 11 Arithmetic Progressions MCQS 34

Question 30.
Solution:
RS Aggarwal Class 10 Solutions Chapter 11 Arithmetic Progressions MCQS 35

Hope given RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions MCQS are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.