CBSE Class 9

Here we are providing Areas of Parallelograms and Triangles Class 9 Extra Questions Maths Chapter 9 with Answers Solutions, Extra Questions for Class 9 Maths was designed by subject expert teachers.

Extra Questions for Class 9 Maths Areas of Parallelograms and Triangles with Answers Solutions

Extra Questions for Class 9 Maths Chapter 9 Areas of Parallelograms and Triangles with Solutions Answers

Areas of Parallelograms and Triangles Class 9 Extra Questions Very Short Answer Type

Question 1.
Two parallelograms are on equal bases and between the same parallels. Find the ratio of their areas.
Solution:
1:1 [∵ Two parallelograms on the equal bases and between the same parallels are equal in
area.]

Question 2.
In ∆XYZ, XA is a median on side YZ. Find ratio of ar(∆XYA) : ar(∆XZA).

Solution:
Here, XA is the median on side YZ.
∴ YA = AZ
Draw XL ⊥ YZ

Question 3.
ABCD is a trapezium with parallel sides AB = a cm and DC = b cm (fig.). E and F are the mid-points of the non parallel sides. Find the ratio of ar(ABFE) and ar(EFCD).

Solution:

Question 4.
ABCD is a parallelogram and Q is any point on side AD. If ar(∆QBC) = 10 cm², find ar(∆QAB) + ar(∆QDC).

Solution:
Here, ∆QBC and parallelogram ABCD are on the same base BC and lie between the same parallels BC || AD.
∴ ar(||^gm ABCD) = 2 ar(∆QBC) ar(∆QAB) + ar(∆QDC) + ar(∆QBC) = 2 ar(∆QBC)
ar(∆QAB) + ar(∆QDC) = ar(∆QBC)
Hence, ar(∆QAB) + ar(∆QDC) = 10 cm² [∵ ar(∆QBC) = 10 cm² (given)]

Question 5.
WXYZ is a parallelogram with XP ⊥ WZ and ZQ ⊥ WX. If WX = 8 cm, XP = 8 cm and ZQ = 2 cm, find YX.

Solution:
ar(||^gm WXYZ) = ar(||^gm WXYZ)
WX × ZQ = WZ × XP
8 × 2 = WZ × 8
⇒ WZ = 2 cm
Now, YX = WZ = 2 cm [∵ opposite sides of parallelogram are equal]

Question 6.
In figure, TR ⊥ PS, PQ || TR and PS || QR. If QR = 8 cm, PQ = 3 cm and SP = 12 cm, find ar(quad. PQRS).

Solution:
Here,
PS || QR [given]
∴ PQRS is a trapezium
Now, TR ⊥ PS and PQ || TR [given]
⇒ PQ ⊥ PS
∴ PQ = TR = 3 cm [given]
Now, ar(quad. PQRS) = \(\frac{1}{2}\) (PS + QR) × PQ = \(\frac{1}{2}\)(12 + 8) × 3 = 30 cm²

Question 7.
In the given figure, ABCD is a parallelogram and L is the mid-point of DC. If ar(quad. ABCL) is 72 cm, then find ar(∆ADC).

Solution:
In ||^gm ABCD, AC is the diagonal

Question 8.
In figure, TR ⊥ PS, PQ || TR and PS || QR. If QR = 8 cm, PQ = 3 cm and SP = 12 cm, find ar (PQRS).

Solution:
Here, PS || QR
∴ PQRS is a trapezium in which PQ = 3 cm, QR = 8 cm and SP = 12 cm
Now, TR I PS and PQ || TR
∴ PQRT is a rectangle [∵ PQ || TR, PT || QR and ∠PTR = 90°]
⇒ PQ = TR = 3 cm
Now, ar(PQRS) = \(\frac{1}{2}\)(PS + QR) × TR = \(\frac{1}{2}\)(12 + 8) × 3 = 30 cm².

Areas of Parallelograms and Triangles Class 9 Extra Questions Short Answer Type 1

Question 1.
ABCD is a parallelogram and O is the point of intersection of its diagonals. If ar(A AOD) = 4 cm\(²\) find area of parallelogram ABCD.

Solution:
Here, ABCD is a parallelogram in which its diagonals AC and BD intersect each other in O.
∴ O is the mid-point of AC as well as BD.
Now, in ∆ADB, AO is its median
∴ ar(∆ADB) = 2 ar(∆AOD)
[∵ median divides a triangle into two triangles of equal areas]
So, ar(∆ADB) = 2 × 4 = 8 cm²
Now, ∆ADB and ||^gm ABCD lie on the same base AB and lie between same parallels AB and CD
∴ ar(ABCD) = 2 ar(∆ADB).
= 2 × 8
= 16cm²

Question 2.
In the given figure of ∆XYZ, XA is a median and AB || YX. Show that YB is also a median.

Solution:
Here, in ∆XYZ, AB || YX and XA is a median.
∴ A is the mid-point of YZ. Now, AB is a line segment from mid-point of one side (YZ) and parallel to another side (AB || YX), therefore, it bisects the third side XZ.
⇒ B is the mid-point of XZ.
Hence, YB is also a median of ∆XYZ.

Question 3.
ABCD is a trapezium. Diagonals AC and BD intersect each other at O. Find the ratio ar (∆AOD) : ar (∆BOC).
Solution:

Here, ABCD is a trapezium in which diagonals AC and BD intersect each other at O. ∆ADC and ABCD are on the same base DC and between the same ‘parallels i.e., AB || DC.
∴ ar(∆ADC) = ar(∆BCD)
⇒ ar(∆AOD) + ar(∆ODC)
= ar(ABOC) + ar(AODC)
⇒ ar(∆AOD) = ar(∆BOC)
⇒ \(\frac { ar(∆AOD) }{ ar(∆BOC) } \) = 1

Question 4.
ABCD is a parallelogram and BC is produced to a point Q such that AD = CQ (fig.). If AQ intersects DC at P, show that ar(∆BPC) = ar(∆DPQ).

Solution:
In ||^gm ABCD,
ar(∆APC) = ar(∆BCP) …(i)
[∵ triangles on the same base and between the same parallels have equal area]
Similarly, ar(∆ADQ) = ar(∆ADC) i …(ii)
Now, ar(∆ADQ) – ar(∆ADP) = ar(∆ADC) – ar(∆ADP)
ar(∆DPQ) = ar(∆ACP) … (iii)
From (i) and (iii), we have
ar(∆BCP) = ar(∆DPQ)
or ar(∆BPC) = ar(∆DPQ)

Question 5.
In the figure, PQRS is a parallelogram with PQ = 8 cm and ar(∆PXQ) = 32 cm². Find the altitude of gm PQRS and hence its area.

Solution:
Since parallelogram PQRS and APXQ are on the same base PQ and lie between the same
parallels PQ || SR
∴ Altitude of the ∆PXQ and ||^gm PQRS is same.
Now, \(\frac{1}{2}\)PQ × altitude = ar(∆PXQ)
⇒ \(\frac{1}{2}\) × 8 × altitude = 32
altitude = 8 cm
ar(||^gm PQRS) = 2 ar(∆PXQ)
= 2 × 32 = 64 cm²
Hence, the altitude of parallelogram PQRS is 8 cm and its area is 64 cm².

Question 6.
In ∆ABC. D and E are points on side BC such that CD = DE = EB. If ar(∆ABC) = 27 cm, find ar(∆ADE)

Solution:
Since in ∆AEC, CD = DE, AD is a median.
∴ ar(∆ACD) = ar(∆ADE)
[∵ median divides a triangle into two triangles of equal areas]
Now, in ∆ABD, DE = EB, AE is a median
ar(∆ADE) = ar(∆AEB)… (ii)
From (i), (ii), we obtain
ar(∆ACD) = ar(∆ADE) = ar(∆AEB)\(\frac{1}{3}\)ar(∆ABC)
∴ ar(∆ADE) = \(\frac{1}{3}\) × 27 = 9 cm²

Areas of Parallelograms and Triangles Class 9 Extra Questions Short Answer Type 2

Question 1.
For the given figure, check whether the following statement is true or false. Also justify your answer. PQRS is a trapezium with PQ || SR, PS || RU and ST || RQ, then ar(PURS) = ar(TQRS)

Solution:
Since ST || RQ and SR || TQ [given]
⇒ STQR is a ||^gm
Similarly, PS || UR and SR || PU [given]
⇒ PSRU is a ||^gm
Also, similarly, ||^gm STQR and ||^gm PSRU lie on same base SR and between same parallels PQ and SR.
∴ ar(||^gm STQR) = ar(||^gm PSRU)
Hence, the given statement is true.

Question 2.
In the given figure, WXYZ is a quadrilateral with a point P on side WX. If ZY || WX, show that :
(i) ar(∆ZPY) = ar(∆ZXY)
(ii) ar(∆WZY) = ar(∆ZPY)
(iii) ar(∆ZWX) = ar(∆XWY)

Solution:
∆ZPY and ∆ZXY lie on same base ZY and between same parallels ZY and WX
∴ ar(∆ZPY) = ar(∆ZXY)
Again, (∆WZY) and (∆ZPY) lie on same base ZY and between same parallels ZY and WX
∴ ar(∆WZY) = ar(∆ZPY)
Also, ∆zwX and ∆XWY lie on same base XW and between same parallels ZY and WX
∴ ar(∆ZWX) = ar(∆XWY)

Question 3.
In ∆ABC ; D, E and F are mid-points of sides BC, AC and AB respectively. A line through C drawn parallel to DE meets FE produced to G. Show that ar(∆FDE) = ar(∆EGC).

Solution:
Here, in ∆ABC; D, E and F are the mid-points of sides BC, AC and AB respectively. A line through C is drawn parallel to DE meets FE produced at G.
Since a line segment drawn through the mid-points of two sides, is parallel to third side and is half of it.
∴ DE || AB, EF || BC and FD || AC
⇒ AEDF, EFDC and EFDB are parallelograms
Also, a diagonal of a parallelogram divides it into two congruent triangles
∴ ∆AFE ≅ ∆DEF
∆DEF ≅ ∆FDB and
∆DEF ≅ ∆EDC
∴ ar(∆FDE) ≅ ar(∆EDC) …(i)
Again, in quad. EGCD, we have
CG || DE and DC || EG [given]
∴ EGCD is a parallelogram .
∴ ∆EDC = ∆CGE
⇒ ar(AEDC) = ar(ACGE) … (ii)
From (i) and (ii), we obtain
ar(∆FDE) = ar(∆EGC)

Question 4.
In ∆PQR, A and B are points on side QR such that they trisect QR. Prove that ar(∆PQB) = 2ar(∆PBR).
Solution:
Here, in ∆PQR, A and B are points on side QR
such that QA = AB = BR.

Through P, draw a line / parallel to QR
Now, APQA, APAB and APBR on the equal bases
and between the same parallels l || QR
⇒ ar(∆PQA) = ar(∆PAB) = ar(∆PBR) …. (i)
Now, ar(∆PQB) = ar(∆PQA) + ar(∆PAB)
= 2ar(∆PQA)[using (i)]
= 2ar(∆PBR) [using (i)]

Question 5.
The side AB of a parallelogram ABCD is produced to any point P. A line through A and parallel to CP meets CB produced at Q, then parallelogram PBQR is completed (see figure). Show that ar(||^gm ABCD) = ar(||^gm PBQR).

Solution:

Join AC and QP, also it is given that ∆Q || CP
∴ ∆ACQ and ∆APQ are on the same base ∆Q and lie between the same parallels ∆Q || CP.
∴ ar(∆ACQ) = ar(∆APQ)
or ar(∆ABC) + ar(∆ABQ) = ar(ABPQ) + ar(∆ABQ)
ar(∆ABC) = ar(ABPQ)
or \(\frac{1}{2}\)ar(||^gm ABCD) = \(\frac{1}{2}\)ar (||^gm PBQR)
or ar(||^gm ABCD) = ar(||^gm PBQR)

Areas of Parallelograms and Triangles Class 9 Extra Questions Long Answer Type

Question 1.
EFGH is a parallelogram and U and T are points on sides EH and GF respectively. If ar(∆EHT) = 16 cm, find ar(∆GUF).

Solution:
∴ ar(∆EHT) = \(\frac{1}{2}\) ar(||^gm EFGH) …..(i)
Similarly, ∆GUF and parallelogram EFGH are on the same base GF and lie between the same parallels GF and HE
∴ ar(∆GUF) = \(\frac{1}{2}\) ar(||^gm EFGH) …..(ii)
From (i) and (ii), we have
ar(∆GUF) = ar(∆EHT)
= 16 cm² [∵ ar(∆EHT) = 16 cmcm²] [given]

Question 2.
ABCD is a parallelogram and P is any point in its interior. Show that :
ar(∆APB) + ar(∆CPD) = ar(∆BPC) + ar(∆APD)

Solution:
Through P, draw a line LM || DA and EF || AB
Since ∆APB and ||^gm ABFE are on the same base AB and lie between the same parallels AB and EF.
∴ ar(∆APB) = \(\frac{1}{2}\) ar(||^gm ABFE) … (i)

Similarly, ACPD and parallelogram DCFE are on the same base DC and between the same parallels DC and EF.
∴ ar(∆CPD) = \(\frac{1}{2}\) ar(||^gm DCFE) … (ii)
Adding (i) and (ii), we have
ar(∆APB) + ar(∆CPD) = \(\frac{1}{2}\) ar (||^gm ABFE) + ar(||^gm DCFE)
= \(\frac{1}{2}\) ar(|lgm ABCD) … (iii)
Since ∆APD and parallelogram ADLM are on the same base AB and between the same parallels AD and ML
∴ ar(∆APD) = \(\frac{1}{2}\) ar(||^gm ADLM) …..(iv)
Similarly, ar(∆BPC) = \(\frac{1}{2}\) = arc||^gm BCLM) ….(v)
Adding (iv) and (u), we have
ar(∆APD) + ar (∆BPC) = \(\frac{1}{2}\) ar(||^gm ABCD) ….(vi)
From (iii) and (vi), we obtain
ar(∆APB) + ar(∆CPD) = ar(∆APD) + ar(ABPC)

Question 3.
In the given figure, ABCD is a square. Side AB is produced to points P and Q in such a way that PA = AB = BQ. Prove that DQ = CP.

Solution:
In ∆PAD, ∠A = 90° and DA = PA = AB
⇒ ∠ADP = ∠APD = \(\frac{90^{\circ}}{2}\) = 45°
Similarly, in ∆QBC, ∠B = 90° and BQ = BC = AB
⇒ ∠BCQ = ∠BQC = \(\frac{90^{\circ}}{2}\) = 45°
In ∆PAD and ∆QBC, we have
PA = BQ [given]
∠A = ∠B [each = 90°]
AD = BC [sides of a square]
⇒ ∠PAD ≅ ∆QBC [by SAS congruence rule]
⇒ PD = QC [c.p.c.t.]
Now, in APDC and ∆QCD
DC = DC [common]
PD = QC [prove above]
∠PDC = ∠QCD [each = 90° + 45° = 135°]
⇒ ∆PDC = ∆QCD [by SAS congruence rule]
⇒ PC = QD or DQ = CP

Areas of Parallelograms and Triangles Class 9 Extra Questions HOTS

Question 1.
In the given figure, PQRS, SRNM and PQNM are parallelograms, Show that :
ar(∆PSM) = ar(∆QRN).

Solution:
Since PQRS is a parallelogram.
∴ PS = QR and PS || QR
Since SRNM is also a parallelogram.
∴ SM = RN and SM || RN
Also, PQNM is a parallelogram
∴ PM || QM and PM = QM
Now, in APSM and ∆QRN
PS = QR
SM = RN
PM = QN
∆PSM ≅ ∆QRN [by SSS congruence axiom]
∴ ar (∆PSM) = ar (∆QRN) [congruent triangles have same areas)

Areas of Parallelograms and Triangles Class 9 Extra Questions Value Based (VBQs)

Question 1.
Naveen was having a plot in the shape of a quadrilateral. He decided to donate some portion of it to construct a home for orphan girls. Further he decided to buy a land in lieu of his donated portion of his plot so as to form a triangle.
(i) Explain how this proposal will be implemented?
(ii) Which mathematical concept is used in it?
(iii) What values are depicted by Naveen?
Solution:

(i) Let ABCD be the plot and Naveen decided to donate some portion to construct a home for orphan girls from one corner say C of plot ABCD. Now, Naveen also purchases equal amount of land in lieu of land CDO, so that he may have triangular form of plot. BD is joined. Draw a line through C parallel to DB to meet AB produced in P.
Join DP to intersect BC at 0.
Now, ABCD and ABPD are on the same base and between same parallels CP || DB.
ar(∆BCD) = ar(∆BPD) ar(∆COD) + ar(∆DBO) = ar(∆BOP) + ar(∆DBO)
ar(ACOD) = ar(ABOP) ar(quad. ABCD)
= ar(quad. ABOD) + ar(∆COD)
= ar(quad. ABOD) + ar(∆BOP)
[∵ ar(ACOD) = ar(ABOP)] (proved above]
= ar(∆APD)
Hence, Naveen purchased the portion ABOP to meet his requirement.
(ii) Two triangles on the same base and between same parallels are equal in area.
(iii) We should help the orphan children.

Question 2.
A flood relief camp was organized by state government for the people affected by the natural calamity near a city. Many school students volunteered to participate in the relief work. In the camp, the food items and first aid centre kits were arranged for the flood victims. The piece of land used for this purpose is shown in the figure.
(a) If EFGH is a parallelogram with P and Q as mid-points of sides GH and EF respectively, then show that area used for first aid is half of the total area.
(b) What can you say about the student volunteers working for the relief work?

Solution:
(a) Here, EFGH is a ||^gm
∴ EF = GH and EF || GH

Hence, area used for first aid is half of the total area.

(b) Students working for the noble cause show compassion towards the affected people. They also realize their social responsibility to work for helping the ones in need.

Here we are providing Quadrilaterals Class 9 Extra Questions Maths Chapter 8 with Answers Solutions, Extra Questions for Class 9 Maths was designed by subject expert teachers.

Extra Questions for Class 9 Maths Quadrilaterals with Answers Solutions

Extra Questions for Class 9 Maths Chapter 8 Quadrilaterals with Solutions Answers

Quadrilaterals Class 9 Extra Questions Very Short Answer Type

Question 1.
If one angle of a parallelogram is twice of its adjacent angle, find the angles of the parallelogram.
Solution:
Let the two adjacent angles be x and 2x.
In a parallelogram, sum of the adjacent angles are 180°
∴ x + 2x = 180°
⇒ 3x = 180°
⇒ x = 60°
Thus, the two adjacent angles are 120° and 60°. Hence, the angles of the parallelogram are 120°, 60°, 120° and 60°.

Question 2.
If the diagonals of a quadrilateral bisect each other at right angles, then name the
quadrilateral.
Solution:
Rhombus.

Question 3.
Three angles of a quadrilateral are equal and the fourth angle is equal to 144o. Find each of the equal angles of the quadrilateral.
Solution:
Let each equal angle of given quadrilateral be x.
We know that, sum of all interior angles of a quadrilateral is 360°
∴ x + x + x + 144° = 360°
3x = 360° – 144°
3x = 216°
x = 72°
Hence, each equal angle of the quadrilateral is of 72o measures.

Question 4.
If ABCD is a parallelogram, then what is the measure of ∠A – ∠C ?
Solution:
∠A – ∠C = 0° (opposite angles of parallelogram are equal]

Question 5.
PQRS is a parallelogram, in which PQ = 12 cm and its perimeter is 40 cm. Find the length of each side of the parallelogram.
Solution:

Here, PQ = SR = 12 cm
Let PS = x and PS = QR
∴ x + 12 + x + 12 = Perimeter
2x + 24 = 40
2x = 16
x= 8
Hence, length of each side of the parallelogram is 12 cm, 8 cm, 12 cm and 8 cm.

Question 6.
Two consecutive angles of a parallelogram are (x + 60)° and (2x + 30)°. What special name can you give to this parallelogram?
Solution:
We know that consecutive interior angles of a parallelogram are supplementary.
∴ (x + 60° + (2x + 30)° = 180°
⇒ 3x° + 90° = 180°
⇒ 3x° = 90°
⇒ x° = 30°
Thus, two consecutive angles are (30 + 60)°, 12 x 30 + 30)”. i.e., 90° and 90°.
Hence, the special name of the given parallelogram is rectangle.

Question 7.
ONKA is a square with ∠KON = 45°. Determine ∠KOA.
Solution:
Since ONKA is a square
∴ ∠AON = 90°
We know that diagonal of a square bisects its ∠s
⇒ ∠AOK = ∠KON = 45°
Hence, ∠KOA = 45°

Question 8.
In quadrilateral PQRS, if ∠P = 60° and ∠Q : ∠R : ∠S = 2 : 3 : 7, then find the measure of ∠S.
Solution:
Let ∠Q = 2x, ∠R = 3x and ∠S = 7x
Now, ∠P + ∠Q + ∠R + ∠S = 360°
⇒ 60° + 2x + 3x + 7x = 360°
⇒ 12x = 300°
x = \(\frac{300^{\circ}}{12}\) = 25°
∠S = 7x = 7 x 25° = 175°

Quadrilaterals Class 9 Extra Questions Short Answer Type 1

Question 1.
ABCD is a parallelogram in which ∠ADC = 75° and side AB is produced to point E as shown in the figure. Find x + y.
Solution:

Here, ∠C and ∠D are adjacent angles of the parallelogram.
∴ ∠C + ∠D = 180°
⇒ x + 75° = 180°
⇒ x = 105°
Also, y = x = 105° [alt. int. angles]
Thus, x + y = 105° + 105° = 210°

Question 2.
If the diagonals of a parallelogram are equal, then show that it is a rectangle.
Solution:

Given: A parallelogram ABCD, in which AC = BD.
To Prove: ΔBCD is a rectangle.
Proof : In ΔABC and ΔBAD
AB = AB (common]
AC = BD (given]
BC = AD(opp. sides of a ||^gm]
⇒ ΔABC ≅ ΔBAD
[by SSS congruence axiom]
⇒ ∠ABC = ∠BAD (c.p.c.t.)
Also, ∠ABC + ∠BAD = 180° (co-interior angles)
∠ABC + ∠ABC = 180° [ ∵ ∠ABC = ∠BAD ]
2∠ABC = 180°
∠ABC = 1/2 x 180° = 90°
Hence, parallelogram ABCD is a rectangle.

Question 3.
In the figure, ABCD is a rhombus, whose diagonals meet at O. Find the values of x and y.

Solution:
Since diagonals of a rhombus bisect each other at right angle.
In ∴ ΔAOB, we have
∠OAB + ∠x + 90° = 180°
∠x = 180° – 90° – 35°
= 55°
Also,
∠DAO = ∠BAO = 35°
∠y + ∠DAO + ∠BAO + ∠x = 180°
⇒ ∠y + 35° + 35° + 55° = 180°
⇒ ∠y = 180° – 125o = 55°
Hence, the values of x amd y are x = 55°, y = 55°.

Question 4.
ABCD is a parallelogram and AP and CQ are perpendiculars from vertices A and C on diagonal BD (see fig.). Show that :
(i) ΔAPB = ΔCQD
(ii) AP = CQ
Solution:

Given : In ||gm ABCD, AP and CQ are perpendiculars from the
vertices A and C on the diagonal BD.
To Prove: (i) ΔAPB ≅ ΔCQD
(ii) AP = CQ
Proof : (i) In ΔAPB and ΔCQD
AB = DC (opp. sides of a ||gm ABCD]
∠APB = ∠DQC (each = 90°)
∠ABP = ∠CDQ (alt. int. ∠s]
⇒ ΔAPB ≅ ΔCQD[by AAS congruence axiom]
(ii) ⇒ AP = CQ [c.p.c.t.]

Quadrilaterals Class 9 Extra Questions Short Answer Type 2

Question 1.
The diagonals of a quadrilateral ABCD are perpendicular to each other. Show that the quadrilateral formed by joining the mid-points of its sides is a rectangle.

Solution:
Given: A quadrilateral ABCD whose diagonals AC and BD are perpendicular to each other at O. P, Q, R and S are mid-points of side AB, BC, CD and DA respectively are joined are formed quadrilateral PQRS.
To Prove: PQRS is a rectangle.
Proof : In ∆ABC, P and Q are mid-points of AB and BC respectively.
∴ PQ || AC and PQ = \(\frac{1}{2}\) AC … (i) (mid-point theorem]
Further, in SACD, R and S are mid-points of CD and DA respectively.
SR || AC and SR = \(\frac{1}{2}\) AC … (ii) (mid-point theorem]
From (i) and (ii), we have PQ || SR and PQ = SR
Thus, one pair of opposite sides of quadrilateral PQRS are parallel and equal.
∴ PQRS is a parallelogram.
Since PQ|| AC PM || NO
In ∆ABD, P and S are mid-points of AB and AD respectively.
PS || BD (mid-point theorem]
⇒ PN || MO
∴ Opposite sides of quadrilateral PMON are parallel.
∴ PMON is a parallelogram.
∠ MPN = ∠ MON (opposite angles of ||gm are equal]
But ∠MON = 90° [given]
∴ ∠MPN = 90° ⇒ ∠QPS = 90°
Thus, PQRS is a parallelogram whose one angle is 90°
∴ PQRS is a rectangle.

Question 2.
In the fig., D, E and F are, respectively the mid-points of sides BC, CA and AB of an equilateral triangle ABC. Prove that DEF is also an equilateral triangle.

Solution:
Since line segment joining the mid-points of two sides of a triangle is half of the third side.
Therefore, D and E are mid-points of BC and AC respectively.
⇒ DE = \(\frac{1}{2}\)AB …(i)
E and F are the mid-points of AC and AB respectively.
∴ EF = \(\frac{1}{2}\)BC … (ii)
F and D are the mid-points of AB and BC respectively.
∴ FD = \(\frac{1}{2}\) AC … (iii)
Now, SABC is an equilateral triangle.
⇒ AB = BC = CA
⇒ \(\frac{1}{2}\)AB = \(\frac{1}{2}\)BC = \(\frac{1}{2}\)CA
⇒ DE = EF = FD (using (i), (ii) and (iii)]
Hence, DEF is an equilateral triangle.

Question 3.
In quadrilateral ABCD of the given figure, X and Y are points on diagonal AC such that AX = CY and BXDY is a parallelogram. Show that ABCD is a parallelogram.

Solution:
Since BXDY is a parallelogram.
XO = YO
DO = BO
[∵ diagonals of a parallelogram bisect each other]
But AX = CY …. (iii) (given]
Adding (i) and (iii), we have
XO + AX = YO + CY
⇒ AO = CO …. (iv)
From (ii) and (iv), we have
AO = CO and DO = BO
Thus, ABCD is a parallelogram, because diagonals AC and BD bisect each other at O.

Question 4.
ABCD is a quadrilateral in which the bisectors of ∠A and ∠C meet DC produced at Y and BA produced at X respectively. Prove that
∠X +∠Y = \(\frac{1}{2}\)(∠A + ∠C)
Solution:

Here, ∠1 = ∠2 and ∠3 = ∠4
In ΔXBC, we have
∠X + ∠B + ∠4 = 180°
∠X + ∠B + \(\frac{1}{2}\)∠C = 180
In ΔADY, we have
∠2 + ∠D + ∠Y= 180°
\(\frac{1}{2}\) ∠A + ∠D + ∠Y = 180°
Adding (i) and (ii), we have
∠X + ∠Y + ∠B + ∠D + \(\frac{1}{2}\) ∠C + \(\frac{1}{2}\) ∠A = 360°
Also, in quadrilateral ABCD
∠A + ∠B + ∠C + ∠D = 360°
∠X + ∠Y + ∠B + ∠D + \(\frac{1}{2}\) ∠C + \(\frac{1}{2}\) ∠A = ∠A + ∠B + ∠C + ∠D
∠X + ∠Y = ∠A – \(\frac{1}{2}\) ∠A + \(\frac{1}{2}\) ∠C – \(\frac{1}{2}\) ∠C
∠X+ ∠Y = \(\frac{1}{2}\) (∠A + ∠C)

Quadrilaterals Class 9 Extra Questions Long Answer Type

Question 1.
In the figure, P, Q and R are the mid-points of the sides BC, AC and AB of ΔABC. If BQ and PR intersect at X and CR and PQ intersect at Y, then show that XY = \(\frac{1}{4}\) BC.

Solution:
Here, in ΔABC, R and Q are the mid-points of AB and AC respectively.
∴ By using mid-point theorem, we have
RQ || BC and RQ = \(\frac{1}{2}\) BC
∴ RQ = BP = PC [∵ P is the mid-point of BC]
∴ RQ || BP and RQ || PC
In quadrilateral BPQR
RQ || BP, RQ = BP (proved above]
∴ BPQR is a parallelogram. [∵ one pair of opp. sides is parallel as well as equal]
∴ X is the mid-point of PR. [∵ diagonals of a ||gm bisect each other]
Now, in quadrilateral PCQR
RQ || PC and RQ = PC [proved above)
∴ PCQR is a parallelogram [∵ one pair of opp. sides is parallel as well as equal]
∴ Y is the mid-point of PQ [∵ diagonals of a ||gm bisect each other]
In ΔPQR
∴ X and Y are mid-points of PR and PQ respectively.

Question 2.
In the given figure, AE = DE and BC || AD. Prove that the points A, B, C and D are concyclic. Also, prove that the diagonals of the quadrilateral ABCD are equal.

Solution:
Since AE = DE
∠D = ∠A …. (i) [∵ ∠s opp. to equal sides of a Δ]
Again, BC || AD
∠EBC = ∠A …. (ii) (corresponding ∠s]
From (i) and (ii), we have
∠D = ∠EBC …. (iii)
But ∠EBC + ∠ABC = 180° (a linear pair]
∠D + ∠ABC = 180° (using (iii)]
Now, a pair of opposite angles of quadrilateral ABCD is supplementary
Thus, ABCD is a cyclic quadrilateral i.e., A, B, C and D’are concyclic. In ΔABD and ΔDCA
∠ABD = ∠ACD [∠s in the same segment for cyclic quad. ABCD]
∠BAD = ∠CDA [using (i)]
AD = AD (common]
So, by using AAS congruence axiom, we have
ΔABD ≅ ΔDCA
Hence, BD = CA [c.p.c.t.]

Question 3.
In ΔABC, AB = 8 cm, BC = 9 cm and AC = 10 cm. X, Y and Z are mid-points of AO, BO and CO respectively as shown in the figure. Find the lengths of the sides of ΔXYZ.

Solution:
Here, in ΔABC, AB = 8 cm, BC = 9 cm, AC = 10 cm.
In ΔAOB, X and Y are the mid-points of AO and BO.
∴ By using mid-point theorem, we have
XY = \(\frac{1}{2}\) AB = \(\frac{1}{2}\) x 8 cm = 4 cm
Similarly, in Δ𝜏BOC, Y and Z are the mid-points of BO and CO.
∴ By using mid-point theorem, we have
YZ = \(\frac{1}{2}\) BC = \(\frac{1}{2}\) x 9cm = 4.5 cm
And, in Δ𝜏COA, Z and X are the mid-points of CO and AO.
∴ ZX = \(\frac{1}{2}\) AC = \(\frac{1}{2}\) x 10 cm = 5 cm
Hence, the lengths of the sides of ΔXYZ are XY = 4 cm, YZ = 4.5 cm and ZX = 5 cm.

Question 4.
PQRS is a square and ∠ABC = 90° as shown in the figure. If AP = BQ = CR, then prove that ∠BAC = 45°

Solution:
Since PQRS is a square.
∴ PQ = QR … (I) [∵ sides of a square are equal]
Also, BQ = CR … (ii) [given]
Subtracting (ii) from (i), we obtain
PQ – BQ = QR – CR
⇒ PB = QC … (iii)
In Δ𝜏APB and Δ𝜏BQC
AP = BQ[given
∠APB = ∠BQC = 90°](each angle of a square is 90°)
PB = QC (using (iii)]

So, by using SAS congruence axiom, we have
ΔAPB ≅ ΔBQC
∴ AB = BC [c.p.c.t.]
Now, in ΔABC
AB = BC [proved above]
∴ ∠ACB = ∠BAC = x° (say) [∠s opp. to equal sides]
Also, ∠B + ∠ACB + ∠BAC = 180°
⇒ 90° + x + x = 180°
⇒ 2x° = 90°
x° = 45°
Hence, ∠BAC = 45°

Question 5.
ABCD is a parallelogram. If the bisectors DP and CP of angles D and C meet at P on side AB, then show that P is the mid-point of side AB.
Solution:

Since DP and CP are angle bisectors of ∠D and ∠C respectively.
: ∠1 = ∠2 and ∠3 = ∠4
Now, AB || DC and CP is a transversal
∴ ∠5 = ∠1 [alt. int. ∠s]
But ∠1 = ∠2 [given]
∴ ∠5 = ∠2

Now, in ABCP, ∠5 = ∠2
⇒ BC = BP … (I) [sides opp. to equal ∠s of a A]
Again, AB || DC and DP is a transversal.
∴ ∠6= ∠3 (alt. int. Δs]
But ∠4 = ∠3 [given]
∴ ∠6 = ∠4
Now, in ΔADP, ∠6 = ∠4
⇒ DA = AP …. (ii) (sides opp. to equal ∠s of a A]
Also, BC = DA… (iii) (opp. sides of parallelogram)
From (i), (ii) and (iii), we have
BP = AP
Hence, P is the mid-point of side AB.

Question 6.
In the figure, ΔBCD is a trapezium in which AB || DC. E and F are the mid-points of AD and BC respectively. DF and AB are produced to meet at G. Also, AC and EF intersect at the point O. Show that :
(i) EO || AB
(ii) AO = CO

Solution:
Here, E and F are the mid-points of AD and BC respectively.
In ΔBFG and ΔCFD
BF = CF [given]
∠BFG = ∠CFD (vert. opp. ∠s]
∠BGF = ∠CDF (alt. int. ∠s, as AB || DC)
So, by using AAS congruence axiom, we have
ΔBFG ≅ ΔCFD
⇒ DF = FG [c.p.c.t.)
Now, in ΔAGD, E and F are the mid-points of AD and GD.
∴ By mid-point theorem, we have
EF || AG
or EO || AB
Also, in ΔADC, EO || DC
∴ EO is a line segment from mid-point of one side parallel to another side.
Thus, it bisects the third side.
Hence, AO = CO

Here we are providing Triangles Class 9 Extra Questions Maths Chapter 7 with Answers Solutions, Extra Questions for Class 9 Maths was designed by subject expert teachers.

Extra Questions for Class 9 Maths Triangles with Answers Solutions

Extra Questions for Class 9 Maths Chapter 7 Triangles with Solutions Answers

Triangles Class 9 Extra Questions Very Short Answer Type

Question 1.
Find the measure of each exterior angle of an equilateral triangle.
Solution:
We know that each interior angle of an equilateral triangle is 60°.
∴ Each exterior angle = 180° – 60° = 120°

Question 2.
If in ∆ABC, ∠A = ∠B + ∠C, then write the shape of the given triangle.
Solution:
Here, ∠A = ∠B + ∠C
And in ∆ABC, by angle sum property, we have
∠A + ∠B + C = 180°
⇒ ∠A + ∠A = 180°
⇒ 2∠A = 180°
⇒ ∠A = 90°
Hence, the given triangle is a right triangle.

Question 3.
In ∆PQR, PQ = QR and ∠R = 50°, then find the measure of ∠Q.
Solution:
Here, in ∆PQR, PQ = QR
⇒ ∠R = ∠P = 50° (given)
Now, ∠P + ∠Q  + ∠R = 180°
50° + ∠Q + 50° = 180°
⇒ ∠Q = 180° – 50° – 50°
= 80°

Question 4.
If ∆SKY ≅ ∆MON by SSS congruence rule, then write three equalities of corresponding angles.
Solution:
Since ∆SKY ≅ ∆MON by SSS congruence rule, then three equalities of corresponding angles
are ∠S = ∠M, ∠K = ∠O and ∠Y = ∠N.

Question 5.
Is ∆ABC possible, if AB = 6 cm, BC = 4 cm and AC = 1.5 cm ?
Solution:
Since 4 + 1.5 = 5.5 ≠ 6
Thus, triangle is not possible.

Question 6.
In ∆MNO, if ∠N = 90°, then write the longest side.
Solution:
We know that, side opposite to the largest angle is longest.
∴ Longest side = MO.

Question 7.
In ∆ABC, if AB = AC and ∠B = 70°, find ∠A.
Solution:
Here, in ∆ABC AB = AC ∠C = ∠B [∠s opp. to equal sides of a ∆)
Now, ∠A + ∠B + ∠C = 180°
⇒ ∠A + 70° + 70° = 180° [∵ ∠B = 70°]
⇒ ∠A = 180° – 70° – 70° = 40°

Question 8.
In ∆ABC, if AD is a median, then show that AB + AC > 2AD.
Solution:

Produce AD to E, such that AD = DE.
In ∆ADB and ∆EDC, we have
BD = CD, AD = DE and ∠1 = ∠2
∆ADB ≅ ∆EDC
AB = CE
Now, in ∆AEC, we have
AC + CE > AE
AC + AB > AD + DE
AB + AC > 2AD [∵ AD = DE]

Triangles Class 9 Extra Questions Short Answer Type 1

Question 1.
In the given figure, AD = BC and BD = AC, prove that ∠DAB = ∠CBA.
Solution:

In ∆DAB and ∆CBA, we have
AD = BC [given]
BD = AC [given]
AB = AB [common]
∴ ∆DAB ≅ ∆CBA [by SSS congruence axiom]
Thus, ∠DAB =∠CBA [c.p.c.t.]

Question 2.
In the given figure, ∆ABD and ABCD are isosceles triangles on the same base BD. Prove that ∠ABC = ∠ADC.
Solution:

In ∆ABD, we have
AB = AD (given)
∠ABD = ∠ADB [angles opposite to equal sides are equal] …(i)
In ∆BCD, we have
CB = CD
⇒ ∠CBD = ∠CDB [angles opposite to equal sides are equal] … (ii)
Adding (i) and (ii), we have
∠ABD + ∠CBD = ∠ADB + ∠CDB
⇒ ∠ABC = ∠ADC

Question 3.
In the given figure, if ∠1 = ∠2 and ∠3 = ∠4, then prove that BC = CD.
Solution:

In ∆ABC and ACDA, we have
∠1 = ∠2 (given)
AC = AC [common]
∠3 = ∠4 [given]
So, by using ASA congruence axiom
∆ABC ≅ ∆CDA
Since corresponding parts of congruent triangles are equal
∴ BC = CD

Question 4.
In the given figure, ∠B < ∠A and ∠C < ∠D. Show that AD < BC.

Solution:
Here, ∠B < ∠A
⇒ AO < BO …..(i)
and ∠C < ∠D
⇒ OD < CO …..(ii)
[∴ side opposite to greater angle is longer]
Adding (i) and (ii), we obtain
AO + OD < BO + CO
AD < BC

Question 5.
In the given figure, AC > AB and D is a point on AC such that AB = AD. Show that BC > CD.
Solution:

Here, in ∆ABD, AB = AD
∠ABD = ∠ADB
[∠s opp. to equal sides of a ∆]
In ∆BAD
ext. ∠BDC = ∠BAD + ∠ABD
⇒ ∠BDC > ∠ABD ….(ii)
Also, in ∆BDC .
ext. ∠ADB > ∠CBD …(iii)
From (ii) and (iii), we have
∠BDC > CD [∵ sides opp. to greater angle is larger]

Question 6.
In a triangle ABC, D is the mid-point of side AC such that BD = \(\frac{1}{2}\) AC. Show that ∠ABC is a right angle.
Solution:

Here, in ∆ABC, D is the mid-point of AC.
⇒ AD = CD = \(\frac{1}{2}\)AC …(i)
Also, BD = \(\frac{1}{2}\)AC… (ii) [given]
From (i) and (ii), we obtain
AD = BD and CD = BD
⇒ ∠2 = ∠4 and ∠1 = ∠3 …..(iii)
In ∆ABC, we have
∠ABC + ∠ACB + ∠CAB = 180°
⇒ ∠1 + ∠2 + ∠3 + ∠4 = 180°
⇒ ∠1 + ∠2 + ∠1 + ∠2 = 180° [using (iii)]
⇒ 2(∠1 + ∠2) = 180°
⇒ ∠1 + ∠2 = 90°
Hence, ∠ABC = 90°

Triangles Class 9 Extra Questions Short Answer Type 2

Question 1.
ABC is an isosceles triangle with AB = AC. P and Q are points on AB and AC respectively such that AP = AQ. Prove that CP = BQ.
Solution:

In ∆ABQ and ∆ACP, we have
AB = AC (given)
∠BAQ = ∠CAP [common]
AQ = AP (given)
∴ By SAS congruence criteria, we have
∆ABQ ≅ ∆ACP
CP = BQ

Question 2.
In the given figure, ∆ABC and ∆DBC are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC, AD is extended to intersect BC at P. Show that : (i) ∆ABD ≅ ∆ACD (ii) ∆ABP ≅ ∆ACP

Solution:
(i) In ∆ABD and ∆ACD
AB = AC [given]
BD = CD [given]
AD = AD [common)]
∴ By SSS congruence axiom, we have
∆ABD ≅ ∆ACD
(ii) In ∆ABP and ∆ACP
AB = AC [given]
∠BAP = ∠CAP [c.p.cit. as ∆ABD ≅ ∆ACD]
AP = AP [common]
∴ By SAS congruence axiom, we have
∆ABP ≅ ∆ACP

Question 3.
In the given figure, it is given that AE = AD and BD = CE. Prove that ∆AEB ≅ ∆ADC.

Solution:
We have AE = AD … (i)
and CE = BD … (ii)
On adding (i) and (ii),
we have AE + CE = AD + BD
⇒ AC = AB
Now, in ∆AEB and ∆ADC,
we have AE = AD [given]
AB = AC [proved above]
∠A = ∠A [common]
∴ By SAS congruence axiom, we have
∆AEB = ∆ADC

Question 4.
In the given figure, in ∆ABC, ∠B = 30°, ∠C = 65° and the bisector of ∠A meets BC in X. Arrange AX, BX and CX in ascending order of magnitude.

Solution:
Here, AX bisects ∠BAC.
∴ ∠BAX = ∠CAX = x (say)
Now, ∠A + ∠B + C = 180° [angle sum property of a triangle]
⇒ 2x + 30° + 65° = 180°
⇒ 2x + 95 = 180°
⇒ 2x = 180° – 95°
⇒ 2x = 85°
⇒ x = \(\frac{85^{\circ}}{2}\) = 42.59
In ∆ABX, we have x > 30°
BAX > ∠ABX
⇒ BX > AX (side opp. to larger angle is greater)
⇒ AX < BX
Also, in ∆ACX, we have 65° > x
⇒ ∠ACX > ∠CAX
⇒ AX > CX [side opp. to larger angle is greater]
⇒ CX > AX … (ii)
Hence, from (i) and (ii), we have
CX < AX < BX

Question 5.
In figure, ‘S’ is any point on the side QR of APQR. Prove that PQ + QR + RP > 2PS.

Solution:
In ∆PQS, we have
PQ + QS > PS …(i)
[∵ sum of any two sides of a triangle is greater than the third side]
In ∆PRS, we have
RP + RS > PS …(ii)
Adding (i) and (ii), we have
PQ + (QS + RS) + RP > 2PS
Hence, PQ + QR + RP > 2PS. [∵ QS + RS = QR]

Question 6.
If two isosceles triangles have a common base, prove that the line joining their vertices bisects them at right angles.
Solution:
Here, two triangles ABC and BDC having the common
base BC, such that AB = AC and DB = DC.
Now, in ∆ABD and ∆ACD

AB = AC [given]
BD = CD [given]
AD = AD [common]
∴ ΔABD ≅ ΔΑCD [by SSS congruence axiom]
⇒ ∠1 = ∠2 [c.p.c.t.]
Again, in ∆ABE and ∆ACE, we have
AB = AC [given]
∠1 = ∠2 [proved above]
AE = AE [common]
∆ABE = ∆ACE [by SAS congruence axiom]
BE = CE [c.p.c.t.]
and ∠3 = ∠4 [c.p.c.t.]
But ∠3 + ∠4 = 180° [a linear pair]
⇒ ∠3 = ∠4 = 90°
Hence, AD bisects BC at right angles.

Triangles Class 9 Extra Questions Long Answer Type

Question 1.
In the given figure, AP and DP are bisectors of two adjacent angles A and D of quadrilateral ABCD. Prove that 2 ∠APD = ∠B + 2C.

Solution:
Here, AP and DP are angle bisectors of ∠A and ∠D
∴ ∠DAP = \(\frac{1}{2}\)∠DAB and ∠ADP = \(\frac{1}{2}\)∠ADC ……(i)
In ∆APD, ∠APD + ∠DAP + ∠ADP = 180°
⇒ ∠APD + \(\frac{1}{2}\) ∠DAB + \(\frac{1}{2}\)∠ADC = 180°
⇒ ∠APD = 180° – \(\frac{1}{2}\)(∠DAB + ∠ADC)
⇒ 2∠APD = 360° – (∠DAB + ∠ADC) ……(ii)
Also, ∠A + ∠B + C + ∠D = 360°
∠B + 2C = 360° – (∠A + ∠D)
∠B + C = 360° – (∠DAB + ∠ADC) ……(iii)
From (ii) and (iii), we obtain
2∠APD = ∠B + ∠C

Question 2.
In figure, ABCD is a square and EF is parallel to diagonal BD and EM = FM. Prove that
(i) DF = BE (i) AM bisects ∠BAD.

Solution:
(i) EF || BD = ∠1 = ∠2 and ∠3 = ∠4 [corresponding ∠s]
Also, ∠2 = ∠4
⇒ ∠1 = ∠3
⇒ CE = CF (sides opp. to equals ∠s of a ∆]
∴ DF = BE [∵ BC – CE = CD – CF)

(ii) In ∆ADF and ∆ABE, we have
AD = AB [sides of a square]
DF = BE [proved above]
∠D = ∠B = 90°
⇒ ∆ADF ≅ ∆ABE [by SAS congruence axiom]
⇒ AF = AE and ∠5 = ∠6 … (i) [c.p.c.t.]
In ∆AMF and ∆AME
AF = AE [proved above]
AM = AM [common]
FM = EM (given)
∴ ∆AMF ≅ ∆AME [by SSS congruence axiom]
∴ ∠7 = ∠8 …(ii) [c.p.c.t.]
Adding (i) and (ii), we have
∠5 + ∠7 = ∠6 + ∠8
∠DAM = ∠BAM
∴ AM bisects ∠BAD.

Question 3.
In right triangle ABC, right-angled at C, M is the mid-point of hypotenuse AB. C is joined to M and produced to a point D such that DM = CM. Point D is joined to point B (see fig.). Show that : (i) ∆AMC ≅ ∆BMD (ii) ∠DBC = 90° (ii) ∆DBC ≅ ∆ACB (iv) CM = \(\frac{1}{2}\)AB

Solution:
Given : ∆ACB in which 4C = 90° and M is the mid-point of AB.
To Prove :
(i) ∆AMC ≅ ∆BMD
(ii) ∠DBC = 90°
(iii) ∆DBC ≅ ∆ACB
(iv) CM = \(\frac{1}{2}\)AB
Proof : Consider ∆AMC and ∆BMD,
we have AM = BM [given]
CM = DM [by construction]
∠AMC = ∠BMD [vertically opposite angles]
∴ ∆AMC ≅ ∆BMD [by SAS congruence axiom]
⇒ AC = DB …(i) [by c.p.c.t.]
and ∠1 = ∠2 [by c.p.c.t.]
But ∠1 and ∠2 are alternate angles.
⇒ BD || CA
Now, BD || CA and BC is transversal.
∴ ∠ACB + ∠CBD = 180°
⇒ 90° + CBD = 180°
⇒ ∠CBD = 90°
In ∆DBC and ∆ACB,
we have CB = BC [common]
DB = AC [using (i)]
∠CBD = ∠BCA
∴ ∆DBC ≅ ∆ACB
⇒ DC = AB
⇒ \(\frac{1}{2}\)AB = \(\frac{1}{2}\)DC
⇒ \(\frac{1}{2}\)AB = CM or CM = \(\frac{1}{2}\)AB (∵ CM = \(\frac{1}{2}\)DC)

Question 4.
In figure, ABC is an isosceles triangle with AB = AC. D is a point in the interior of ∆ABC such that ∠BCD = ∠CBD. Prove that AD bisects ∠BAC of ∆ABC.

Solution:
In ∆BDC, we have ∠DBC = ∠DCB (given).
⇒ CD = BD (sides opp. to equal ∠s of ∆DBC)
Now, in ∆ABD and ∆ACD,
we have AB = AC [given]
BD = CD [proved above]
AD = AD [common]
∴ By using SSS congruence axiom, we obtain
∆ABD ≅ ∆ACD
⇒ ∠BAD = ∠CAD [c.p.ç.t.]
Hence, AD bisects ∠BAC of ∆ABC.

Question 5.
Prove that two triangles are congruent if any two angles and the included side of one triangle is equal to any two angles and the included side of the other triangle.
Solution:

Given : Two As ABC and DEF in which
∠B = ∠E,
∠C = ∠F and BC = EF
To Prove : ∆ABC = ∆DEF
Proof : We have three possibilities
Case I. If AB = DE,
we have AB = DE,
∠B = ∠E and BC = EF.
So, by SAS congruence axiom, we have ∆ABC ≅ ∆DEF

Case II. If AB < ED, then take a point Mon ED
such that EM = AB.
Join MF.
Now, in ∆ABC and ∆MEF,
we have
AB = ME, ∠B = ∠E and BC = EF.
So, by SAS congruence axiom,
we have ΔΑΒC ≅ ΔΜEF
⇒ ∠ACB = ∠MFE
But ∠ACB = ∠DFE
∴ ∠MFE = ∠DFE

Which is possible only when FM coincides with B FD i.e., M coincides with D.
Thus, AB = DE
∴ In ∆ABC and ∆DEF, we have
AB = DE,
∠B = ∠E and BC = EF
So, by SAS congruence axiom, we have
∆ABC ≅ ∆DEF
Case III. When AB > ED
Take a point M on ED produced
such that EM = AB.
Join MF
Proceeding as in Case II, we can prove that
∆ABC = ∆DEF
Hence, in all cases, we have
∆ABC = ∆DEF.

Question 6.
In the given figure, side QR is produced to the point S. If the bisectors of ∠PQR and ∠PRS meet at T,
prove that ∠QTR = \(\frac{1}{2}\) ∠QPR.

Solution:
Here, QT is angle bisector of ∠PQR

Triangles Class 9 Extra Questions HOTS

Question 1.
Show that the difference of any two sides of a triangle is less than the third side.
Solution:

Consider a triangle ABC
To Prove :
(i) AC – AB < BC
(ii) BC – AC < AB
(iii) BC – AB < AC
Construction : Take a point D on AC
such that AD = AB.
Join BD.
Proof : In ∆ABD, we have ∠3 > ∠1 …(i)
[∵ exterior ∠ is greater than each of interior opposite angle in a ∆]
Similarly, in ∆BCD, we have
∠2 > ∠4 …..(ii) [∵ ext. ∠ is greater then interior opp. angle in a ∆]
In ∆ABD, we have
AD = AB [by construction]
∠1 = ∠2 …(iii) [angles opp. to equal sides are equal in a triangle]
From (i), (ii) and (iii), we have
⇒ ∠3 > ∠4 =
⇒ BC > CD
⇒ CD < BC
AC – AD < BC
AC – AB < BC [∵ AD = AB]
Hence, AC – AB < BC
Similarly, we can prove
BC – AC < AB
and BC – AB < AC

Question 2.
In the figure, O is the interior point of ∆ABC. BO meets AC at D. Show that OB + OC < AB + AC.

Solution:
In ∆ABD, AB + AD > BD …(i)
∵ The sum of any two sides of a triangle is greater than the third side. Also, we have
BD = BO + OD
AB + AD > BO + OD ….(ii)
Similarly, in ∆COD, we have
OD + DC > OC … (iii)
On adding (ii) and (iii), we have
AB + AD + OD + DC > BO + OD + OC
⇒ AB + AD + DC > BO + OC
⇒ AB + AC > OB + OC
or OB + OC < AB + AC
Hence, proved.

Triangles Class 9 Extra Questions Value Based (VBQs)

Question 1.
A campaign is started by volunteers of mathematical club to boost school and its surrounding under Swachh Bharat Abhiyan. They made their own logo for this campaign. What values are acquired by mathematical club ?
If it is given that ∆ABC ≅ ∆ECD, BC = AE.
Prove that ∆ABC ≅ ∆CEA.

Solution:
Here, it is given that
∆ABC ≅ ∆ECD
AB = CE [c.p.c.t.]
BC = CD [c.p.c.t.]
AC = ED [c.p.c.t.]
Now, in ∆ABC and ∆CEA
BC = AE [given]
AB = EC [proved above]
AC = AC [common]
∴ By using SSS congruence axiom, we have
∆ABC ≅ ∆CEA
Value : Cleanliness and social concerning.

Question 2.
Rajiv, a good student and actively involved in applying knowledge A of mathematics in daily life. He asked his classmate Rahul to make triangle as shown by choosing one of the vertex as common. Rahul tried but not correctly. After sometime Rajiv hinted Rahul about congruency of triangle. Now, Rahul fixed vertex C as common vertex and locate point D, E such that AC = CD and BC = CE. Was the triangle made by Rahul is congruent ? Write the condition satisfying congruence.
What value is depicted by Rajiv’s action?

Solution:
In ∆ABC and ∆DEC, we have
AC = DC [by construction]
BC = EC [by construction]
∠ACB = ∠ECD [vert. opp. ∠s]
By using SAS congruence axiom, we have
∆ABC ≅ ∆DEC
Value : Cooperative learning, use of concept and friendly nature.

Here we are providing Lines and Angles Class 9 Extra Questions Maths Chapter 6 with Answers Solutions, Extra Questions for Class 9 Maths was designed by subject expert teachers.

Extra Questions for Class 9 Maths Lines and Angles with Answers Solutions

Extra Questions for Class 9 Maths Chapter 6 Lines and Angles with Solutions Answers

Lines and Angles Class 9 Extra Questions Very Short Answer Type

Question 1.
If an angle is half of its complementary angle, then find its degree measure.
Solution:
Let the required angle be x
∴ Its complement = 90° – x
Now, according to given statement, we obtain
x = \(\frac{1}{2}\)(90° – x)
⇒ 2x = 90° – x
⇒ 3x = 90°
⇒ x = 30°
Hence, the required angle is 30°.

Question 2.
The two complementary angles are in the ratio 1 : 5. Find the measures of the angles.
Solution:
Let the two complementary angles be x and 5x.
∴ x + 5x = 90°
⇒ 6x = 90°
⇒ x = 15°
Hence, the two complementary angles are 15° and 5 × 15° i.e., 15° and 75°.

Question 3.
In the given figure, if PQ || RS, then find the measure of angle m.

Solution:
Here, PQ || RS, PS is a transversal.
⇒ ∠PSR = ∠SPQ = 56°
Also, ∠TRS + m + ∠TSR = 180°
14° + m + 56° = 180°
⇒ m = 180° – 14 – 56 = 110°

Question 4.
If an angle is 14o more than its complement, then find its measure.
Solution:
Let the required angle be x
∴ Its complement = 90° – x
Now, according to given statement, we obtain
x = 90° – x + 14°
⇒ 2x = 104°
⇒ x = 52°
Hence, the required angle is 52o.

Question 5.
If AB || EF and EF || CD, then find the value of x.

Solution:
Since EF || CD ∴ y + 150° = 180°
⇒ y = 180° – 150° = 30°
Now, ∠BCD = ∠ABC
x + y = 70°
x + 30 = 70
⇒ x = 70° – 30° = 40°
Hence, the value of x is 40°.

Question 6.
In the given figure, lines AB and CD intersect at O. Find the value of x.

Solution:
Here, lines AB and CD intersect at O.
∴ ∠AOD and ∠BOD forming a linear pair
⇒ ∠AOD + ∠BOD = 180°
⇒ 7x + 5x = 180°
⇒ 12x = 180°
⇒ x = 15°

Question 7.
In the given figure, PQ || RS and EF || QS. If ∠PQS = 60°, then find the measure of ∠RFE.

Solution:
Since PQ || RS
∴ ∠PQS + ∠QSR = 180°
⇒ 60° + ∠QSR = 180°
⇒ ∠QSR = 120°
Now, EF || QS ⇒ ∠RFE = ∠QSR [corresponding ∠s]
⇒ ∠RFE = 120°

Question 8.
In the given figure, if x°, y° and z° are exterior angles of ∆ABC, then find the value of x° + y° + z°.

Solution:
We know that, an exterior angle of a triangle is equal to sum of two opposite interior angles.
⇒ x° = ∠1 + ∠3
⇒ y° = ∠2 + ∠1
⇒ z° = ∠3 + ∠2
Adding all these, we have
x° + y° + z° = 2(∠1 + ∠2 + ∠3)
= 2 × 180°
= 360°

Lines and Angles Class 9 Extra Questions Short Answer Type 1

Question 1.
In the given figure, AB || CD, ∠FAE = 90°, ∠AFE = 40°, find ∠ECD.

Solution:
In AFAE,
ext. ∠FEB = ∠A + F
= 90° + 40° = 130°
Since AB || CD
∴ ∠ECD = FEB = 130°
Hence, ∠ECD = 130°.

Question 2.
In the fig., AD and CE are the angle bisectors of ∠A and ∠C respectively. If ∠ABC = 90°, then find ∠AOC.

Solution:
∵ AD and CE are the bisector of ∠A and ∠C

In ∆AOC,
∠AOC + ∠OAC + ∠OCA = 180°
⇒ ∠AOC + 45o = 180°
⇒ ∠AOC = 180° – 45° = 135°.

Question 3.
In the given figure, prove that m || n.

Solution:
In ∆BCD,
ext. ∠BDM = ∠C + ∠B
= 38° + 25° = 63°
Now, ∠LAD = ∠MDB = 63°
But, these are corresponding angles. Hence,
m || n

Question 4.
In the given figure, two straight lines PQ and RS intersect each other at O. If ∠POT = 75°, find the values of a, b, c.

Solution:
Here, 4b + 75° + b = 180° [a straight angle]
5b = 180° – 75° = 105°
b – \(\frac{105^{\circ}}{5}\) = 21°
∴ a = 4b = 4 × 21° = 84° (vertically opp. ∠s]
Again, 2c + a = 180° [a linear pair]
⇒ 2c + 84° = 180°
⇒ 2c = 96°
⇒ c = \(\frac{96^{\circ}}{2}\) = 48°
Hence, the values of a, b and c are a = 84°, b = 21° and c = 48°.

Question 5.
In figure, if AB || CD. If ∠ABR = 45° and ∠ROD = 105°, then find ∠ODC.

Solution:

Through O, draw a line ‘l’ parallel to AB.
⇒ line I will also parallel to CD, then
∠1 = 45°[alternate int. angles]
∠1 + ∠2 + 105° = 180° [straight angle]
∠2 = 180° – 105° – 45°
⇒ ∠2 = 30°
Now, ∠ODC = ∠2 [alternate int. angles]
= ∠ODC = 30°

Question 6.
In the figure, ∠X = 72°, ∠XZY = 46°. If YO and ZO are bisectors of ∠XYZ and ∠XZY respectively of ∆XYZ, find ∠OYZ and ∠YOZ.

Solution:
In ∆XYZ, we have
∠X + XY + ∠Z = 180°
⇒ ∠Y + ∠Z = 180° – ∠X
⇒ ∠Y + ∠Z = 180° – 72°
⇒ Y + ∠Z = 108°
⇒ \(\frac{1}{2}\) ∠Y + \(\frac{1}{2}\)∠Z = \(\frac{1}{2}\) × 108°
∠OYZ + ∠OZY = 54°
[∵ YO and ZO are the bisector of ∠XYZ and ∠XZY]
⇒ ∠OYZ + \(\frac{1}{2}\) × 46° = 54°
∠OYZ + 23° = 54°
⇒ ∠OYZ = 549 – 23° = 31°
In ∆YOZ, we have
∠YOZ = 180° – (∠OYZ + ∠OZY)
= 180° – (31° + 23°) 180° – 54° = 126°

Lines and Angles Class 9 Extra Questions Short Answer Type 2

Question 1.
Prove that if two lines intersect each other, then the bisectors of vertically opposite angles are in the same line.

Solution:
Let AB and CD be two intersecting lines intersecting each other in O.
OP and OQ are bisectors of ∠AOD and ∠BOC.
∴ ∠1 = ∠2 and ∠3 = ∠4 …(i)
Now, ∠AOC = ∠BOD [vertically opp. ∠s] ……(ii)
⇒ ∠1 + ∠AOC + ∠3 = ∠2 + ∠BOD + ∠4 [adding (i) and (ii)]
Also, ∠1 + ∠AOC + ∠3 + ∠2 + ∠BOD + ∠4 = 360° (∵ ∠s at a point are 360°]
⇒ ∠1 + ∠AOC + ∠3 + ∠1 + ∠AOC + ∠3 = 360° [using (i), (ii)]
⇒ ∠1 + ∠AOC + ∠3 = 180°
or ∠2 + ∠BOD + ∠4 = 180°
Hence, OP and OQ are in the same line.

Question 2.
In figure, OP bisects ∠BOC and OQ bisects ∠AOC. Prove that ∠POQ = 90°

Solution:
∵ OP bisects ∠BOC
∴ ∠BOP = ∠POC = x (say)
Also, OQ bisects. ∠AOC
∠AOQ = ∠COQ = y (say) .
∵ Ray OC stands on ∠AOB
∴ ∠AOC + ∠BOC = 180° [linear pair]
⇒ ∠AOQ + ∠QOC + ∠COP + ∠POB = 180°
⇒ y + y + x + x = 180°.
⇒ 2x + 2y = 180°
⇒ x + y = 90°
Now, ∠POQ = ∠POC + ∠COQ
= x + y = 90°

Question 3.
In given figure, AB || CD and EF || DG, find ∠GDH, ∠AED and ∠DEF.

Solution:
Since AB || CD and HE is a transversal.
∴ ∠AED = ∠CDH = 40° [corresponding ∠s]
Now, ∠AED + ∠DEF + ∠FEB = 180° [a straight ∠]
40° + CDEF + 45° = 180°
∠DEF = 180° – 45 – 40 = 95°
Again, given that EF || DG and HE is a transversal. .
∴ ∠GDH = ∠DEF = 95° [corresponding ∠s]
Hence, ∠GDH = 95°, ∠AED = 40° and ∠DEF = 95°

Question 4.
In figure, DE || QR and AP and BP are bisectors of ∠EAB and ∠RBA respectively. Find ∠APB.

Solution:

Here, AP and BP are bisectors of ∠EAB and ∠RBA respectively.
⇒ ∠1 = ∠2 and ∠3 = ∠4
Since DE || QR and transversal n intersects DE and QR at A and B respectively.
⇒ ∠EAB + ∠RBA = 180°
[∵ co-interior angles are supplementary]
⇒ (∠1 + ∠2) + (∠3 + ∠4) = 180°
⇒ (∠1 + ∠1) + (∠3 + ∠3) = 180° (using (i)
⇒ 2(∠1 + ∠3) = 180°
⇒ ∠1 + ∠3 = 90°
Now, in ∆ABP, by’angle sum property, we have
∠ABP + ∠BAP + ∠APB = 180°
⇒ ∠3 + ∠1 + ∠APB = 180°
⇒ 90° + ∠APB = 180°
⇒ ∠APB = 90°

Question 5.
In the given figure, POQ is a line. Ray OR is perpendicular to line PQ. OS is another ray lying between rays OP and OR. Prove that ∠ROS = \(\frac{1}{2}\) (∠QOS – ∠POS).

Solution:
Given that OR is perpendicular to PQ
⇒ ∠POR = ∠ROQ = 90°
∴ ∠POS + ∠ROS = 90°
⇒ ∠ROS = 90° – ∠POS
Adding ∠ROS to both sides, we have
∠ROS + ∠ROS = (90° + ∠ROS) – ∠POS
⇒ 2∠ROS = ∠QOS – ∠POS
⇒ ∠ROS = \(\frac{1}{2}\) (∠QOS – ∠POS).

Question 6 .
If two parallel lines are intersected by a transversal, then prove that the bisectors of any two corresponding angles are parallel.
Solution:
Given : AB || CD and transversal PQ meet these lines at E and F respectively. EG and FH are
the bisectors of pair of corresponding angles ∠PEB and ∠EFD.
To Prove : EG || FH Proof
∵ EG and FH are bisectors of ∠PEB respectively.

∠PEG = ∠EFH
Which are corresponding angles of EG and FH
∴ EG || FH.

Question 7.
In the given figure, m and n are two plane mirrors perpendicular to each other. Show that incident rays CA is parallel to reflected ray BD.

Solution:
Let normals at A and B meet at P.
As mirrors are perpendicular to each other, therefore, BP || OA and AP || OB.
So, BP ⊥ PA i.e., ∠BPA = 90°
Therefore, ∠3 + ∠2 = 90° [angle sum property] …(i)
Also, ∠1 = ∠2 and ∠4 = ∠3 [Angle of incidence = Angle of reflection]
Therefore, ∠1 + ∠4 = 90° [from (i)) …(ii]
Adding (i) and (ii), we have
∠1 + ∠2 + ∠3 + ∠4 = 180°
i.e., ∠CAB + ∠DBA = 180°
Hence, CA || BD

Lines and Angles Class 9 Extra Questions Long Answer Type

Question 1.
If two parallel lines are intersected by a transversal, prove that the bisectors of two pairs of interior angles form a rectangle.
Solution:
Given : AB || CD and transversal EF cut them at P and Q respectively and the bisectors of
pair of interior angles form a quadrilateral PRQS.

To Prove : PRQS is a rectangle.
Proof : ∵ PS, QR, QS and PR are the bisectors of angles
∠BPQ, ∠CQP, ∠DQP and ∠APQ respectively.
∴∠1 =\(\frac{1}{2}\) ∠BPQ, ∠2 = \(\frac{1}{2}\)∠CQP,
∠3 = \(\frac{1}{2}\)∠DQP and ∠4 = \(\frac{1}{2}\)∠APQ
Now, AB || CD and EF is a transversal
∴ ∠BPQ = ∠CQP
⇒ ∠1 = ∠2 (∵∠1 \(\frac{1}{2}\) ∠BPQ and ∠2 = \(\frac{1}{2}\)∠QP)
But these are pairs of alternate interior angles of PS and QR
∴ PS || QR
Similarly, we can prove ∠3 = ∠4 = QS || PR
∴ PRQS is a parallelogram.
Further ∠1 + ∠3 = \(\frac{1}{2}\)∠BPQ + \(\frac{1}{2}\)∠DQP = \(\frac{1}{2}\) (∠BPQ + ∠DQP)
= \(\frac{1}{2}\) × 180° = 90° (∵ ∠BPQ + ∠DQP = 180°)
∴ In ∆PSQ, we have ∠PSQ = 180° – (∠1 + ∠3) = 180° – 90° = 90°
Thus, PRRS is a parallelogram whose one angle ∠PSQ = 90°.
Hence, PRQS is a rectangle.

Question 2.
If in ∆ABC, the bisectors of ∠B and ∠C intersect each other at O. Prove that ∠BOC = 90° + \(\frac{1}{2}\)∠ A.

Solution:
Let ∠B = 2x and ∠C = 2y
∵OB and OC bisect ∠B and ∠C respectively.
∠OBC = \(\frac{1}{2}\)∠B = \(\frac{1}{2}\) × 2x = x
and ∠OCB = \(\frac{1}{2}\)∠C = \(\frac{1}{2}\) × 2y = y
Now, in ∆BOC, we have
∠BOC + ∠OBC + ∠OCB = 180°
⇒ ∠BOC + x + y = 180°
⇒ ∠BOC = 180° – (x + y)
Now, in ∆ABC, we have
∠A + 2B + C = 180°
⇒ ∠A + 2x + 2y = 180°
⇒ 2(x + y) = \(\frac{1}{2}\)(180° – ∠A)
⇒ x + y = 90° – \(\frac{1}{2}\)∠A …..(ii)
From (i) and (ii), we have
∠BOC = 180° – (90° – \(\frac{1}{2}\)∠A) = 90° + \(\frac{1}{2}\) ∠A

Question 3.
In figure, if I || m and ∠1 = (2x + y)°, ∠4 = (x + 2y)° and ∠6 = (3y + 20)°. Find ∠7 and ∠8.

Solution:
Here, ∠1 and ∠4 are forming a linear pair
∠1 + ∠4 = 180°
(2x + y)° + (x + 2y)° = 180°
3(x + y)° = 180°
x + y = 60
Since I || m and n is a transversal
∠4 = ∠6
(x + 2y)° = (3y + 20)°
x – y = 20
Adding (i) and (ii), we have
2x = 80 = x = 40
From (i), we have
40 + y = 60 ⇒ y = 20
Now, ∠1 = (2 x 40 + 20)° = 100°
∠4 = (40 + 2 x 20)° = 80°
∠8 = ∠4 = 80° [corresponding ∠s]
∠1 = ∠3 = 100° [vertically opp. ∠s]
∠7 = ∠3 = 100° [corresponding ∠s]
Hence, ∠7 = 100° and ∠8 = 80°

Question 4.
In the given figure, if PQ ⊥ PS, PQ || SR, ∠SQR = 28o and ∠QRT = 65°. Find the values of x, y and z.

Solution:
Here, PQ || SR .
⇒ ∠PQR = ∠QRT
⇒ x + 28° = 65°
⇒ x = 65° – 28° = 37°
Now, in it. ∆SPQ, ∠P = 90°
∴ ∠P + x + y = 180° [angle sum property]
∴ 90° + 37° + y = 180°
⇒ y = 180° – 90° – 37° = 53°
Now, ∠SRQ + ∠QRT = 180° [linear pair]
z + 65° = 180°
z = 180° – 65° = 115°

Question 5.
In figure, AP and DP are bisectors of two adjacent angles A and D of a quadrilateral ABCD. Prove that 2∠APD = ∠B + ∠C.

Solution:
In quadrilateral ABCD, we have
∠A + ∠B.+ ∠C + ∠D = 360°

Question 6.
In figure, PS is bisector of ∠QPR ; PT ⊥ RQ and Q > R. Show that ∠TPS = \(\frac{1}{2}\)(∠Q – ∠R).

Solution:
Since PS is the bisector of ∠QPR
∴∠QPS = ∠RPS = x (say)
In ∆PRT, we have
∠PRT + ∠PTR + ∠RPT = 180°
⇒ ∠PRT + 90° + ∠RPT = 180°
⇒ ∠PRT + ∠RPS + ∠TPS = 90°
⇒ ∠PRT + x + ∠TPS = 90°
⇒ ∠PRT or ∠R = 90° – ∠TPS – x
In ΔΡQT, we have
∠PQT + ∠PTQ + ∠QPT = 180°
⇒ ∠PQT + 90° + ∠QPT = 180°
⇒ ∠PQT + ∠QPS – TPS = 90°
⇒ ∠PQT + x – ∠TPS = 90° [∵∠QPS = x]
⇒ ∠PQT or ∠Q = 90° + ∠TPS – x …(ii)
Subtracting (i) from (ii), we have
⇒ ∠Q – ∠R = (90° + ∠TPS – x) – 190° – ∠TPS – x)
⇒ ∠Q – ∠R = 90° + ∠TPS – X – 90° + ∠TPS + x
⇒ 2∠TPS = 2Q- ∠R
⇒ ∠TPS = \(\frac{1}{2}\)(Q – ∠R)

Lines and Angles Class 9 Extra Questions HOTS

Question 1.
In the given figure, p ll q, find the value of x.

Solution:

Extend line p to meet RT at S.
Such that MS || QT
Now, in ARMS, we have
∠RMS = 180° – ∠PMR (linear pair]
= 180° – 120°
= 60°
∠RMS + ∠MSR + ∠SRM = 180° [by angle sum property of a ∆]
⇒ 60° + ∠MSR + 30o = 180°
⇒ MSR = 90°
Now, PS || QT – ∠MSR = ∠RTQ
⇒ ∠RTQ = x = MSR = 90° (corresponding ∠s]

Question 2.
In a triangle, if ∠A = 2∠B = 6∠C, find the measures of ∠A, ∠B and ∠C and find the value of \(\frac{\angle A+2 \angle B}{3 \angle C}\).
Solution:
Let us consider ∠A = 2∠B = 6∠C = x
∴∠A = x
2∠B = x = ∠B = \(\frac{1}{2}\)

Lines and Angles Class 9 Extra Questions Value Based (VBQs)

Question 1.
Students in a school are preparing banner for a rally to make people aware for saving electricity. What value are they exhibiting by doing so ? Parallel lines I and m are cut by transversal t, if ∠4 = ∠5 and ∠6 = ∠7, what is measure of angle 8 ?

Solution:
Here, given that ∠4 = ∠5 and ∠6 = ∠7
Now, I || m and t is a transversal
∴ ∠4 + ∠5 + ∠6 + ∠7 = 180° [∵ co-interior angles are supplementary]
∠5 + ∠5 + ∠6 + ∠6 = 180° [using (i)]
2(∠5 + ∠6) = 180°
∠5 + ∠6 = 90°
We know that, sum of all interior angles of a triangle is 180°
∴ ∠8 + ∠5 + ∠6 = 180°
⇒ ∠8 + 90° = 180° [using (ii)]
⇒ ∠8 = 180° – 90° = 90°
Save electricity, save energy.

Question 2.
To protect poor people from cold weather, Ram Lal. has given his land to make a shelter home for them. What value is being exhibited by him? In the given figure, ‘sides QP and RQ of ∆PQR are produced to point S and T respectively. If ∠PQT = 110° and ∠SPR = 135°, find ∠PRQ.

Solution:
Here,
∠SPR + ∠QPR = 180° [a linear pair]
135° + ∠QPR = 180° [∵ ∠SPR = 135°]∠
⇒ ∠QPR = 180° – 135° = 45°
In ∆PQR, by exterior angle property, we have
∠QPR + ∠PRQ = ∠PQT
45° + ∠PRQ = 110°
∠PRQ = 110° – 45° = 65°
Helpful nature, service of mankind and helping the needy people.

Question 3.
In an activity of mathematics, a teacher ask students to divide a circular sheet of radius 15 cm into 6 equal parts to write 6 values they like. What is the central angle subtended by each part ? Give the six values of your preference.
Solution:

Draw a circle of radius 15 cm and divide the circular region into six equal parts by constructing an angle of measure 60° at the centre.
Central angle subtended by each part
= \(\frac{360^{\circ}}{6}\) = 60°
The six values are : Honesty, Truth, Regularity, Tolerance, Punctuality and Sincerity.

Here we are providing Introduction to Euclid’s Geometry Class 9 Extra Questions Maths Chapter 5 with Answers Solutions, Extra Questions for Class 9 Maths was designed by subject expert teachers.

Extra Questions for Class 9 Maths Introduction to Euclid’s Geometry with Answers Solutions

Extra Questions for Class 9 Maths Chapter 5 Introduction to Euclid’s Geometry with Solutions Answers

Introduction to Euclid’s Geometry Class 9 Extra Questions Very Short Answer Type

Question 1.
Give a definition of parallel lines. Are there other terms that need to be defined first ? What are they and how might you define them?
Solution:
Two coplanar lines in a plane) which are not intersecting are called parallel lines. The other term intersecting is undefined.

Question 2.
Give a definition of perpendicular lines. Are there other terms that need to be defined first ? What are they and how might you define them?
Solution:
Two coplanar (in a plane) lines are perpendicular if the angle between them at the point
of intersection is one right angle. The other terms point of intersection and one right angle are undefined.

Question 3.
Give a definition of line segment. Are there other terms that need to be defined first ? What are they and how might you define them ?
Solution:
A line segment PQ of a line ‘l is the continuous part of the line I with end points P and Q.

Here, continuous part of the line ‘l is undefined.

Question 4.
Solve the equation a – 15 = 25 and state which axiom do you use here.
Solution:
a – 15 = 25
On adding 15 to both sides, we obtain
a – 15 + 15 = 25 + 15 [using Euclid’s second axiom]
a = 40

Question 5.
Ram and Ravi have the same weight. If they each gain weight by 2 kg, how will their new weights be compared ?
Solution:
Let x kg be the weight each of Ram and Ravi.
On adding 2 kg,
Weight of Ram and Ravi will be (x + 2) kg each.
According to Euclid’s second axiom, when equals are added to equals, the wholes are equal.
So, weight of Ram and Ravi are again equal.

Question 6.
If a point C be the mid-point of a line segment AB, then write the relation among AC, BC and AB.
Solution:
Here, C is the mid-point of AB
⇒ AC = BC
⇒ AC = BC = \(\frac{1}{2}\)AB

Question 7.
If a point P be the mid-point of MN and C is the mid-point of MP, then write the relation between MC and MN.
Solution:
Here, P is the mid-point of MN and C is the mid-point of MP.

∴ MC = \(\frac{1}{4}\)MN

Question 8.
How many lines does pass through two distinct points ?
Solution:
One and only one.

Question 9.
In the given figure, if AB = CD, then prove that AC = BD. Also, write the Euclid’s axiom used for proving it.

Solution:
Here, given that
AB = CD
By using Euclid’s axiom 2, if equals are added to equals, then the wholes are equal, we have
AB + BC = CD + BC
⇒ AC = BD

Introduction to Euclid’s Geometry Class 9 Extra Questions Short Answer Type 2

Question 1.
Define :
(a) a square (b) perpendicular lines.
Solution:
(a) A square : A square is a rectangle having same length and breadth. Here, undefined terms are length, breadth and rectangle.
(b) Perpendicular lines : Two coplanar (in a plane) lines are perpendicular, if the angle between them at the point of intersection is one right angle. Here, the term one right angle is undefined.

Question 2.
In the given figure, name the following :
(i) Four collinear points
(ii) Five rays
(iii) Five line segments
(iv) Two-pairs of non-intersecting line segments.

Solution:
(i) Four collinear points are D, E, F, G and H, I, J, K
(ii) Five rays are DG, EG, FG, HK, IK.
(iii) Five line segments are DH, EI, FJ; DG, HK.
(iv) Two-pairs of non-intersecting line segments are (DH, EI) and (DG, HK).

Question 3.
In the given figure, AC = DC and CB = CE. Show that AB = DE. Write the Euclid’s axiom to support this.

Solution:
We have
AC = DC
CB = CE
By using Euclid’s axiom 2, if equals are added to equals, then wholes are equal.
⇒ AC + CB = DC + CE
⇒ AB = DE.

Question 4.
In figure, it is given that AD=BC. By which Euclid’s axiom it can be proved that AC = BD?

Solution:
We can prove it by Euclid’s axiom 3. “If equals are subtracted from equals, the remainders are equal.”
We have AD = BC
⇒ AD – CD = BC – CD
⇒ AC = BD

Question 4.
If A, B and C are any three points on a line and B lies between A and C (see figure), then prove that AB + BC = AC.

Solution:
In the given figure, AC coincides with AB + BC. Also, Euclid’s axiom 4, states that things
which coincide with one another are equal to one another. So, it is evident that:
AB + BC = AC.

Question 5.
In the given figure, AB = BC, BX = BY, show that
AX = CY.

Solution:
Given that AB = BC
and BX = BY
By using Euclid’s axiom 3, equals subtracted from equals, then the remainders are equal, we have
AB – BX = BC – BY
AX = CY

Question 6.

In the above figure, if AB = PQ, PQ = XY, then AB = XY. State True or False. Justify your answer.
Solution:
True. ∵ By Euclid’s first axiom “Things which are equal to the same thing are equal to one another”.
∴ AB = PQ and XY = PQ ⇒ AB = XY

Question 7.
In the given figure, if ∠1 = ∠3, ∠2 = ∠4 and ∠3 = ∠4, write the relation between ∠1 and ∠2, using an Euclid’s axiom.

Solution:
Here, ∠3 = ∠4, ∠1 = ∠3 and ∠2 = ∠4. Euclid’s first axiom says, the things which are equal to equal thing are equal to one another. So ∠1 = ∠2.,

Question 8.
In the given figure, we have ∠1 = ∠2, ∠3 = ∠4. Show that ∠ABC = ∠DBC. State the Euclid’s Axiom used.
Solution:
Here, we have 1 = ∠2 and ∠3 = ∠4. By using Euclid’s Axiom 2. If equals are added to
equals, then the wholes are equal..
∠1 + ∠3 = ∠2 + ∠4
∠ABC = ∠DBC.

Question 9.
In the figure, we have BX and \(\frac{1}{2}\) AB =\(\frac{1}{2}\) BC. Show that BX = BY.

Solution:
Here, BX = \(\frac{1}{2}\) AB and BY = \(\frac{1}{2}\) BC …(i) [given]
Also, AB = BC [given]
⇒ \(\frac{1}{2}\)AB = \(\frac{1}{2}\)BC …(ii)
[∵ Euclid’s seventh axiom says, things which are halves of the same thing are equal to one another]
From (i) and (ii), we have BX = BY

Question 10.
In the given figure, AC = XD, C is mid-point of AB and D is mid-point of XY. Using an Euclid’s axiom, show that AB = XY.

Solution:
∵ C is the mid-point of AB
AB = 2AC
Also, D is the mid-point of XY
XY = 2XD
By Euclid’s sixth axiom “Things which are double of same things are equal to one another.”
∴ AC = XD = 2AC = 2XD ⇒ AB = XY

Introduction to Euclid’s Geometry Class 9 Extra Questions HOTS

Question 1.
For given four distinct points in a plane, find the number of lines that can be drawn through :
(i) When all four points are collinear.
(ii) When three of the four points are collinear.
(iii) When no three of the four points are collinear.

Solution:

(i) Consider the points given are A, B, C and D.
When all the four points are collinear :
One line \(\overrightarrow{\mathrm{AD}}\).

(ii) When three of the four points are collinear :
4 lines
Here, we have four lines \(\stackrel{\leftrightarrow}{A B}, \stackrel{\leftrightarrow}{B C}, \stackrel{\leftrightarrow}{B D}, \stackrel{\leftrightarrow}{A D}\) (four).

(iii) When no three of the four points are collinear :
6 lines Here, we have
\(\stackrel{\leftrightarrow}{A B}, \stackrel{\leftrightarrow}{B C}, \stackrel{\leftrightarrow}{A C}, \stackrel{\leftrightarrow}{A D}, \stackrel{\leftrightarrow}{B D}, \stackrel{\leftrightarrow}{C D}\) (six).

Question 2.
Show that : length AH > sum of lengths of AB + BC + CD.

Solution:
We have
AH = AB + BC + CD + DE + EF + FG + GH
Clearly, AB + BC + CD is a part of AH.
⇒ AH > AB + BC + CD
Hence, length AH > sum of lengths AB + BC + CD.

Introduction to Euclid’s Geometry Class 9 Extra Questions Value Based (VBQs)

Question 1.
Rohan’s maid has two children of same age. Both of them have equal number of dresses. Rohan on his birthday plans to give both of them same number of dresses. What can you say about the number of dresses each one of them will have after Rohan’s birthday? Which Euclid’s axiom is used to answer this question? What value is Rohan depicting by doing so ? Write one more Euclid’s axiom.
Solution:
Here, Rohan’s maid has two children of same age group and both of them have equal number of dresses. Rohan on his birthday plans to give both of them same number of dresses.
∴ By using Euclid’s Axiom 2, if equals are added to equals, then the whole are equal. Thus, again both of them have equal number of dresses. Value depicted by Rohan are caring and other social values. According to Euclid’s Axiom 3, if equals are subtracted from equals, then the remainders
are equal.

Question 2.
Three lighthouse towers are located at points A, B and C on the section of a national forest to protect animals from hunters by the forest department as shown in figure. Which value is department exhibiting by locating extra towers ? How many straight lines can be drawn from A to C? State the Euclid Axiom which states the required result. Give one more. Postulate.

Solution:
One and only one line can be drawn from A to C. According to Euclid’s Postulate, “A straight line may be drawn from any point to any other point:” An other postulate : “A circle may be described with any centre and any radius.” Wildlife is a part of our environment and conservation of each of its element is important for ecological balance.