# Frank Solutions for Class 9 Maths Chapter 11 Triangles and Their Congruency

Frank Solutions for Class 9 Maths Chapter 11 Triangles and Their Congruency contain answers in a step-by-step manner, as per the studentâ€™s understanding abilities. Students who aspire to enhance their problem-solving skills are suggested to practise these solutions on a daily basis, which will help them excel in the annual examination with good marks.

Chapter 11 has problems based on determining congruent triangles, as per the current ICSE Board syllabus. Students can refer to these solutions to clear their doubts during the self-study process. The main purpose of providing the solutions is to help students boost their exam preparation. They can download Frank Solutions for Class 9 Maths Chapter 11 Triangles and Their Congruency from the link provided below.

## Access Frank Solutions for Class 9 Maths Chapter 11 Triangles and Their Congruency

1. In the given figure, âˆ Q: âˆ R = 1: 2. Find:

(a) âˆ Q

(b) âˆ R

Solution:

Given

âˆ Q: âˆ R = 1: 2

Let us consider âˆ Q = x0

âˆ R = 2x0

Now,

âˆ RPX = âˆ Q + âˆ R [by exterior angle property]

1050 = x0 + 2x0

1050 = 3x0

We get,

x0 = 350

Therefore,

âˆ Q = x0 = 350 and

âˆ R = 2x0 = 700

2. The exterior angles, obtained on producing the side of a triangle both ways, are 1000 and 1200. Find all the angles of the triangle.

Solution:

âˆ ABP + âˆ ABC = 1800 (Linear pair)

1000 + âˆ ABC = 1800

âˆ ABC = 1800 â€“ 1000

âˆ ABC = 800

âˆ ACQ + âˆ ACB = 1800 (Linear pair)

1200 + âˆ ACB = 1800

âˆ ACB = 1800 – 1200

âˆ ACB = 600

Now,

In â–³ABC,

âˆ A + âˆ B + âˆ C = 1800 (Angle sum property of a triangle)

âˆ A + 800 + 600 = 1800

âˆ A = 1800 â€“ 800 – 600

We get,

âˆ A = 400

Therefore, the angles of a triangle are 400, 600 and 800

3. Use the given figure to find the value of x in terms of y. Calculate x, if y = 150.

Solution:

(2x â€“ y)0 = (x + 50) + (2y + 25)0 (Exterior angle property)

2x0 â€“ y0 = x0 + 50 + 2y0 + 250

2x0 â€“ x0 = 2y0 + y0 + 300

x0 = 3y0 + 300

When y = 150,

We have,

x0 = 3 Ã— 150 + 300

x0 = 450 + 300

We get,

x0 = 750

4. In a triangle PQR, âˆ P + âˆ Q = 1300 and âˆ P + âˆ R = 1200. Calculate each angle of the triangle.

Solution:

Given

In â–³PQR,

âˆ P + âˆ Q = 1300

WKT

âˆ P + âˆ Q = âˆ PRY (Exterior angle property)

âˆ PRY = 1300

âˆ PRY + âˆ R = 1800 (Linear pair)

1300 + âˆ R = 1800

âˆ R = 1800 – 1300

We get,

âˆ R = 500

Also,

Given

âˆ P + âˆ R = 1200

Now,

âˆ P + âˆ R = âˆ PQX (Exterior angle property)

âˆ PQX = 1200

âˆ PQX + âˆ Q = 1800 (Linear pair)

1200 + âˆ Q = 1800

âˆ Q = 1800 â€“ 1200

We get,

âˆ Q = 600

In â–³PQR,

âˆ P + âˆ Q + âˆ R = 1800 (Angle sum property of a triangle)

âˆ P + 600 + 500 = 1800

âˆ P = 1800 â€“ 1100

We get,

âˆ P = 700

Therefore, the angles of â–³PQR are,

âˆ P = 700

âˆ Q = 600 and

âˆ R = 500

5. The angles of a triangle are (x + 10)0, (x + 30)0 and (x â€“ 10)0. Find the value of â€˜xâ€™. Also, find the measure of each angle of the triangle.

Solution:

For any triangle,

Sum of measures of all three angles = 1800

Hence,

We have,

(x + 10)0 + (x + 30)0 + (x â€“ 10)0 = 1800

x0 + 100 + x0 + 300 + x0 â€“ 100 = 1800

3x0 + 300 = 1800

3x0 = 1800 – 300

We get,

3x0 = 1500

x0 = 500

Now,

(x + 10)0 = (50 + 10)0

(x + 10)0 = 600

(x + 30)0 = (50 + 30)0

(x + 30)0 = 800

(x â€“ 10)0 = (50 â€“ 10)0

(x â€“ 10)0 = 400

Therefore, the angles of a triangle are 600, 800 and 400

6. Use the given figure to find the value of y in terms of p, q and r

Solution:

Here, SR is produced to meet PQ at point E

In â–³PSE,

âˆ P + âˆ S + âˆ PES = 1800 â€¦â€¦ (Angle sum property of a triangle)

p0 + y0 + âˆ PES = 1800

âˆ PES = 1800 â€“ p0 â€“ y0 â€¦.. (1)

In â–³RQE,

âˆ R + âˆ Q + âˆ REQ = 1800 â€¦â€¦.. (Angle sum property of a triangle)

(1800 â€“ q0) + r0 + âˆ REQ = 1800

âˆ REQ = 1800 â€“ (1800 â€“ q0) â€“ r0

âˆ REQ = q0 â€“ r0 â€¦â€¦â€¦.(2)

Now,

âˆ PES + âˆ REQ = 1800 â€¦â€¦â€¦â€¦ (Linear pair)

(1800 â€“ p0 â€“ y0) + (q0 â€“ r0) = 1800 â€¦ [from (1) and (2)]

-p0 â€“ y0 + q0 â€“ r0 = 0

– y0 = -q0 + p0 + r0

We get,

y0 = q0 â€“ p0 â€“ r0

7. In the figure given below, if RS is parallel to PQ, then find the value of âˆ y.

In â–³PQR,

âˆ P + âˆ Q + âˆ R = 1800 â€¦.. (angle sum property)

4x0 + 5x0 + 9x0 = 1800

18x0 = 1800

x = 10

âˆ P = 4x0 = 4 Ã— 100

âˆ P = 400

âˆ Q = 5x0 = 5 Ã— 100

âˆ Q = 500

âˆ QPR = âˆ PRS â€¦â€¦. (Alternate angles)

And,

âˆ QPR = 400

âˆ PRS = 400

By exterior angle property,

âˆ PQR + âˆ QPR = âˆ PRS + y0

400 + 500 = 400 + y0

We get,

y = 500

8. Use the given figure to show that: âˆ p + âˆ q + âˆ r = 3600

Solution:

By exterior angle property,

âˆ p = âˆ PQR + âˆ PRQ

âˆ q = âˆ QPR + âˆ PRQ

âˆ r = âˆ PQR + âˆ QPR

Now,

âˆ p + âˆ q + âˆ r = âˆ PQR + âˆ PRQ + âˆ QPR + âˆ PRQ + âˆ PQR + âˆ QPR

On further calculation, we get,

âˆ p + âˆ q + âˆ r = 2 âˆ PQR + 2âˆ PRQ + 2 âˆ QPR

âˆ p + âˆ q + âˆ r = 2 (âˆ PQR + âˆ PRQ + âˆ QPR)

âˆ p + âˆ q + âˆ r = 2 Ã— 1800 (Angle Sum property: âˆ PQR + âˆ PRQ + âˆ QPR = 1800)

We get,

âˆ p + âˆ q + âˆ r = 3600

Hence,

âˆ p + âˆ q + âˆ r = 3600

9. In â–³ABC and â–³PQR, AB = PQ, BC = QR and CB and RQ are extended to X and Y, respectively and âˆ ABX = âˆ PQY. Prove that â–³ABC â‰… â–³PQR.

Solution:

In â–³ABC and â–³PQR

AB = PQ

BC = QR

âˆ ABX + âˆ ABC = âˆ PQY + âˆ PQR = 1800

So,

âˆ ABX = âˆ PQY

âˆ ABC = âˆ PQR

Therefore,

â–³ABC â‰… â–³PQR (SAS criteria)

Hence, proved

10. In the figure, âˆ CPD = âˆ BPD and AD is the bisector of âˆ BAC. Prove that â–³CAP â‰… BAP and CP = BP.

Solution:

In â–³BAP and â–³CAP

âˆ BAP = âˆ CAP (AD is the bisector of âˆ BAC)

AP = AP

âˆ BPD + âˆ BPA = âˆ CPD + âˆ CPA = 1800

We get,

âˆ BPD = âˆ CPD

âˆ BPA = âˆ CPA

Therefore,

â–³CAP â‰… â–³BAP (ASA criteria)

So,

CP = BP

Hence, proved

11. In the figure, BC = CE and âˆ 1 = âˆ 2. Prove that â–³GCB â‰… â–³DCE.

Solution:

In â–³GCB and â–³DCE

âˆ 1 + âˆ GBC = âˆ 2 + âˆ DEC = 1800

âˆ 1 = âˆ 2

âˆ GBC = âˆ DEC

So,

BC = CE

âˆ GCB = âˆ DCE (vertically opposite angles)

Therefore,

â–³GCB â‰… â–³DCE (ASA criteria)

Hence, proved.

12. In the figure, AB = EF, BC = DE, AB and FE are perpendicular on BE. Prove that â–³ABD â‰… â–³FEC

Solution:

Given that,

In â–³ABD and â–³FEC

AB = FE and

BD = CE (âˆµ BC = DE; CD is common)

Therefore,

âˆ B = âˆ E

â–³ABD â‰… â–³FEC (SAS criteria)

Hence, proved

13. In the figure, BM and DN are both perpendiculars on AC and BM = DN. Prove that AC bisects BD.

Solution:

In â–³BMR and â–³DNR

BM = DN

âˆ BMR = âˆ DNR = 900

âˆ BRM = âˆ DRN (vertically opposite angles)

Hence,

âˆ MBR = âˆ NDR (Sum of angles of a triangle = 1800)

â–³BMR â‰… â–³DNR (ASA criteria)

Therefore,

BR = DR

So,

AC bisects BD

Hence, proved

14. AD and BE are altitudes of an isosceles triangle ABC with AC = BC. Prove that AE = BD.

Solution:

CA = CB (Isosceles triangle)

âˆ CDA = âˆ CEB = 900

âˆ ACD = âˆ BCE (common)

Therefore,

Hence,

CE = CD

But,

CA = CB

AE + CE = BD + CD

AE = BD

Hence, proved

15. In â–³ABC, X and Y are two points on AB and AC such that AX = AY. If AB = AC, prove that CX = BY.

Solution:

In â–³ABC

AB = AC

AX = AY

BX = CY

In â–³BXC and â–³CYB

BX = CY

BC = BC

âˆ B = âˆ C â€¦â€¦ (Angles opposite to equal sides are equal)

Therefore,

â–³BXC â‰… â–³CYB (SAS criteria)

So,

CX = BY

Hence, proved