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 student’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.

Frank Solutions for Class 9 Maths Chapter 11 Triangles and Their Congruency Download Free PDF

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

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

(a) ∠Q

(b) ∠R

Solution:

Given

∠Q: ∠R = 1: 2

Let us consider ∠Q = x⁰

∠R = 2x⁰

Now,

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

105⁰ = x⁰ + 2x⁰

105⁰ = 3x⁰

We get,

x⁰ = 35⁰

Therefore,

∠Q = x⁰ = 35⁰ and

∠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:

∠ABP + ∠ABC = 180⁰ (Linear pair)

100⁰ + ∠ABC = 180⁰

∠ABC = 180⁰ – 100⁰

∠ABC = 80⁰

∠ACQ + ∠ACB = 180⁰ (Linear pair)

120⁰ + ∠ACB = 180⁰

∠ACB = 180⁰ – 120⁰

∠ACB = 60⁰

Now,

In △ABC,

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

∠A + 80⁰ + 60⁰ = 180⁰

∠A = 180⁰ – 80⁰ – 60⁰

We get,

∠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 – y)⁰ = (x + 5⁰) + (2y + 25)⁰ (Exterior angle property)

2x⁰ – y⁰ = x⁰ + 5⁰ + 2y⁰ + 25⁰

2x⁰ – x⁰ = 2y⁰ + y⁰ + 30⁰

x⁰ = 3y⁰ + 30⁰

When y = 15⁰,

We have,

x⁰ = 3 × 15⁰ + 30⁰

x⁰ = 45⁰ + 30⁰

We get,

x⁰ = 75⁰

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

Solution:

Given

In △PQR,

∠P + ∠Q = 130⁰

WKT

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

∠PRY = 130⁰

∠PRY + ∠R = 180⁰ (Linear pair)

130⁰ + ∠R = 180⁰

∠R = 180⁰ – 130⁰

We get,

∠R = 50⁰

Also,

Given

∠P + ∠R = 120⁰

Now,

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

∠PQX = 120⁰

∠PQX + ∠Q = 180⁰ (Linear pair)

120⁰ + ∠Q = 180⁰

∠Q = 180⁰ – 120⁰

We get,

∠Q = 60⁰

In △PQR,

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

∠P + 60⁰ + 50⁰ = 180⁰

∠P = 180⁰ – 110⁰

We get,

∠P = 70⁰

Therefore, the angles of △PQR are,

∠P = 70⁰

∠Q = 60⁰ and

∠R = 50⁰

5. The angles of a triangle are (x + 10)⁰, (x + 30)⁰ and (x – 10)⁰. Find the value of ‘x’. 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 – 10)⁰ = 180⁰

x⁰ + 10⁰ + x⁰ + 30⁰ + x⁰ – 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 – 10)⁰ = (50 – 10)⁰

(x – 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 △PSE,

∠P + ∠S + ∠PES = 180⁰ …… (Angle sum property of a triangle)

p⁰ + y⁰ + ∠PES = 180⁰

∠PES = 180⁰ – p⁰ – y⁰ ….. (1)

In △RQE,

∠R + ∠Q + ∠REQ = 180⁰ …….. (Angle sum property of a triangle)

(180⁰ – q⁰) + r⁰ + ∠REQ = 180⁰

∠REQ = 180⁰ – (180⁰ – q⁰) – r⁰

∠REQ = q⁰ – r⁰ ……….(2)

Now,

∠PES + ∠REQ = 180⁰ ………… (Linear pair)

(180⁰ – p⁰ – y⁰) + (q⁰ – r⁰) = 180⁰ … [from (1) and (2)]

-p⁰ – y⁰ + q⁰ – r⁰ = 0

– y⁰ = -q⁰ + p⁰ + r⁰

We get,

y⁰ = q⁰ – p⁰ – r⁰

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

In △PQR,

∠P + ∠Q + ∠R = 180⁰ ….. (angle sum property)

4x⁰ + 5x⁰ + 9x⁰ = 180⁰

18x⁰ = 180⁰

x = 10

∠P = 4x⁰ = 4 × 10⁰

∠P = 40⁰

∠Q = 5x⁰ = 5 × 10⁰

∠Q = 50⁰

∠QPR = ∠PRS ……. (Alternate angles)

And,

∠QPR = 40⁰

∠PRS = 40⁰

By exterior angle property,

∠PQR + ∠QPR = ∠PRS + y⁰

40⁰ + 50⁰ = 40⁰ + y⁰

We get,

y = 50⁰

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

Solution:

By exterior angle property,

∠p = ∠PQR + ∠PRQ

∠q = ∠QPR + ∠PRQ

∠r = ∠PQR + ∠QPR

Now,

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

On further calculation, we get,

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

∠p + ∠q + ∠r = 2 (∠PQR + ∠PRQ + ∠QPR)

∠p + ∠q + ∠r = 2 × 180⁰ (Angle Sum property: ∠PQR + ∠PRQ + ∠QPR = 180⁰)

We get,

∠p + ∠q + ∠r = 360⁰

Hence,

∠p + ∠q + ∠r = 360⁰

9. In △ABC and △PQR, AB = PQ, BC = QR and CB and RQ are extended to X and Y, respectively and ∠ABX = ∠PQY. Prove that △ABC ≅ △PQR.

Solution:

In △ABC and △PQR

AB = PQ

BC = QR

∠ABX + ∠ABC = ∠PQY + ∠PQR = 180⁰

So,

∠ABX = ∠PQY

∠ABC = ∠PQR

Therefore,

△ABC ≅ △PQR (SAS criteria)

Hence, proved

10. In the figure, ∠CPD = ∠BPD and AD is the bisector of ∠BAC. Prove that △CAP ≅ BAP and CP = BP.

Solution:

In △BAP and △CAP

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

AP = AP

∠BPD + ∠BPA = ∠CPD + ∠CPA = 180⁰

We get,

∠BPD = ∠CPD

∠BPA = ∠CPA

Therefore,

△CAP ≅ △BAP (ASA criteria)

So,

CP = BP

Hence, proved

11. In the figure, BC = CE and ∠1 = ∠2. Prove that △GCB ≅ △DCE.

Solution:

In △GCB and △DCE

∠1 + ∠GBC = ∠2 + ∠DEC = 180⁰

∠1 = ∠2

∠GBC = ∠DEC

So,

BC = CE

∠GCB = ∠DCE (vertically opposite angles)

Therefore,

△GCB ≅ △DCE (ASA criteria)

Hence, proved.

12. In the figure, AB = EF, BC = DE, AB and FE are perpendicular on BE. Prove that △ABD ≅ △FEC

Solution:

Given that,

In △ABD and △FEC

AB = FE and

BD = CE (∵ BC = DE; CD is common)

Therefore,

∠B = ∠E

△ABD ≅ △FEC (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 △BMR and △DNR

BM = DN

∠BMR = ∠DNR = 90⁰

∠BRM = ∠DRN (vertically opposite angles)

Hence,

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

△BMR ≅ △DNR (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:

In △CAD and △CBE

CA = CB (Isosceles triangle)

∠CDA = ∠CEB = 90⁰

∠ACD = ∠BCE (common)

Therefore,

△CAD ≅ △CBA (AAS criteria)

Hence,

CE = CD

But,

CA = CB

AE + CE = BD + CD

AE = BD

Hence, proved

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

Solution:

In △ABC

AB = AC

AX = AY

BX = CY

In △BXC and △CYB

BX = CY

BC = BC

∠B = ∠C …… (Angles opposite to equal sides are equal)

Therefore,

△BXC ≅ △CYB (SAS criteria)

So,

CX = BY

Hence, proved