Frank Solutions for Class 9 Maths Chapter 11 Triangles has answers in a step by step manner, as per the studentâ€™s intelligence quotient. Students, who aspire to speed up their problem solving skills, are suggested to use these solutions on a daily basis. The solutions also help them to excel with good marks in the examination. Students can access Frank Solutions for Class 9 Maths Chapter 11 Triangles and their congruency, from the links available below.

Chapter 11 has problems based on determining the congruent triangles. Students can make use of these solutions to clear their doubts during the self study process. The main purpose of providing the solutions is to help students in boosting their exam preparation, as per the current ICSE board.

## Frank Solutions for Class 9 Maths Chapter 11 Triangles and their Congruency Download PDF

## 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 = x^{0}

âˆ R = 2x^{0}

Now,

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

105^{0} = x^{0} + 2x^{0}

105^{0} = 3x^{0}

We get,

x^{0} = 35^{0}

Therefore,

âˆ Q = x^{0} = 35^{0} and

âˆ R = 2x^{0} = 70^{0}

**2. The exterior angles, obtained on producing the side of a triangle both ways, are 100 ^{0} and 120^{0}. Find all the angles of the triangle.**

**Solution:**

âˆ ABP + âˆ ABC = 180^{0} (Linear pair)

100^{0} + âˆ ABC = 180^{0}

âˆ ABC = 180^{0} â€“ 100^{0}

âˆ ABC = 80^{0}

âˆ ACQ + âˆ ACB = 180^{0} (Linear pair)

120^{0} + âˆ ACB = 180^{0}

âˆ ACB = 180^{0} – 120^{0}

âˆ ACB = 60^{0}

Now,

In â–³ABC,

âˆ A + âˆ B + âˆ C = 180^{0} (Angle sum property of a triangle)

âˆ A + 80^{0} + 60^{0} = 180^{0}

âˆ A = 180^{0} â€“ 80^{0} – 60^{0}

We get,

âˆ A = 40^{0}

Therefore, the angles of a triangle are 40^{0}, 60^{0} and 80^{0}

**3. Use the given figure to find the value of x in terms of y. Calculate x, if y = 15 ^{0}. **

**Solution:**

(2x â€“ y)^{0} = (x + 5^{0}) + (2y + 25)^{0} (Exterior angle property)

2x^{0} â€“ y^{0} = x^{0} + 5^{0} + 2y^{0} + 25^{0}

2x^{0} â€“ x^{0} = 2y^{0} + y^{0} + 30^{0}

x^{0} = 3y^{0} + 30^{0}

When y = 15^{0},

We have,

x^{0} = 3 Ã— 15^{0} + 30^{0}

x^{0} = 45^{0} + 30^{0}

We get,

x^{0} = 75^{0}

**4. In a triangle PQR, âˆ P + âˆ Q = 130 ^{0} and âˆ P + âˆ R = 120^{0}. Calculate each angle of the triangle.**

**Solution:**

Given

In â–³PQR,

âˆ P + âˆ Q = 130^{0}

WKT

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

âˆ PRY = 130^{0}

âˆ PRY + âˆ R = 180^{0} (Linear pair)

130^{0} + âˆ R = 180^{0}

âˆ R = 180^{0} – 130^{0}

We get,

âˆ R = 50^{0}

Also,

Given

âˆ P + âˆ R = 120^{0}

Now,

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

âˆ PQX = 120^{0}

âˆ PQX + âˆ Q = 180^{0} (Linear pair)

120^{0} + âˆ Q = 180^{0}

âˆ Q = 180^{0} â€“ 120^{0}

We get,

âˆ Q = 60^{0}

In â–³PQR,

âˆ P + âˆ Q + âˆ R = 180^{0} (Angle sum property of a triangle)

âˆ P + 60^{0} + 50^{0} = 180^{0}

âˆ P = 180^{0} â€“ 110^{0}

We get,

âˆ P = 70^{0}

Therefore, the angles of â–³PQR are,

âˆ P = 70^{0}

âˆ Q = 60^{0} and

âˆ R = 50^{0}

**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 = 180^{0}

Hence,

We have,

(x + 10)^{0} + (x + 30)^{0} + (x â€“ 10)^{0} = 180^{0}

x^{0} + 10^{0} + x^{0} + 30^{0} + x^{0} â€“ 10^{0} = 180^{0}

3x^{0} + 30^{0} = 180^{0}

3x^{0} = 180^{0} – 30^{0}

We get,

3x^{0} = 150^{0}

x^{0} = 50^{0}

Now,

(x + 10)^{0} = (50 + 10)^{0}

(x + 10)^{0} = 60^{0}

(x + 30)^{0} = (50 + 30)^{0}

(x + 30)^{0} = 80^{0}

(x â€“ 10)^{0} = (50 â€“ 10)^{0}

(x â€“ 10)^{0} = 40^{0}

Therefore, the angles of a triangle are 60^{0}, 80^{0} and 40^{0}

**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 = 180^{0} â€¦â€¦ (Angle sum property of a triangle)

p^{0} + y^{0} + âˆ PES = 180^{0}

âˆ PES = 180^{0} â€“ p^{0} â€“ y^{0} â€¦.. (1)

In â–³RQE,

âˆ R + âˆ Q + âˆ REQ = 180^{0} â€¦â€¦.. (Angle sum property of a triangle)

(180^{0} â€“ q^{0}) + r^{0} + âˆ REQ = 180^{0}

âˆ REQ = 180^{0} â€“ (180^{0} â€“ q^{0}) â€“ r^{0}

âˆ REQ = q^{0} â€“ r^{0} â€¦â€¦â€¦.(2)

Now,

âˆ PES + âˆ REQ = 180^{0} â€¦â€¦â€¦â€¦ (Linear pair)

(180^{0} â€“ p^{0} â€“ y^{0}) + (q^{0} â€“ r^{0}) = 180^{0} â€¦ [from (1) and (2)]

-p^{0} â€“ y^{0} + q^{0} â€“ r^{0} = 0

– y^{0} = -q^{0} + p^{0} + r^{0}

We get,

y^{0} = q^{0} â€“ p^{0} â€“ r^{0}

**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 = 180^{0} â€¦.. (angle sum property)

4x^{0} + 5x^{0} + 9x^{0} = 180^{0}

18x^{0} = 180^{0}

x = 10

âˆ P = 4x^{0} = 4 Ã— 10^{0}

âˆ P = 40^{0}

âˆ Q = 5x^{0} = 5 Ã— 10^{0}

âˆ Q = 50^{0}

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

And,

âˆ QPR = 40^{0}

âˆ PRS = 40^{0}

By exterior angle property,

âˆ PQR + âˆ QPR = âˆ PRS + y^{0}

40^{0} + 50^{0} = 40^{0} + y^{0}

We get,

y = 50^{0}

**8. Use the given figure to show that: âˆ p + âˆ q + âˆ r = 360 ^{0}**

**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— 180^{0} (Angle Sum property: âˆ PQR + âˆ PRQ + âˆ QPR = 180^{0})

We get,

âˆ p + âˆ q + âˆ r = 360^{0}

Hence,

âˆ p + âˆ q + âˆ r = 360^{0}

**9. In â–³ABC and â–³PQR and, 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 = 180^{0}

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 = 180^{0}

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 = 180^{0}

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 = 90^{0}

âˆ BRM = âˆ DRN (vertically opposite angles)

Hence,

âˆ MBR = âˆ NDR (Sum of angles of a triangle = 180^{0})

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:**

In â–³CAD and â–³CBE

CA = CB (Isosceles triangle)

âˆ CDA = âˆ CEB = 90^{0}

âˆ ACD = âˆ BCE (common)

Therefore,

â–³CAD â‰… â–³CBA (AAS criteria)

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