Frank Solutions for Class 9 Maths Chapter 12 Isosceles triangle provides comprehensive answers with the intention to help students in solving the most tricky questions with ease. These solutions are helpful in clearing their doubts quickly, which arise while revising for the examination. For more conceptual knowledge, students can refer to Frank Solutions as a valuable aid. Download Frank Solutions for Class 9 Maths Chapter 12 Isosceles triangle from the below provided links.

Chapter 12 consists of problems based on the Isosceles triangle. Students who want to come out with flying colours in exams are suggested to practise Frank Solutions diligently. The aim of preparing the solutions is to boost self-confidence among students in solving complex problems effortlessly.

## Frank Solutions for Class 9 Maths Chapter 12 Isosceles triangle Download PDF

## Access Frank Solutions for Class 9 Maths Chapter 12 Isosceles triangle

**1. Find the angles of an isosceles triangle whose equal angles and the non-equal angles are in the ratio 3:4.**

**Solution:**

Given that,

The equal angles and the non-equal angles are in the ratio 3:4

Let the equal angles be 3x each,

So, non- equal angle is 4x

We know that,

Sum of angles of a triangle = 180^{0}

Hence,

3x + 3x + 4x = 180^{0}

10x = 180^{0}

We get,

x = 18^{0}

Therefore,

3x = 3 Ã— 18^{0} = 54^{0} and

4x = 4 Ã— 18^{0} = 72^{0}

Hence,

The angles of a triangle are 54^{0}, 54^{0} and 72^{0}

**2. Find the angles of an isosceles triangle which are in the ratio 2:2:5**

**Solution:**

The equal angles and the non-equal angle are in the ratio 2:2:5

Let equal angles be 2x each

So, non-equal angle is 5x

We know that,

Sum of angles of a triangle = 180^{0}

2x + 2x + 5x = 180^{0}

9x = 180^{0}

x = 20^{0}

Therefore,

2x = 2 Ã— 20^{0} = 40^{0}

5x = 5 Ã— 20^{0} = 100^{0}

Hence, the angles of a triangle are 40^{0}, 40^{0} and 100^{0}

**3. Each equal angle of an isosceles triangle is less than the third angle by 15 ^{0}. Find the angles.**

**Solution:**

Let equal angles of the isosceles triangle be x each

Therefore, non-equal angle is x + 15^{0}

We know that,

Sum of angles of a triangle = 180^{0}

x + x + (x + 15^{0}) = 180^{0}

3x + 15^{0} = 180^{0}

3x = 180^{0} – 15^{0}

3x = 165^{0}

We get,

x = 55^{0}

So,

(x + 15^{0}) = 55^{0} + 15^{0} = 70^{0}

Hence, the angles of a triangle are 55^{0}, 55^{0} and 70^{0}

**4. Find the interior angles of the following triangles**

**(a)**

** **

**(b) **

** **

**(c) **

**(d)**

**Solution:**

(a)

** **

In â–³ABC,

âˆ A = 110^{0}

AB = AC

âˆ C = âˆ B (angles opposite to two equal sides are equal)

Now,

By angle sum property,

âˆ A + âˆ B + âˆ C = 180^{0}

âˆ A + âˆ B + âˆ B = 180^{0}

110^{0} + 2âˆ B = 180^{0}

2âˆ B = 180^{0} – 110^{0}

2âˆ B = 70^{0}

We get,

âˆ B = 35^{0}

âˆ C = 35^{0}

Hence,

The interior angles are âˆ B = 35^{0} and âˆ C = 35^{0}

(b)

In â–³ABC,

AB = AC

âˆ ACB = âˆ ABC â€¦â€¦â€¦(1) [âˆµ angles opposite to two equal sides are equal]

Now,

âˆ ACB + âˆ ACD = 180^{0} [linear pair]

âˆ ACB = 180^{0} â€“ âˆ ACD

âˆ ACB = 180^{0} – 105^{0}

âˆ ACB = 75^{0}

So,

âˆ ABC = 75^{0} [from equation (1)]

Now, in â–³ABC,

By angle sum property,

âˆ ABC + âˆ ACB + âˆ BAC = 180^{0}

75^{0} + 75^{0} + âˆ BAC = 180^{0}

150^{0} + âˆ BAC = 180^{0}

âˆ BAC = 180^{0} – 150^{0}

We get,

âˆ BAC = 30^{0}

Hence,

In â–³ABC,

âˆ A = 30^{0}

âˆ B = 75^{0}

âˆ C = 75^{0}

(c)

In â–³ABD,

Given that,

AD = BD

âˆ ABD = âˆ BAD â€¦ (angles opposite to two equal sides are equal)

Now,

âˆ ABD = 37^{0} â€¦. (given)

Hence,

âˆ BAD = 37^{0}

By exterior angle property,

âˆ ADC = âˆ ABD + âˆ BAD

âˆ ADC = 37^{0} + 37^{0}

We get,

âˆ ADC = 74^{0}

In â–³ADC,

AC = DC â€¦. (given)

âˆ ADC = âˆ DAC â€¦. (angles opposite to two equal sides are equal)

âˆ DAC = 74^{0}

Now,

âˆ BAC = âˆ BAD + âˆ DAC

âˆ BAC = 37^{0} + 74^{0}

We get,

âˆ BAC = 111^{0}

In â–³ABC,

âˆ BAC + âˆ ABC + âˆ ACB = 180^{0}

111^{0} + 37^{0} + âˆ ACB = 180^{0}

âˆ ACB = 180^{0} – 111^{0} – 37^{0}

We get,

âˆ ACB = 32^{0}

Therefore,

The interior angles of â–³ABC are 37^{0}, 111^{0} and 32^{0}

(d)

In â–³ACD,

AD = CD â€¦â€¦ (given)

âˆ ACD = âˆ CAD â€¦ (angles opposite to two equal sides are equal)

Now,

âˆ ACD = 50^{0} â€¦â€¦ (given)

âˆ CAD = 50^{0}

By exterior angle property,

âˆ ADB = âˆ ACD + âˆ CAD

âˆ ADB = 50^{0} + 50^{0}

âˆ ADB = 100^{0}

In â–³ADB,

AD = BD â€¦â€¦ (given)

âˆ DBA = âˆ DAB â€¦.. (angles opposite to two equal sides are equal)

Also,

âˆ ADB + âˆ DBA + âˆ DAB = 180^{0}

100^{0} + 2âˆ DBA = 180^{0}

2âˆ DBA = 180^{0} â€“ 100^{0}

2âˆ DBA = 80^{0}

We get,

âˆ DBA = 40^{0}

âˆ DAB = 40^{0}

âˆ BAC = âˆ DAB + âˆ CAD

âˆ BAC = 40^{0} + 50^{0}

âˆ BAC = 90^{0}

Therefore, the interior angles of â–³ABC are 50^{0}, 90^{0} and 40^{0}

**5. Side BA of an isosceles triangle ABC is produced so that AB = AD. If AB and AC are the equal sides of the isosceles triangle, prove that âˆ BCD is a right angle.**

**Solution:**

Let âˆ ABC = x

Hence,

âˆ BCA = x (since AB = AC)

In â–³ABC,

âˆ ABC + âˆ BCA + âˆ BAC = 180^{0} â€¦â€¦. (1)

But

âˆ BAC + âˆ DAC = 180^{0} â€¦â€¦. (2)

From equations (1) and (2)

âˆ ABC + âˆ BCA + âˆ BAC = âˆ BAC + âˆ DAC

âˆ DAC = âˆ ABC + âˆ BCA

âˆ DAC = x + x

We get,

âˆ DAC = 2x

Let âˆ ADC = y,

Hence,

âˆ DCA = y (since AD = AC)

Now,

In â–³ADC,

âˆ ADC + âˆ DCA + âˆ DAC = 180^{0} â€¦ (3)

But âˆ BAC + âˆ DAC = 180^{0} â€¦â€¦. (4)

From equations (3) and (4), we get,

âˆ ADC + âˆ DCA + âˆ DAC = âˆ BAC + âˆ DAC

âˆ BAC = âˆ ADC + âˆ DCA

âˆ BAC = y + y

âˆ BAC = 2y

Now, substituting the value of âˆ BAC and âˆ DAC in equation (2)

2x + 2y = 180^{0}

x + y = 90^{0}

âˆ BCA + âˆ DCA = 90^{0}

Therefore,

âˆ BCD is a right angle

**6. The bisectors of the equal angles of an isosceles triangle PQR meet at O. If PQ = PR, prove that PO bisects âˆ P.**

**Solution:**

Join PO and produce to meet QR at point S

In â–³PQS and â–³PRS

PS = PS (common)

PQ = PR (given)

So,

âˆ Q = âˆ R (angles opposite to two equal sides are equal)

Hence,

â–³PQS â‰…Â â–³PRS

Thus,

âˆ QPS = âˆ RPS

Therefore,

PO bisects âˆ P

**7. Prove that the medians corresponding to equal sides of an isosceles triangle are equal.**

**Solution:**

Let â–³ABC be an isosceles triangle with AB = AC

Let D and E be the mid points of AB and AC respectively

Now,

Join BE and CD

Then BE and CD become the medians of this isosceles triangle

In â–³ABE and â–³ACD

AB = AC (given)

AD = AE (D and E are mid points of AB and AC)

âˆ A = âˆ A (common angle)

Hence,

â–³ABE â‰… â–³ACD (SAS criteria)

Therefore,

The medians BE and CD are equal i.e BE = CD

**8. DPQ is an isosceles triangle with DP = DQ. A straight line CD bisects the exterior âˆ QDR. Prove that DC is parallel to PQ**

**Solution:**

In â–³QDP,

DP = DQ

Hence,

âˆ Q = âˆ P (angles opposite to two equal sides are equal)

âˆ QDR = âˆ Q + âˆ P

2âˆ QDC = âˆ Q + âˆ P (DC bisects angle QDR)

2âˆ QDC = âˆ Q + âˆ Q

We get,

2âˆ QDC = 2âˆ Q

Hence,

âˆ QDC = âˆ Q

But these angles are alternate angles

Therefore,

DC || PQ

Hence, proved

**9. In a quadrilateral PQRS, PQ = PS and RQ = RS. If âˆ P = 50 ^{0} and âˆ R = 110^{0}, find âˆ PSR.**

**Solution:**

In â–³PQS,

PQ = PS

Therefore,

âˆ PQS = âˆ PSQ (angles opposite to two equal sides are equal)

âˆ P + âˆ PQS + âˆ PSQ = 180^{0}

50^{0} + 2âˆ PQS = 180^{0}

2âˆ PQS = 180^{0} – 50^{0}

We get,

2âˆ PQS = 130^{0}

âˆ PQS = 65^{0}

So,

âˆ PQS = âˆ PSQ = 65^{0} â€¦â€¦.. (1)

In â–³SRQ,

SR = RQ

Hence,

âˆ RQS = âˆ RSQ (angles opposite to two equal sides are equal)

âˆ R + âˆ RQS + âˆ RSQ = 180^{0}

110^{0} + 2âˆ RQS = 180^{0}

2âˆ RQS = 180^{0} – 110^{0}

We get,

2âˆ RQS = 70^{0}

âˆ RQS = 35^{0}

So,

âˆ RQS = âˆ RSQ = 35^{0} â€¦.. (2)

Adding equations (1) and (2), we get,

âˆ PSQ + âˆ RSQ = 65^{0} + 35^{0}

âˆ PSR = 100^{0}

**10. â–³ABC is an isosceles triangle with AB = AC. Another triangle BDC is drawn with base BC = BD in such a way that BC bisects âˆ B. If the measure of âˆ BDC is 70 ^{0}, find the measures of âˆ DBC and âˆ BAC.**

**Solution:**

In â–³BDC,

âˆ BDC = 70^{0}

BD = BC

Hence,

âˆ BDC = âˆ BCD (angles opposite to two equal sides are equal)

âˆ BCD = 70^{0}

Now,

âˆ BCD + âˆ BDC + âˆ DBC = 180^{0}

70^{0} + 70^{0} + âˆ DBC = 180^{0}

âˆ DBC = 180^{0} – 140^{0}

We get,

âˆ DBC = 40^{0}

âˆ DBC = âˆ ABC (BC is the angle bisector)

Hence,

âˆ ABC = 40^{0}

In â–³ABC,

Since AB = AC, âˆ ABC = âˆ ACB

Hence,

âˆ ACB = 40^{0}

âˆ ACB + âˆ ABC + âˆ BAC = 180^{0}

40^{0} + 40^{0} + âˆ BAC = 180^{0}

âˆ BAC = 180^{0} – 80^{0}

âˆ BAC = 100^{0}

Therefore, the measure of âˆ BAC = 100^{0} and âˆ DBC = 40^{0}

**11. â–³PQR is isosceles with PQ = PR. T is the mid-point of QR, and TM and TN are perpendiculars on PR and PQ respectively. Prove that,**

**(i) TM = TN**

**(ii) PM = PN and **

**(iii) PT is the bisector of âˆ P**

**Solution:**

(i) In â–³PQR,

PQ = PR

Hence,

âˆ R = âˆ Q â€¦â€¦ (1)

Now,

In â–³QNT and â–³RMT

âˆ QNT = âˆ RMT = 90^{0}

âˆ Q = âˆ R [from equation (1)]

QT = TR (given)

Hence,

â–³QNT â‰… â–³RMT (AAS criteria)

Therefore,

TM = TN

(ii) Since, â–³QNT â‰… â–³RMT

NQ = MR â€¦â€¦â€¦ (2)

But,

PQ = PR â€¦.. (3) [given]

Now, subtracting (2) from (3), we get,

PQ â€“ NQ = PR – MR

PN = PM

(iii) In â–³PNT and â–³PMT

TN = TM (proved)

PT = PT (common)

âˆ PNT = âˆ PMT = 90^{0}

Hence,

â–³PNT â‰… â–³PMT

So,

âˆ NPT = âˆ MPT

Therefore,

PT is the bisector of âˆ P

**12. â–³PQR is isosceles with PQ = QR. QR is extended to S so that â–³PRS becomes isosceles with PR = PS. Show that âˆ PSR: âˆ QPS = 1:3 **

**Solution:**

In â–³PQR,

PQ = QR (given)

âˆ PRQ = âˆ QPR â€¦â€¦ (1)

In â–³PRS,

PR = RS (given)

âˆ PSR = âˆ RPS â€¦â€¦.. (2)

Now,

Adding equations (1) and (2), we get,

âˆ QPR + âˆ RPS = âˆ PRQ + âˆ PSR

âˆ QPS = âˆ PRQ + âˆ PSR â€¦â€¦.. (3)

Now,

In â–³PRS,

âˆ PRQ = âˆ RPS + âˆ PSR

âˆ PRQ = âˆ PSR + âˆ PSR [from equation(2)]

âˆ PRQ = 2âˆ PSR â€¦â€¦ (4)

Now,

âˆ QPS = 2âˆ PSR + âˆ PSR [from equation (3) and (4)]

âˆ QPS = 3âˆ PSR

âˆ PSR / âˆ QPS = 1 / 3

Therefore,

âˆ PSR: âˆ QPS = 1: 3

Hence, proved

**13. In â–³KLM, KT bisects âˆ LKM and KT = TM. If âˆ LTK is 80 ^{0}, find the value of âˆ LMK and âˆ KLM.**

**Solution:**

In â–³KTM,

KT = TM (given)

Hence,

âˆ TKM = âˆ TMK â€¦â€¦. (1)

Now,

âˆ KTL = âˆ TKM + âˆ TMK

80^{0} = âˆ TKM + âˆ TKM â€¦â€¦.. [from (1)]

80^{0} = 2âˆ TKM

We get,

âˆ TKM = 40^{0} = âˆ TMK = âˆ LMK â€¦â€¦ (2)

But,

âˆ TKM = âˆ TKL (KT is the angle bisector)

Hence,

âˆ TKL = 40^{0}

In â–³KTL,

âˆ TKL + âˆ KTL + âˆ KLT = 180^{0}

40^{0} + 80^{0} + âˆ KLT = 180^{0}

âˆ KLT = 180^{0} â€“ 40^{0} – 80^{0}

We get,

âˆ KLT = 60^{0} = âˆ KLM

Therefore,

âˆ KLM = 60^{0} and âˆ LMK = 40^{0}

**14. Equal sides QP and RP of an isosceles â–³PQR are produced beyond P to S and T such that â–³PST is an isosceles triangle with PS = PT. Prove that TQ = SR.**

**Solution:**

In** **â–³PTQ and â–³PSR

PQ = PR (given)

PT = PS (given)

âˆ TPQ = âˆ SPR (vertically opposite angles)

Hence,

â–³PTQ â‰… â–³PSR

Therefore,

TQ = SR

Hence, proved

**15. Prove that the bisector of the vertex angle of an isosceles triangle bisects the base perpendicularly.**

**Solution:**

In â–³ADB and â–³ADC

AB = AC (given)

AD = AD (common)

âˆ BAD = âˆ CAD (AD bisects âˆ BAC)

Hence,

â–³ADB â‰… â–³ADC

Therefore,

BD = DC and âˆ BDA = âˆ CDA

But,

âˆ BDA + âˆ CDA = 180^{0}

âˆ BDA = âˆ CDA = 90^{0}

Therefore,

AD bisects BC perpendicularly

Hence, proved