# Frank Solutions for Class 9 Maths Chapter 12 Isosceles triangle

Frank Solutions for Class 9 Maths Chapter 12 Isosceles Triangle provide comprehensive answers with the intention of helping students in solving the most tricky questions with ease. These solutions are helpful in clearing their doubts quickly while revising for the final examination. To obtain in-depth conceptual knowledge, students can practise Frank Solutions on a daily basis.

Chapter 12 consists of problems based on the Isosceles Triangle. Students who want to come out with flying colours in the annual exams are suggested to practise Frank Solutions regularly. The main aim of preparing these solutions is to boost the self-confidence of students in solving complex problems effortlessly. Download Frank Solutions for Class 9 Maths Chapter 12 Isosceles Triangle from the link provided below.

## 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,

The sum of angles of a triangle = 1800

Hence,

3x + 3x + 4x = 1800

10x = 1800

We get,

x = 180

Therefore,

3x = 3 Ã— 180 = 540 and

4x = 4 Ã— 180 = 720

Hence,

The angles of a triangle are 540, 540 and 720

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, the non-equal angle is 5x

We know that,

The sum of angles of a triangle = 1800

2x + 2x + 5x = 1800

9x = 1800

x = 200

Therefore,

2x = 2 Ã— 200 = 400

5x = 5 Ã— 200 = 1000

Hence, the angles of a triangle are 400, 400 and 1000

3. Each equal angle of an isosceles triangle is less than the third angle by 150. Find the angles.

Solution:

Let equal angles of the isosceles triangle be x each

Therefore, the non-equal angle is x + 150

We know that,

The sum of angles of a triangle = 1800

x + x + (x + 150) = 1800

3x + 150 = 1800

3x = 1800 – 150

3x = 1650

We get,

x = 550

So,

(x + 150) = 550 + 150 = 700

Hence, the angles of a triangle are 550, 550 and 700

4. Find the interior angles of the following triangles

(a)

(b)

(c)

(d)

Solution:

(a)

In â–³ABC,

âˆ A = 1100

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 = 1800

âˆ A + âˆ B + âˆ B = 1800

1100 + 2âˆ B = 1800

2âˆ B = 1800 – 1100

2âˆ B = 700

We get,

âˆ B = 350

âˆ C = 350

Hence,

The interior angles are âˆ B = 350 and âˆ C = 350

(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 = 1800 [linear pair]

âˆ ACB = 1800 â€“ âˆ ACD

âˆ ACB = 1800 – 1050

âˆ ACB = 750

So,

âˆ ABC = 750 [from equation (1)]

Now, in â–³ABC,

By angle sum property,

âˆ ABC + âˆ ACB + âˆ BAC = 1800

750 + 750 + âˆ BAC = 1800

1500 + âˆ BAC = 1800

âˆ BAC = 1800 – 1500

We get,

âˆ BAC = 300

Hence,

In â–³ABC,

âˆ A = 300

âˆ B = 750

âˆ C = 750

(c)

In â–³ABD,

Given that,

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

Now,

âˆ ABD = 370 â€¦. (given)

Hence,

By exterior angle property,

âˆ ADC = 370 + 370

We get,

AC = DC â€¦. (given)

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

âˆ DAC = 740

Now,

âˆ BAC = âˆ BAD + âˆ DAC

âˆ BAC = 370 + 740

We get,

âˆ BAC = 1110

In â–³ABC,

âˆ BAC + âˆ ABC + âˆ ACB = 1800

1110 + 370 + âˆ ACB = 1800

âˆ ACB = 1800 – 1110 – 370

We get,

âˆ ACB = 320

Therefore,

The interior angles of â–³ABC are 370, 1110 and 320

(d)

In â–³ACD,

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

Now,

âˆ ACD = 500 â€¦â€¦ (given)

By exterior angle property,

âˆ ADB = 500 + 500

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

Also,

âˆ ADB + âˆ DBA + âˆ DAB = 1800

1000 + 2âˆ DBA = 1800

2âˆ DBA = 1800 â€“ 1000

2âˆ DBA = 800

We get,

âˆ DBA = 400

âˆ DAB = 400

âˆ BAC = âˆ DAB + âˆ CAD

âˆ BAC = 400 + 500

âˆ BAC = 900

Therefore, the interior angles of â–³ABC are 500, 900 and 400

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 = 1800 â€¦â€¦. (1)

But

âˆ BAC + âˆ DAC = 1800 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

Hence,

âˆ DCA = y (since AD = AC)

Now,

âˆ ADC + âˆ DCA + âˆ DAC = 1800 â€¦ (3)

But âˆ BAC + âˆ DAC = 1800 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 = 1800

x + y = 900

âˆ BCA + âˆ DCA = 900

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 midpoints 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 midpoints 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 = 500 and âˆ R = 1100, 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 = 1800

500 + 2âˆ PQS = 1800

2âˆ PQS = 1800 – 500

We get,

2âˆ PQS = 1300

âˆ PQS = 650

So,

âˆ PQS = âˆ PSQ = 650 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 = 1800

1100 + 2âˆ RQS = 1800

2âˆ RQS = 1800 – 1100

We get,

2âˆ RQS = 700

âˆ RQS = 350

So,

âˆ RQS = âˆ RSQ = 350 â€¦.. (2)

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

âˆ PSQ + âˆ RSQ = 650 + 350

âˆ PSR = 1000

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 700, find the measures of âˆ DBC and âˆ BAC.

Solution:

In â–³BDC,

âˆ BDC = 700

BD = BC

Hence,

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

âˆ BCD = 700

Now,

âˆ BCD + âˆ BDC + âˆ DBC = 1800

700 + 700 + âˆ DBC = 1800

âˆ DBC = 1800 – 1400

We get,

âˆ DBC = 400

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

Hence,

âˆ ABC = 400

In â–³ABC,

Since AB = AC, âˆ ABC = âˆ ACB

Hence,

âˆ ACB = 400

âˆ ACB + âˆ ABC + âˆ BAC = 1800

400 + 400 + âˆ BAC = 1800

âˆ BAC = 1800 – 800

âˆ BAC = 1000

Therefore, the measure of âˆ BAC = 1000 and âˆ DBC = 400

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 = 900

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 = 900

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 800, 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

800 = âˆ TKM + âˆ TKM â€¦â€¦.. [from (1)]

800 = 2âˆ TKM

We get,

âˆ TKM = 400 = âˆ TMK = âˆ LMK â€¦â€¦ (2)

But,

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

Hence,

âˆ TKL = 400

In â–³KTL,

âˆ TKL + âˆ KTL + âˆ KLT = 1800

400 + 800 + âˆ KLT = 1800

âˆ KLT = 1800 â€“ 400 – 800

We get,

âˆ KLT = 600 = âˆ KLM

Therefore,

âˆ KLM = 600 and âˆ LMK = 400

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:

AB = AC (given)

Hence,

Therefore,

BD = DC and âˆ BDA = âˆ CDA

But,

âˆ BDA + âˆ CDA = 1800

âˆ BDA = âˆ CDA = 900

Therefore,