# If an isosceles ΔABC in which AB = AC = 6cm, is inscribed in a circle of radius 9 cm, find the area of the triangle.

Given : In a circle, ΔABC is inscribed in which AB = AC = 6 cm and radius of circle is 9 cm

To Find : Area of triangle ABC

Construction : Join OB and OC In AOB and AOC

OB = OC [Radii of same circle]

AO = AO [Common]

AB = AC = 6 cm [Given]

AOB AOC [By Side Side Side Criterion]

OAB = OBC [Corresponding parts of congruent triangles are equal ]

Implies MAB = MBC

Now in ABM and AMC

AB = AC = 6 cm [Given]

AM = AM [Common]

MAB = MBC [Proved Above]

ABM AMC [By Side Angle Side Criterion]

AMB = AMC [Corresponding parts of congruent triangles are equal ]

Now,

AMB + AMC = 180°

AMB + AMB = 180°

2 ∠AMB = 180

AMB = 90°

We know that a perpendicular from center of circle bisects the chord.

So, OA is perpendicular bisector of BC.

Let OM = x

Then, AM = OA - OM = 9 - x

[ As OA = radius = 9 cm]

In right angled ΔAMC, By Pythagoras theorem

(AM)2 + (MC)2 = (AC)2

(9 - x)2 + (MC)2 = (6)2

81 + x2- 18x + (MC)2 = 36

(MC)2 = 18x - x2 - 45 

In OMC , By Pythagoras Theorem

(MC)2 + (OM)2 = (OC)2

18x - x2 - 45 + (x)2 = (9)2

18x - 45 = 81

18x = 126

x = 7

AM = 9 - x = 9 - 7 = 2 cm

In AMC, By Pythagoras Theorem

(AM)2 + (MC)2 = (AC)2

(2)2+ (MC)2 = (6)2

4 + (MC)2= 36

(MC)2 = 32

MC = 4√2 cm

Now,

As MC = BM

BC = 2MC = 2(4√2) = 8√2 cm

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