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°

∠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 + x^{2}- 18x + (MC)^{2} = 36

(MC)^{2} = 18x - x^{2} - 45 [1]

In △OMC , By Pythagoras Theorem

(MC)^{2} + (OM)^{2} = (OC)^{2}

18x - x^{2} - 45 + (x)^{2} = (9)^{2}

18x - 45 = 81

18x = 36

x = 2

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

In △AMC, By Pythagoras Theorem

(AM)^{2} + (MC)^{2} = (AC)^{2}

(7)^{2}+ (MC)^{2} = (6)^{2}

49 + (MC)^{2}= 36

(MC)^{2} = 25

MC = 5 cm

Now,

As MC = BM

BC = 2MC = 2(5) = 10 cm

And

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