Q. 134.0( 21 Votes )

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

Answer :

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 [1]


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


And


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