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
