Question

An isosceles triangle with equal side 6 cm inscribed in a circle with radius 9 cm then find area of triangle

Answer 1

As shown in the figure, let O be the centre of the circle.OA bisects chord BC at M.

Let AM = x cm, then OM = (9-x)cm

From right ∆AMB,we have

= $\sf{{BM}^{2} = {OB}^{2} - {AM}^{2} }$

= $\sf{{6}^{2}-{x}^{2}}$

From right ∆OMB,we have

= $\sf{{BM}^{2} = {OB}^{2} - {OM}^{2} }$

= $\sf{ {9}^{2} - {(9 - x)}^{2} }$

From (i) and (ii), we get

=$\sf{18x - {x}^{2} = 36 - {x}^{2}}$

=$\sf{x = \frac{36}{18} = 2cm \: or \: AM = 2cm}$

Also, BM = $\sf{\sqrt{ {6}^{2} - {2}^{2} } = \sqrt{32} = 4 \sqrt{2} cm}$

$\therefore$ $\sf{BC = 2BM = 8 \sqrt{2} cm}$

Area of ∆ABC

= $\sf{\frac{1}{2}BC×AM=\frac{1}{2}×8√2×2}$

= $\sf{{8 \sqrt{2}cm }^{2} }$