Question

If an isosceles $△ABC$ in which $AB=AC=6$ cm is inscribed in a circle of radius $9$cm. Find area of the triangle.

• A. $8c{m}^2$
• B. $8\sqrt{2}c{m}^2$
• C. $6c{m}^2$
• D. none

Answer 1

Answer: (B)

$8\sqrt{2}c{m}^2$

Let $AP=xcm.$

$PO=(9-x)cm$

In $△APC$

$A{P}^2=36-P{C}^2$

$\Rightarrow AP=\sqrt{36-P{C}^2}$

Similarly, $P{C}^2=36=x^2$

In $△OPC$

$\Rightarrow O{C}^2=O{P}^2+P{C}^2$

$\Rightarrow g^2=(g-x{)}^2+36-x^2$

$\Rightarrow g^2=81+x^2-18x+36-x^2$

$\Rightarrow 18x=36$

$\Rightarrow x=2cm$

$\therefore PC=\sqrt{36-x^2}=\sqrt{36-4}$

$=\sqrt{32}=4\sqrt{2}$

$\therefore BC=2\times PC=8\sqrt{2}cm$

$\because a\perp$ from centre of circle on any cheared bisects the chord.

Area$=\frac{1}{2}\times BC\times AP$

$=\frac{1}{2}\times 8\sqrt{2}\times 2$

$=\frac{1}{2}\times 8\sqrt{2}\times 2=8\sqrt{2}c{m}^2$