Question

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

• A. $4\sqrt{2}c{m}^2$
• B. $8\sqrt{3}c{m}^2$
• C. $4c{m}^2$
• D. $8c{m}^2$

Answer 1

Answer: (B)

$8\sqrt{3}c{m}^2$

Given: $\Delta ABC$ is isosceles inscribed in a circle of radius $9cm$ & center O.

$AB=AC=6cm$

To find out-$Area\left(\Delta ABC\right)$

Solution-

We join AO which intersects BC at P.

In $\Delta ABC$ it is given that $AB=BC$

$⟹\angle ABC=\angle ACB$ ....(i)

Now, in $\Delta ABP$ & $\Delta ACP$ we have,

$AB=AC$ ...(given),

side $AP$ common

And, $\angle ABC=\angle ACB$ ...(by i)

$\therefore \Delta ABP$ & $\Delta ACP$ are congruent.

i.e $\angle APC=\angle APB.$

But they are adjacent angles.

So, $\angle APC=\angle APB={90}^o$.......(ii)

and $BP=PC⟹BP=\frac{1}{2}\times BC$ ...........(iii).

$BP\perp AO$ .....(iv)

Also, $\Delta ABP$ is a right one with hypotenuse AB .........(v)
Let $AP=xcm$ and $PB=PC=ycm$ then

From Pythagorean theorem, we have $A{B}^2=A{P}^2+B{P}^2$
$\Rightarrow 36=x^2+y^2$.........(vi)

And also $\Delta BPO$ is also a right angled triangle, so $B{O}^2=P{B}^2+O{P}^2$
$\Rightarrow 81=y^2+(9-x{)}^2$
$\Rightarrow 81=y^2+81-18x+x^2$........(vii)
Subtracting equation (vi) from equation(vii), we get
$\Rightarrow 81-36=y^2+81-18x+x^2-x^2-y^2$
$\Rightarrow 81-81-36=-18x$
$\Rightarrow 36=18x$
$\therefore x=2cm$
Thus, $y^2=36-2^2$
$=36-4$
$=32$
$\therefore y=4\sqrt{2}cm$
In $\Delta ABC$, Base=BC=2(PB)=2y=2$\left(4\sqrt{2}\right)=8\sqrt{2}cm$
and Height=$x=2cm$
Therefore, area =$\frac{1}{2}\times 8\sqrt{2}\times 2$
$=8\sqrt{2}$sq. cm.