Question

A $△ABC$ is drawn to circumscribe a circle of radius $3$ cm. such that the segments $BD$ and $DC$ are respectively of length $6$ cm and $9$ cm. If the area of $△ABC$ is $54{cm}^2$, then find the lengths of sides $AB$ and $AC$

Answer 1

Area of $△ABC=$area of $△OBC+$area of $△OAC+$area of $△OAB$

We have $BD=6$cm, $BE=6$cm(equal tangents)

$DC=9$cm, $CF=9$cm(equal tangents)

Area of $△OBC=\frac{1}{2}\times b\times h=\frac{1}{2}\times 15\times 3=\frac{45}{2}{cm}^2$

Area of $△OAC=\frac{1}{2}\times (x+9)\times 3=\frac{3(x+9)}{2}{cm}^2$

Area of $△OAB=\frac{1}{2}\times (x+6)\times 3=\frac{3(x+6)}{2}{cm}^2$

By Heron's formula,area of $△ABC=\sqrt{s(s-a)(s-b)(s-c)}{cm}^2$

$s=\frac{x+9+x+6+15}{2}=\frac{2x+30}{2}=x+15$

$\therefore \Delta =\sqrt{(x+15)(x+15-15)(x-15-x-4)(x+15-x-6)}=\sqrt{x(x+15)\times 6\times 9}$

$\Rightarrow \sqrt{54x(x+15)}=\frac{3(x+9)}{2}+\frac{3(x+16)}{2}+\frac{45}{2}$

$\Rightarrow \sqrt{54x(x+15)}=\frac{3}{2}[x+9+x+6+15]$

$\Rightarrow \sqrt{54x(x+15)}=\frac{3}{2}(2x+30)$

$\Rightarrow \sqrt{54x(x+15)}=3x+15$

Squaring on both sides we get

$\Rightarrow 54x(x+15)=9{(x+15)}^2$

$\Rightarrow 6x=x+15$

$\Rightarrow 6x-x=15$

$\Rightarrow 5x=15$

$\therefore x=\frac{15}{5}=3$cm

Hence the sides are $15$cm, $12$cm and $9$cm