Question

In Figure $5$, a $△ABC$ is drawn to circumscribe a circle of radius $3cm$, such that the segments $BD$ and $DC$ are respectively of lengths $6cm$ and $9cm$. If the area of $△ABC$ is $54c{m}^2$, then find the lengths of sides $AB$ and $AC$.

Answer 1

Given $OD=3cm$

$BD$$=BE=6cm$ (equal tangent)

$DC=CF=9cm$ (equal tangent)

Area of triangle $ABC=54{cm}^2$ (Given)

Area of triangle $OBC=$ $\frac{1}{2}\times 15\times 3=\frac{45}{2}c{m}^2$

Area of triangle $OAC=$$\frac{1}{2}(x+9)\times 3=\frac{3(x+9)}{2}c{m}^2$

Area of triangle $OAB=$$\frac{1}{2}(x+6)\times 3=\frac{3(x+6)}{2}c{m}^2$

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

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

$\Rightarrow 54=3(x+15)$

$\Rightarrow 17=x+15$

$\Rightarrow x=2$

Then side are $x+9=2+9=11,x+6=2+6=8$

So sides of triangle $ABC$ are $11,8$ and $15$.