Question

A triangle $ABC$ 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 $\Delta ABC$ is $54c{m}^2$ , find the lengths of sides $AB$ and $AC$ .

Answer 1

Since $BE$ and $BD$ are tangents from same point, therefore $BD=BE=6cm$.

Similarly, $CD=CF=9cm$ and $AE=AF=xcm$ (say)

Now,

$ar\Delta ABC=ar\Delta AOB+ar\Delta BOC+ar\Delta AOC$

$\Rightarrow 54=\frac{1}{2}\times OE\times AB+\frac{1}{2}\times OD\times BC+\frac{1}{2}\times OF\times AC$

$\Rightarrow 54=\frac{1}{2}[3\times (x+6)+3\times (6+9)+3\times (x+9\left)\right]$

$\Rightarrow 54=\frac{1}{2}\times 3\times \left[\right(x+6)+(6+9)+(x+9\left)\right]$

$\Rightarrow 54=\frac{1}{2}\times 3\times [2x+30]$

$\Rightarrow 54=3\times [x+15]$

$\Rightarrow 18=x+15$

$\therefore x=3cm$

Hence, $AB=6+3=9cm$ and $AC=9+3=12cm$