Question

A triangle $ABC$ is drawn to circumscribe a circle of radius $4cm$ such that the segments $BD$ and $DC$ into which $BC$ is divided by the point of contact $D$ are of the lengths $8cm$ and $6cm$ respectively. Find the sides $AB$ and $AC$.

Answer 1

Let the circle touch the side AB and AC of the triangle at point E and F.

In $△ABC$,

$CF=CD=6$ cm ....Tangents from an external point to a circle are equal

$BE=BD=8$cm ....Tangents from an external point to a circle are equal

$AE=AF=y$ ....Tangents from an external point to a circle are equal

Now, $AB=AE+EB$ ....$[\because A-E-B]$

$\therefore AB=(x+8)$ cm

Similarly, $BC=BD+DC=14$ cm

and $CA=CF+FA=(6+x)$ cm

Perimeter of $\left(\Delta ABC\right)=AB+BC+AC$

$=x+8+14+x+6$

$=2x+28$

$s=\frac{2x+28}{2}=x+14$

By Heron's formula,

Area of $\Delta ABC=\sqrt{s(s-a)(s-b)(s-c)}$

$A\left(\Delta ABC\right)=\sqrt{(x+14)\left[\right(x+14)-14]\left[\right(x+14)-(6+x\left)\right]\left[\right(x+14)-(8+x\left)\right]}$

$=\sqrt{(x+14)x\times 8\times 6}$

$=4\sqrt{3(x^2+14x)}$

Area of $\Delta OBC=\frac{1}{2}\times OD\times BC=\frac{1}{2}\times 4\times 14=28$

Area of $\Delta OCA=\frac{1}{2}\times OF\times AC=\frac{1}{2}\times 4(x+6)=2x+12$

Area of $\Delta OAB=\frac{1}{2}\times OE\times AB=\frac{1}{2}\times 4(x+8)=2x+16$

$A\left(\Delta ABC\right)=A\left(\Delta OBC\right)+A\left(\Delta OCA\right)+A\left(\Delta OAB\right)$

$\therefore 4\sqrt{3(x^2+14x)}=28+2x+12+2x+16$

$\Rightarrow \sqrt{3(x^2+14x)}=x+14$

Squaring on both sides we get,

$\Rightarrow 3(x^2+14x)=(x+14{)}^2$

$\Rightarrow 3x=x+14$

$\Rightarrow 2x=14$

$\Rightarrow x=7$

So, $AB=x+8=15$ cm.

and $AC=x+6=13$ cm.