Question

$A△ABC$ is drawn to circumscribe a circle of radius $4cm.$ such that the segment $BD$ and $DC$ in which $BC$ is divided by the point of contact $D$ are of contact $D$ are of length $8cm$ and $6cm$ respectively. Find the sides $AB$.

Answer 1

REF. Image.

To find: length of side AB

BD = BE = 8cm ; CD = CF = 6 cm ; AE = AF = x

[Length of tangents drawn from an external point to the same circle are equal in length]

arc (ABC) = arc (AOB) + arc (BOC) + arc (AOC)

semi - perimeter $=\frac{28+2x}{2}=14+x$

using Heuon's formula, area of $\Delta$ ABC

$\sqrt{s(s-a)(s-b)(s-c)}$ where a,b and c are sides of $\Delta$

are (ABC)=$\sqrt{(14+x)\left(x\right)\left(6\right)\left(8\right)}$

$\sqrt{(14+x)\left(x\right)\left(48\right)}=\frac{1}{2}\left[\right(14\left)\right(4\left)\right]+\frac{1}{2}\left(4\right)(6+x)+\frac{1}{2}\left(4\right)(8+x)$

$\sqrt{(14+x)\left(x\right)\left(48\right)}=28+12+2x+16+2x$

$\sqrt{\left(x\right)\left(48\right)(14+x)}=56+4x$

Squaring Both Sides

$48x(14+x)=16(x^2+196+28x)$

$42x+3x^2=x^2+28x+196$

$2x^2+14x-196=0$

$x^2+7x-98=0$

By Splitting Middle Team.

$x^2+14x-7x-98=0$

$x(x+14)-7(x+14)=0$

x = 7 and $x\ne -14$ ( As length can't be negative)

Length of AB = 15 cm