Question

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

Answer 1

$CE=CD=6cm$

[$\because$ length of tangent drawn from the external point are equal]

$BF=BD=8cm$

$AE=AF=x$

$CB=14cm=c$

$AC=AE+EC=(x+6)cm=a$

$AB=AF+FC=(x+8)cm=b$

Area of $\Delta ABC$ by Heron's formula

$\sqrt{s(s-a)(s-b)(s-c)}$

$s=\frac{a+b+c}{2}=\frac{x+6+x+8+14}{2}=\frac{2(x+14)}{2}=x+14$

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

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

$=\sqrt{48x^2+672x}$

Area of $\Delta ABC$ = area of $\Delta AOC+$ area of $\Delta AOB+$ area of $\Delta BOC$

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

$=\frac{1}{2}\times 4(x+6)+\frac{1}{2}\times 4(x+8)+\frac{1}{2}\times 4\times 14$

$=2x\times 12+2x+16+28=4x+56$

$\sqrt{48x^2+672x}=4x+56$

$48x^2+672x=(4x+56{)}^2$

$48x^2+672x=16x^2+448x+3136$

$32x^2+224x-3136=0$

$32(x^2+7x+98)=0$

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

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

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

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

$x=7$ and $x=-14$

$x$ can't be negative

$\therefore AC=x+6=7+6=13cm$

$AB=x+8=7+8=15cm$