Question

A triangle $ABC$ is drawn to circumscribe a circle of radius $3cm$. such that the segment $BD$ and $DC$ into which $BC$ is divided by the point of contact $D$ are of length $9cm$ and $3cm$. respectively (See adjacent figure). Find the sides $AB$ and $AC$.

Answer 1

$△ABC$ circumscribe the circle with centre $O$

$BD=8cm$

$CD=6cm$

$OD=4cm$

Since tangents drawn from extend point are equal

So, $CE=CD=6cm$

$BF=BD=8cm$

$AE=AF=kcm$

$\therefore CB=CD+DB=6+8=14cmc$

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

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

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

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

Area of $△ABC$

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

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

$=\sqrt{46x^2+672x}$

$\therefore OD=OF=OEradius=4cm$

And $OD\perp BC,OF\perp AB$ and $OE\perp AC(\because tangentO\perp$ to radius)

$ar\left(△ABC\right)=ar\left(△AOC\right)+ar\left(△AOB\right)+ar\left(△BOC\right)......\left(1\right)$

$ar\left(△AOC\right)=\frac{1}{2}\times OE\times AC$

$=\frac{1}{2}\times 4\times (x+6)=2x+12$

$ar\left(△AOB\right)=\frac{1}{2}\times OF\times AB=\frac{1}{2}\times 4(x+8)=2x+16$

$ar\left(△BOC\right)=\frac{1}{2}\times OD\times BC=\frac{1}{2}\times 4\times 14=28$

$\therefore$ from $\left(1\right)$

$(\sqrt{46x^2+672x{)}^2}=(4x+56{)}^2$

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

$41x^2-16x+672x-448x-3136=0$

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

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

as $x$ correct be $-19$ (negative acceptable)

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

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