Question

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

Answer 1

Let the given circle touch the sides AB and AC of the triangle at point E and F respectively and the length of the line segment AF be $x.$

In $△ABC,$ we have

$CF=CD=6cm$ (Tangents on the circle from point C)

$BE=BD=8cm$ (Tangents on the circle from point B)

$AE=AF=x$ (Tangents on the circle from point A)

$AB=AE+EB=x+8$

$BC=BD+DC=8+6=14$

$CA=CF+FA=6+x$

$2s=AB+BC+CA$

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

$=28+2x$

$\therefore s=14+x$

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

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

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

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

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

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

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

Area of $\Delta ABC=$ Area of $\Delta OBC$ + Area of $\Delta OCA$ + Area of $\Delta OAB$

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

$\Rightarrow 4\sqrt{3(14x+x^2)}=56+4x$

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

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

$\Rightarrow 42x+3x^2)=(196+x^2+28x)$

$\Rightarrow 42x+3x^2=(196+x^2+28x)$

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

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

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

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

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

$\Rightarrow x=-14or,x=-7$

Either $x+14=0$ or $x-7=0$

Therefore, $x=-14$ and $7$

However, $x=-14$ is not possible as the length of the sides will be negative.

Therefore, $x=7$

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

$CA=6+x=6+7=13cm$​