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. Find the sides AB and AC.

• A. $13$ and $15$
  
• B. $15$ and $13$
  
• C. $16$ and $12$
  
• D. None of These
  

Answer 1

The correct option is B $15$ and $13$

Let there is a circle having center $O$ touches the sides $AB$ and $AC$ of the triangle at point $E$ and $F$ respectively.

Let the length of the line segment $AE$ is $x$.

Now in $△ABC$,

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

$BE=BD=6$ (tangents on the circle from point $B$)

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

Now $AB=AE+EB$

$⟹AB=x+8=c$

$BC=BD+DC$

$⟹BC=8+6=14=a$

$CA=CF+FA$

$⟹CA=6+x=b$

Now

Semi-perimeter, $s=\frac{(AB+BC+CA)}{2}$

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

$s=\frac{(2x+28)}{2}$

$⟹s=x+14$

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

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

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

$=\sqrt{(14+x)\left(x\right)(2\times 4)(2\times 3)}$

$Areaofthe△ABC=4\sqrt{3x(14+x)}$ .............................(1)

$Areaof△OBC=\frac{1}{2}\times OD\times BC$

$=\frac{1}{2}\times 4\times 14=28$

$Areaof△OBC=\frac{1}{2}\times OF\times AC$

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

$=12+2x$

$Areaof\times OAB=\frac{1}{2}\times OE\times AB$

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

$=16+2x$

Now, $Areaofthe△ABC=Areaof△OBC+Areaof△OBC+Areaof△OAB$

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

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

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

On squaring both sides, we get

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

$3x=14+x$ ------------- $(14+x=0⟹x=-14$ is not possible$)$

$3x-x=14$

$2x=14$

$x=\frac{14}{2}$

$x=7$

Hence

$AB=x+8$

$AB=7+8$

$AB=15$

$AC=6+x$

$AC=6+7$

$AC=13$

So, the value of $AB$ is $15cm$ and that of $AC$ is $13cm$. Option B.