Question

In the given figure, a triangle $ABC$ is drawn to circumscribe a circle of radius 3 cm such that the segments BD and DC into which BC is divided by the point of contact D, are of lengths 6 cm and 9 cm respectively. If the area of $\Delta ABC=54c{m}^2$ then find the lengths of sides AB and AC.

Answer 1

In the given figure, $\Delta ABC$ circumscribed the circle with centre O.

Radius $OD=3cm$

$BD=6cm,DC=9cm$

Area of $\Delta ABC=54c{m}^2$

To find : Length of $AB$ and $AC$.

$AF$ and $EA$ are tangents to the circle at point A.

Let $AF=EA=x$

$BD$ and $BF$ are tangents to the circle at point B.

$BD=BF=6cm$

$CD$ and $CE$ are tangents to the circle at point C.

$CD=CE=9cm$

Now, new sides of the triangle are:

$AB=AF+FB=x+6cm$

$AC=AE+EC=x+9cm$

$BC=BD+DC=6+9=15cm$

Now, using Heron's formula:

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

Where $S=\frac{a+b+c}{2}$

$S=1/2(x+6+x+9+15)=x+15$

Area of $ABC=\sqrt{(x+15)(x+15-(x+6\left)\right)(x+15-(x-9\left)\right)(x+15-15)}$

Or

$54=\sqrt{(x+15)\left(9\right)\left(6\right)\left(x\right)}$

Squaring both sides, we have

${54}^2=54x(x+15)$

$x^2+15x-54=0$

Solve this quadratic equation and find the value of x.

$x^2+18x-3x-54=0$

$x(x+18)-3(x+18)=0$

$(x-3)(x+18)=0$

Either $x=3$ or $x=-18$

But x cannot be negative.

So, $x=3$

Answer :-

$AB=x+6=3+6=9cm$

$AC=x+9=3+9=12cm$