Question

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

Answer 1

Hint: First we make the diagram for the question. Using the theorem that tangents drawn on a circle from a point outside the circle are of equal length, we will calculate the length of CE and BF. In the same way AE = AF and let it be x and finally we will equate the area in terms of x.

Complete step-by-step answer:
The diagram is given below:

Given : A circle whose radius is 4cm and is inscribed in a triangle ABC. BD =6cm and CD = 8cm.
We know that the length of tangents drawn from the external point to a circle are of equal lengths.
Hence, BE = BD =6cm and CF = CD = 8cm.
Let the length of AE = AF = x cm.
Now the sides of three sides are:
AB = AE+BE = x+8cm
BC = CD + BD = 8+6 = 14cm
AC = AE +CF = x+8cm.
Now, we will use heron's formula to find the area of triangle ABC.
Area of triangle = $\sqrt {s(s - a)(s - b)(s - c)} $ .
Where, S = $\dfrac{{AB + BC + CA}}{2} = \dfrac{{x + 6 + x + 8 + 14}}{2} = x + 14cm$
Putting the value in the formula, we get:
Area = $\sqrt {(x + 14)(x + 14 - x - 6)(x + 14 - x - 8)(x + 15 - 14)} = \sqrt {48{x^2} + 672x} $ (1)
Now, we know that tangent is perpendicular to the radius. Therefore, we will find the area of $\vartriangle BOC,\vartriangle AOC{\text{ and }}\vartriangle {\text{AOB}}$and add them to get the area of triangle ABC.
${\text{Area of }}\vartriangle {\text{ABC = Area of }}\vartriangle BOC{\text{ + Area of }}\vartriangle {\text{AOB + Area of }}\vartriangle {\text{AOC}}$
                                           =$\dfrac{1}{2} \times BC \times OD + \dfrac{1}{2} \times AB \times OE + \dfrac{1}{2} \times AC \times OF$
                                           =$\dfrac{1}{2} \times 14 \times 4 + \dfrac{1}{2} \times (x + 6) \times 4 + \dfrac{1}{2} \times (x + 8) \times 4 = 4x + 56$ (2)
Equating the two area, we get:
$\sqrt {48{x^2} + 672x} = 4x + 56$
On squaring both sides, we get:
$48{x^2} + 672x = {\left( {4x + 56} \right)^2} = 16{x^2} + 448x + 3136$
$ \Rightarrow 48{x^2} - 16{x^2} - 448x - 3136 + 672x = 0$
$ \Rightarrow 32({x^2} + 7x - 98) = 0$
$ \Rightarrow ({x^2} + 7x - 98) = 0$
$ \Rightarrow $ (x-7)(x+14)=0
$\therefore $ x= 7 and x =-14.
Since x cannot be negative. So, x= 7.
Therefore. AC = x+6= 7+6 = 13cm and AB = x+8 = 7+8 = 15cm.

Note: In this type of question, you should remember the theorem related to the length of tangent drawn from an external point on the circle. Heron’s formula is generally used to find the area of the scalene triangle.