Question

In Figure a $\Delta$ ABC is drawn to circumscribe a circle of radius 4 cm such that the segments BD and DC are of lengths 8 cm and 6 cm respectively. Find the lengths of sides AB and AC, when area of $\Delta$ ABC is 84 $c{m}^2$.

Answer 1

Firstly, consider that the given circle will touch the sides AB and AC of the triangle at point E and F respectively.

Let AF = x

Now, in ABC,

CF = CD = 6cm (Tangents drawn from an exterior point to a circle are equal. Here, tangent is drawn from exterior point C)

BE = BD = 8cm (Tangents drawn from an exterior point to a circle are equal. Here, tangent is drawn from exterior point B)

AE = AF = x (Again, tangents drawn from an exterior point to a circle are equal. Here, tangent is drawn from exterior point A)

Now, AB = AE + EB

= x + 8

Also, BC = BD + DC = 8 + 6 = 14 and CA = CF + FA = 6 + x

Now, we get all the sides of a triangle so its area can be find out by using Heron's formula as:

2s = AB + BC + CA

= x + 8 + 14 + 6 + x

= 28 + 2x

​​ Semi-perimeter = s = $\frac{(28+2x)}{2}$ = 14 + x

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

$\to$$\sqrt{(14+x)\left[\right(14+x)-(14\left)\right]\left[\right(14+x)-(6+x\left)\right]\left[\right(14+x)–(8+x\left)\right]}$

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

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

Again, area of triangle is also equal to the $\frac{1}{2}$ base x height

Therefore,

Area of $△OBC$ = $\frac{1}{2}$ OD . BC

= $\frac{1}{2}$ (4) (14)

= 28

Area of $△OCA$ = $\frac{1}{2}$ OF. AC

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

= 6 + 2x

Area of $△OAB$ =$\frac{1}{2}$ OE . AB

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

= 16 + x

Area of $△ABC$ = Area of $△OBC$+Area of $△OCA$ + Area of $△OAB$

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

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

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

On squaring both sides, we get

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

$\to$ 43 x + 3$x^2$ = 196 + $x^2$ + 28 x

$\to$ 2 $x^2$ + 4x +-196 = 0

Dividing LHS and RHS byv2 we have

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

$x^2$ + 14 x - &x -98 = 0

( x = 14) ( x - 7) = 0

Either x = 14 = 0 Or x -7 = 0

$\to$ x = -14 $\to$ x = 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 = 15 cm

CA = 6 + x = 6 + 7 = 13 cm