Question

In the given figure, 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 6 cm and 8 cm respectively. If the area of ΔABC is 84 cm², find the lengths of sides AB and AC.

Answer 1

Given, BD = 6 cm, DC = 8 cm

Here, BD = BF and DC = CE .....[Tangents drawn from an external point to a circle are equal]

∴ BF = 6 cm and CE = 8 cm

Let AF = x = AE .....[Tangents drawn from external point A to the circle are equal]

In ΔABC,

a = BC = BD + DC = 6 + 8 = 14 cm

b = AC = CE + AE = (8 + x) cm

c = AB = BF + AF = (6 + x) cm

Now, s = `("a" + "b" + "c")/2`

= `(14 + (8 + "x") + (6 + "x"))/2`

= `(28 + 2"x")/2`

= (14 + x) cm

∴ Area of ΔABC = `sqrt("s"("s" - "a")("s" - "b")("s" - "c"))`

84 = `sqrt((14 + "x")(14 + "x" - 14)(14 + "x" - 8 - "x")(14 + "x" - 6 - "x"))`

84 = `sqrt("x"(14 + "x")(6)(8)`

84 = `sqrt(48"x"("x" + 14)) "cm"^2`  ...(i)

`sqrt(48"x"("x" + 14))` = 84

On squaring both sides, we get

48x(x + 14) = 84 × 84

⇒ 4x(x + 14) = 84 × 7

⇒ x² + 14x – 147 = 0

⇒ x² + 21x – 7x – 147 = 0

x(x + 21) – 7(x + 21) = 0

(x + 21)(x – 7) = 0

So, x = 7, or x = –21 (rejected as - ve)

Hence, x = 7

Therefore, AB = c = 6 + x = 6 + 7 = 13 cm

AC = b = 8 + x = 8 + 7 = 15 cm