Question

The given figure shows a circle with centre O and radius 4 cm circumscribed by ΔABC. BС touches the circle at D such that BD = 6 cm, DC = 10 cm. Find the length of AE.

Answer 1

Given, BD = 6 cm

DC = 10 cm

Since tangents are drawn from an external points are equal.

BF = BD = 6 cm

CD = CE = 10 cm

Let AE = AF = x

Now, c = AB = AF + BF

= x + 6

a = BC = BD + DC

= 6 + 10

= 16

b = AC = AE + EC

= x + 10

Perimeter = AB + BC + AC

Semi-perimeter = `(AB + BC + AC)/2`

S = `(x + 6 + 16 + 10 + x)/2`

S = `(2x + 32)/2`

S = x + 16

Area of Δ = `sqrt(S(S - a)(S - b)(S - c))`

`Δ = sqrt((x + 16)(x + 16 - 16)(x + 16 - (10 + x))(x + 16 - (x + 6))`

`Δ = sqrt((x + 16)x(6)(10))`

Area (ΔABC) = `sqrt(60x(x + 16))`   ...(1)

Area of ΔABC = Area (ΔBOC) + Area (ΔAOC) + Area (ΔAOB) 

= `(1/2 xx OD xx BC) + (1/2 xx OE xx AC) + (1/2 xx OF xx AB)`

= `(1/2 xx 4 xx 16) + (1/2 xx 4 xx (x + 10)) + (1/2 xx 4 xx (x + 6))`

= 32 + 2x + 20 + 2x + 12

Area (ΔABC) = 4x + 64   ...(2)

From equations (1) and (2)

`sqrt(60x(x + 16)) = 4(x + 16)`

Squaring on both sides,

60x(x + 16) = 16(x + 16)²

60x = 16(x + 16)

60x = 16x + 256

44x = 256

x = `256/44`

x = `64/11`

As x = `64/11`

Length of AE = `64/11` = 5.8 cm