(Q8) 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 length 9cm and 3 cm. respectively. Find the sides AB and AC.

ABCO9cm3cm

The given figure can also be drawn as

ABCDOEF

Given: let △ABC be the given triangle circumscribing the given circle with center 'O' and radius 3 cm.

The circle touches the sides BC, CA and AB at D,E,F respectively

It is given that BD = 9cm

CD = 3cm

Lengths of two tangents drawn from an external point to a circle are equal.

BF = BD = 9cm

CD=CE = 3cm

AF = AE = xcm say

The sides of the triangle are

12cm, (9+x)cm , (3+x) cm

Perimeter = 2S = 12+9+x+3+x

2S = 24+2X

S = 12+x

S-a = 12+x-=

S-b = 12+x-3-x=

S-c = 12+x-9-x =

Area of the triangle

△ABC = √S √(S-a)√(S-b)√(S-c)

   = √(12+x)√()√()√()

    = √√(x2+12x)--------(1)

But , △ABC = △OBC+△OCA+△OAB

=
1
2
 BC×OD+
1
2
 ×CA×OE+
1
2
 AB ×OF
=
1
2
 (×3)+
1
2
 (+x)×3+
1
2
 (+x)×3
=
1
2
 [++3x++3x]
=
1
2
 [+6x] = +3x--------(2)

From (1) and (2),

√√(x2+12x) = + 3x

Squaring on both sides we get,

(x2+12x) = (+3x)2

x2+324x=+9x2+216x

x2+108x-=0

x2+6x-=0

x(x+12)-(x+12)=0

(x-)(x+12)=0

x = or -12

But 'x' cant be negative hence x=

AB = 9+ = cm

AC = 3+ = cm