Question

A triangle ABC is drawn to circumscribing a circle of radius 3cm such that segments BD and DC into BC is divided by the point of contact D are of length 9cm. And 3 cm. ... Find AB and AC

Answer 1

Answer:-

• AB = 9

• AC=15

Explanation:-

Let us assume that the circle touches AB in F, BC in D and AC in E.

Also given is BD = 9cm and DC = 3cm.

Let us consider AF = x.

For ΔABC,

★AF = AE = x   (∵ tangents drawn from an external point to a circle are congruent i.e. AE and AF are tangent drawn from external point A.)

Similarly we have,

★BE = BD = 3cm (∵ congruent tangents from point B)

★And CF = CD = 9cm (∵ congruent tangents from point C)

BC = BD + DC = 12

Now,

• AB = AE + EB = x + 3
• AC = AF + FC = x + 9

Then,

2s = AB + BC +CA

= x + 3 + 12 + 1 + x + 9

= 2x + 24

∴ s = x + 12

Using Heron’s formula,

Area of triangle ABC = 3√3(12-x²) ....(1)

Total area = Area(OBC) + Area(OAB) + Area(OAC)

=18 + 9+3x/2 + 27+3x/2

=12 + x

On solving further,

$\sf\implies\:x^2 + 6x - 72=0$

$\sf\implies\:(x+12)(x-6)=0$

$\sf\implies\:x=-12\: and\: x=6$

Hence,

• AB = x+3=9
• BC =12
• AC=x+9=15