Question

A triangle ABC is drawn to circumscribe a circle of radius 3cm, such that the segments BD and DC are
respectively of lengths 6cm and 9cm. If the area of AABC is 54cm', then find the lengths of sides AB
and AC
(March, 2015)
O
Всh
Bc
D​

Answer 1

Answer:

Step-by-step explanation:

Given OD=3cm

Construction join OA,OB and x

Proof : area of the ΔABC = area of ΔOBC + area of ΔOAC +are of ∠OAB

BD=6cm:BE=6cm(equal tangents)

DC=9cm:CF=9cm(equal tangents)

area of ΔOBC=12×b×h

=12×15×3

=452cm2

area of ΔOAC=12×(x+9)×3

=3(x+9)12cm2

area of ΔOAB=12×(x+6)×3

=3(x+6)2cm2

area of the ΔABC=s(s−a)(s−b)(s−c)−√cm2

s=(x+9)+(x+6)+152=2x+302=x+15

∴Δ=(x+15)(x+15−15)(x−15−x−4)(x+15−x−6)−√

=(x+15)(x)(6)(9)−√

x(x+15)(54)−√=3(x+9)2+3(x+16)2+452

=> x(x+15)(54)−√=32[(x+9)+(x+6)+15]

x(x+15)(54)−√=32(2x+30)

x(x+15)(54)−√=(3x+15)

Squaring on both sides we get

x(x+15)(54)=9(x+15)2

6x=x+15

5x=15

x=3cm

Hence the sides are 15cm,12cm,9cm