Question

In the given fig. if radius of circle is 3 cm. Find the perimeter of ABC.

Answer 1

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