1條f.4 maths

(a)(i)
BQ=BP=x (property of tangent)
CQ=CR=y (property of tangent)
Let O be the center of PQR
∠OPA=∠ORA=90o (property of tangent)
AP=AR=3cm

BQ+BP+CQ+CR+AP+AR=36
2x+2y+2(3)=36
2(x+y)=30
x+y=15
(ii)
Area of ∆OPB +Area of ∆OQB+Area of ∆OQC+Area of ∆ORC+Area of ∆OPA+Area of ∆ORA=Area of ∆ABC
3(BQ)+3(QC)+3(AP)=(AB)(AC)/2
3(x+y)+9=(3+x)(3+y)/2
putting y=15-x
3(15)+9=(3+x)(3+15-x)/2
108=(3+x)(18-x)
54+15x-x^2-108=0
x^2-15x+54=0
(b)
From (a)(ii)
x^2-15x+54=0
(x-6)(x-9)=0
x=6 (rejected) or x=9
and y=6
AB=3+x=12
AC=3+y=9
BC=x+y=15