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Let r = radius of the inscribed circle. Then,
Area of Δ ABC = Area of Δ OBC + Area of Δ OCA + area of Δ OAB
=(1/2 × r× BC)+(1/2 × r × CA)+(1/2 × r × AB)
=1/2 x r x(BC + CA + AB)
=1/2 x r x (24 +24 + 24)
=36r cm2 ----- 1st
Also Area of Δ ABC = 1/2 x BC x PA
BC= 24
and PA^2=(BA)^2 - (BP)^2 = 24^2 - 12^2
PA^2= 576 -144 =432
PA=√432 = 12√3
So Area of Δ ABC = 1/2 x 24 x 12√3 = 144√3 -----2nd
now putting 1st = 2nd
36r = 144√3
r=4√3
Area of a circle = π(4√3)^2 = 48π
area of the remaining portion of the triangle= 144√3 - 48π
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