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since it is given that pq=12cm and pr=9cm
by phythagorous triplet we have
QR^2 = PQ^2+PR^2
QR = √[12^2+16^2] = √144+256 =√400=20cm
so qr = diameter =20
radius =10
area of shaded region = area of semicircle – area of the triangle
area of shaded region = π(10)^2–1/2*12*16 = 100π–96 = 218.15cm^2
hence the area of shaded region = 218.15cm^2
by phythagorous triplet we have
QR^2 = PQ^2+PR^2
QR = √[12^2+16^2] = √144+256 =√400=20cm
so qr = diameter =20
radius =10
area of shaded region = area of semicircle – area of the triangle
area of shaded region = π(10)^2–1/2*12*16 = 100π–96 = 218.15cm^2
hence the area of shaded region = 218.15cm^2
vipulanand:
My answer is right
no 20.8125 is not a right answer
and can you explain how you got it
Answered by
0
the answer is 20.8125
please add me brainlist please please please
My answer is right
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