ABCD is 20×40m rectangle. P is a point on BC such that semicircles are drawn AD, CP, BP. i) Find perimeter of shaded region. ii) find area of unshaded region
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perimeter of shaded region is 2(40+20)
=160m-22/7×10×2×22/7×5
160-440/7
=680/7m
now area of shaded region=
800-1650/7
3950/7m^2
=160m-22/7×10×2×22/7×5
160-440/7
=680/7m
now area of shaded region=
800-1650/7
3950/7m^2
Kudrat11:
thnx a lot... ßut r u sure abt the answer
ya
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