In a house , there is a semicircular corridor, a circular hall and a circular garden. Sum of perimeter of corridor and hall is 102 m.If radius of hall is 50% more than radius of corridor then find area of circular garden, given radius of garden is 300% more than sum of radius of corridor and radius of hall together.
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Given In a house , there is a semi circular corridor, a circular hall and a circular garden. Sum of perimeter of corridor and hall is 102 m.If radius of hall is 50% more than radius of corridor then find area of circular garden, given radius of garden is 300% more than sum of radius of corridor and radius of hall together.
- Let the radius of the hall be r1 and corridor be r2
- Given perimeter of corridor and hall is 102 m
- According to question
- πr1 + 2r1 + 2πr2 = 102
- So r2 = r1 + 0.5 r1
- Or r2 = 1.5 r1
- Now πr1 + 2r1 + 2π(1.5 r1) = 102
- So πr1 + 2r1 + 3πr1 = 102
- 2r1 + 4πr1 = 102
- So r1 = 51 / 1 + 2π
- According to question let radius of garden be r3
- So r3 = (r1 + r2) + 3(r1 + r2)
- So r3 = 4 (r1 + r2)
- = 4(r1 + 1.5 r1)
- = 4(2.5 r1)
- So r3 = 10r1
- Now r3 = 10 (51 / 1 + 2π)
- Now area of a circular garden = π r3^2
- = π 10(510 / 1 + 2π)^2
- So r3 = 510 x 510 x π / 1 + 4π^2 + 4π
- So r3 = 4907.69 π m^2
- Therefore area of circular garden is 4907 π sq m
- Reference link will be
- https://brainly.in/question/13240454
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