Question

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.

Answer 1

Step-by-step explanation:

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