A rectangular swimming pool ABCE has a quadrant DEF with centre E where ED = x/2. Given that the remaining swimming area is `63*pi` cm² and the length of BC is twice the length of arc FD, find...

A rectangular swimming pool ABCE has a quadrant DEF with centre E where ED = x/2. Given that the remaining swimming area is `63*pi` cm² and the length of BC is twice the length of arc FD, find the value of x and y where BC = y and CE = x.

 

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najm1947 | Elementary School Teacher | (Level 1) Valedictorian

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Given:

A rectangular swimming pool ABCE where:

BC = y and CE = x

Has a quadrant DEF with centre E and ED = x/2

Area of rectangular pool excluding quadrant = 63.pi cm2

BC = y = Twice length of Arc

Solution:

Length of Arc = 1/4*(2*pi*x/2) = pi*x/4

Therefore y = 2*pi*x/4 = pi*x/2

Area of the quadrant = 1/4*pi*(x/4)^2 = 1/64*pi*x^2

Total area of rectangular pool = x*y = x*pi*x^2 = pi*x^2

Remaining area of the pool = pi*x^2 - 1/64*pi*x^2

                                       = 63/64*pi*x^2

thus 63/64*pi*x^2 = 63*pi

or     x^2 = 64

or     x = 8 cm

and   y = pi*x = 8.pi cm

The values of x and y (2 sides of the rectangular swimming pool are: 8 cm and 8.pi cm

 

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