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a circular hall has a hemispherical roof. the greatest height is equal to the internal diameter .if the capacity of the hall is 43510 m^3 ,then the area of the floor is​

User Marcus Tik
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This building is basically a cylinder with a hemisphere on top. Let d be the diameter of the hemispherical roof. This is also the height of the entire building as the greatest height is this value (which is located at the center of the building)

The hemisphere itself is r = d/2 meters high, which is the radius of the roof. The cylinder portion's height must be d - r = d - d/2 = 2d/2 - d/2 = d/2 = r meters. So basically the cylinder and the hemisphere both have the same height of r.

At the same time, the cylinder and hemisphere also have the same radius r, or else the roof wouldn't fit on the cylinder.

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Volume of cylinder = pi*r^2*h = pi*r^2*r = pi*r^3

Note how I replaced h with r and simplified.

Also,

Volume of hemisphere = (1/2)*(volume of sphere)

Volume of hemisphere = (1/2)*( (4/3)*pi*r^3 )

Volume of hemisphere = (2/3)*pi*r^3

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Add the two volume expressions

total volume = (volume of cylinder) + (volume of hemisphere)

total volume = (pi*r^3) + ( (2/3)*pi*r^3 )

total volume = (3/3)pi*r^3 + (2/3)pi*r^3

total volume = (5/3)pi*r^3

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Set this equal to 43510 and solve for r

(5/3)pi*r^3 = 43510

pi*r^3 = (3/5)*43510

pi*r^3 = 26106

r^3 = 26106/pi

r = (26106/pi)^(1/3)

r = 20.2549023450428

The building has a radius of approximately 20.2549023450428 meters

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We can now compute the area of the circular floor

area = pi*r^2

area = pi*(20.2549023450428)^2

area = 1,288.87316044698

Answer: approximately 1288.87316044698 square meters

Round this however you need to

User Nikhil Bhatia
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