Question

A training field is formed by joining a rectangle and two semicircles, as shown below. The rectangle is 96m long and 74m wide.

Find the area of the training field. Use the value 3,14 for pie , and do not round your answer. Be sure to include the correct unit in your answer.

Answer 1

`11402.66 m^(2)`

Explanation:

The width of rectangle is the diameter of the semi-circle part

Area of one semicircle is given by
`\frac {0.5\pi d^(2)}{4}`

Total area of semi circle will be
`2*\frac {0.5\pi d^(2)}{4}`

Substituting 74 m for d and
`\pi` as 3.14 we obtain

Total area semi-circle=
`2*\frac {0.5*3.14* 74^(2)}{4}=4298.66 m^(2)`

Area of rectangle is given by the product of length and width

Rectangular area=
`96 m*74 m=7104 m^(2)`

Total area of rectangular and semi-circles will be

`4298.66 m^(2)+7104 m^(2)=11402.66 m^(2)`

Therefore, area of training field is
`11402.66 m^(2)`