Question

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

Answer 1

Answer:

$11402.66 m^{2}$

Step-by-step 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 $2\times\frac {0.5\pi d^{2}}{4}$

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

Total area semi-circle=$2\times\frac {0.5*3.14\times 74^{2}}{4}=4298.66 m^{2}$

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 $11402.66 m^{2}$