Question

Been going back and forth with this task examples for the past 40 minutes because of the end part

Answer 1

Hello

To solve this question, we would solve for the area of the semi-circle and area of the rectangle.

The dimensions of the rectangle is

`\begin{gathered} l=100m \\ w=66m \end{gathered}`

Area of a rectangle is given as

`\begin{gathered} A_{\text{rectangle}}=L* w \\ A=100*66=6600m^2 \end{gathered}`

The area of the rectangle is 6600m^2

The formula of area of a semi-circle is

`A_(s-c)=(\pir^2)/(2)`

But the radius would be half the value of the width of the training field

`r=(66)/(2)=33m`

Area of the semi-circle is

`\begin{gathered} A_c=(\pi r^2)/(2) \\ A_c=(3.14*33^2)/(2)=1709.73m^2 \end{gathered}`

The area of the semi-circle is 1709.73m^2

The area of the training field is area of the rectangle + 2(area of the semi-circle)

NB: We are multiplying the area of the semi-circle by 2 because we have two semi-circle attached to the rectangle

The area of the training pitch is

`6600+1709.73+1709.73=10019.46m^2`

From the calculations above, the area of the training pitch is 10019.46m^2