Chapter 11.8, Problem 6CYU

a..

To determine

Find the area of the trampoline.

Answer to Problem 6CYU

  $\boxed{A=1017.36inc{h}^2}$

Explanation of Solution

Given information:

Find the area of the trampoline whose diameter is $36inch$ . Compare to the estimate.

Calculation:

Consider the diameter of trampoline having circular shape, is $36inch$

Area of circle,

  $\begin{array}{l}A=\pi r^2\\ d=36in\\ r=\frac{d}{2}=36/2=18inches\\ A=\pi \times {18}^2\\ =324\pi \\ A=1017.36inc{h}^2\end{array}$

Hence the area of the given trampoline is $\boxed{A=1017.36inc{h}^2}$

b.

To determine

What is the area of jumping surface?

Answer to Problem 6CYU

  $\boxed{A\approx 615.44i{n}^2}$

Explanation of Solution

Given information:

The trampoline whose diameter is $36inch$ . The safety padding around the jumping surface is $4inches$ wide. What is the area of jumping surface?

Calculation:

Consider the figure,

Diameter of trampoline having circular shape, is $36inch$

The diameter of jumping surface is,

  $\begin{array}{l}36-8=28inches\\ radius=28/2=14inches\end{array}$

Now calculate area of jumping surface,

  $\begin{array}{l}A=\pi r^2\\ A=\pi \times {14}^2\\ =196\pi \\ A=196\times 3.14\\ A\approx 615.44i{n}^2\end{array}$

Hence the area of the jumping surface is, $\boxed{A\approx 615.44i{n}^2}$