Chapter 8.4, Problem 46PPS

a.

To determine

Find the expression for the area of the top of the flying disks.

Answer to Problem 46PPS

The new area of the top will be $22.3inc{h}^2$ .

Explanation of Solution

Given: It is given in the question that the flying disk has a radius of $4.9$ inches. Jolanda discovers that adding a hole with a radius of $3.75$ inches to the center of the disk reduces the weight of the disk so it travel farther.

Concept Used:

In this, use the concept of circle and the formula used is $A=\pi r^2$ .

Calculation: Here, the Bigger radius, $R=4.6$ and the smaller radius, $r=3.75$

Now, the net area will be the $A=\pi R^2-\pi r^2$ .

  $\begin{array}{l}A=\pi R^2-\pi r^2\\ A=\pi (R^2-r^2)\\ $ A=3.14[{(4.6)}^2-{(3.75)}^2]$A=3.14[21.1-14.0]\\ A=3.14[7.1]\\ A=22.3inc{h}^2\end{array}$

Conclusion:

The new area will be $22.3inc{h}^2$ .

b.

To determine

Explain the solution process.

Answer to Problem 46PPS

The area of the bigger circle subtract to the area of the smaller circle will give the net area of circle.

Explanation of Solution

Given:

It is given in the question that the flying disk has a radius of $4.9$ inches. Jolanda discovers that adding a hole with a radius of $3.75$ inches to the center of the disk reduces the weight of the disk so it travel farther.

In this case, in a flying disk there would be a bigger radius and also a smaller radius such that it has to find the net area you have to subtract the outer circle to the inner circle to get the rest of the area i.e. $A=\pi R^2-\pi r^2$ . And as soon as you lowers the inner radius the flying disk will become more weightless and fly very efficiently.