Question

Please respond asap!!!

Answer 1

ANSWER

$4\pi - 8$

EXPLANATION

The diagonal of the square can be found

using Pythagoras Theorem.

${d}^{2} = {(2 \sqrt{2} )}^{2} + {(2 \sqrt{2} )}^{2}$

${d}^{2} = 4 \times 2+ 4 \times 2$

${d}^{2} = 8+ 8$

${d}^{2} = 16$

Take positive square root

$d = \sqrt{16} = 4$

The radius is half the diagonal because the diagonal formed the diameter of the circle.

Hence r=2 units.

Area of circle is

$\pi {r}^{2} =\pi \times {2}^{2} = 4\pi$

The area of the square is

${l}^{2} = {(2 \sqrt{2)} }^{2} = 4 \times 2 = 8$

The difference in area is

$4\pi - 8$

Answer 2

Hello!

The answer is:

The difference between the circle and the square is:

$Difference=4\pi -8$

Why?

To solve the problem, we need to find the area of the circle and the area of the square, and then, subtract them.

For the square we have:

$side=2\sqrt{2}$

We can calculate the diagonal of a square using the following formula:

$diagonal=side*\sqrt{2}$

So,

$diagonal=2\sqrt{2}*\sqrt{2}=2*(\sqrt{2})^{2}=2*2=4units$

The area will be:

$Area_{square}=side^{2}= (2\sqrt{2})^{2} =4*2=8units^{2}$

For the circle we have:

$radius=\frac{4units}{2}=2units$

The area will be:

$Area_{Circle}=\pi *radius^{2}=\pi *2^{2}=\pi *4=4\pi units^{2}$

$Area_{Circle}=4\pi units^{2}$

Then, the difference will be:

$Difference=Area_{Circle}-Area{Square}=4\pi -8$

Have a nice day!