# A square is inscribed in a circle with radius 4 inches, how do you find the area of the region inside the circle and outside the square?

Nov 8, 2015

$16 \pi - 32$

#### Explanation:

To get the area inside the circle but outside the square, we simply get the difference of their areas.

${A}_{c} = {r}^{2} \pi = {4}^{2} \pi = 16 \pi$

A square is also a rhombus (with equal diagonals), so we can use the formula for the area of the rhombus. What do we use as the value of the diagonal? The diameter of the circle!

${A}_{s} = \frac{1}{2} {d}_{1} {d}_{2} = \frac{1}{2} {d}^{2} = \frac{1}{2} \left({8}^{2}\right) = \frac{1}{2} \left(64\right) = 32$

Hence, $A = 16 \pi - 32$

If you think you won't be able to remember the formula of the rhombus, we can go the usual way of getting the area of the square.

${A}_{s} = {s}^{2}$

Now, we need to find the length of the side. To do this, we can form an imaginary right isosceles triangle with the radius of the circle as legs and the side as the hypotenuse. How do we know the triangle is right? It's a property of a square, the diagonals form a right angle.

Now, we use the Pythagorean theorem to get the side,

${c}^{2} = {a}^{2} + {b}^{2} = 2 {a}^{2} = 2 \left({4}^{2}\right) = 32$
$\implies c = 4 \sqrt{2}$

Now we can get the area of the square.

${A}_{s} = {s}^{2} = {\left(4 \sqrt{2}\right)}^{2} = 32$

Hence, $A = 16 \pi - 32$