The inner circle is the largest one that can be drawn inside the square. The outer circle is the smallest one that can be drawn with the square inside it. Prove that the shaded area between the 2 circles is the same as the area enclosed by inner circle?

The diameter of the larger circle is given by pythagoras, because it can be found by drawing a diagonal through the square.

Or the same as the inner circle.

So we've proved that this is the case.

Hopefully this helps!

To prove that the shaded area between the two circles is the same as the area enclosed by the inner circle, we first need to establish the relationship between the radii of the circles and the side length of the square.

Let the side length of the square be $s$. The diameter of the inner circle is equal to the side length of the square, so its radius is $\frac{s}{2}$.

The diameter of the outer circle is equal to the diagonal of the square. Using the Pythagorean theorem, the diagonal of a square with side length $s$ is $s\sqrt{2}$, so the radius of the outer circle is $\frac{s\sqrt{2}}{2}$.

Now, we can calculate the areas of the circles:

• The area of the inner circle is $\pi \left( \frac{s}{2} \right)^2 = \frac{\pi s^2}{4}$.
• The area of the outer circle is $\pi \left( \frac{s\sqrt{2}}{2} \right)^2 = \frac{\pi s^2}{2}$.

The shaded area between the two circles is the difference between the areas of the outer circle and the inner circle:

$\text{Shaded area} = \frac{\pi s^2}{2} - \frac{\pi s^2}{4} = \frac{\pi s^2}{4}$

This is exactly the same as the area enclosed by the inner circle, which proves that the shaded area between the two circles is equal to the area of the inner circle.

To prove that the shaded area between the two circles is the same as the area enclosed by the inner circle, we need to show that the shaded area is equal to the area of the inner circle.

Let's denote the radius of the inner circle as $r$, the radius of the outer circle as $R$, and the side length of the square as $s$.

The radius $R$ of the outer circle is equal to half the diagonal of the square. Since the diagonal of a square is $s\sqrt{2}$, we have $R = \frac{s\sqrt{2}}{2}$.

The radius $r$ of the inner circle is equal to half the side length of the square, so $r = \frac{s}{2}$.

The area of the shaded region between the circles is the difference between the areas of the outer circle and the inner circle. Thus, the shaded area is:

$\text{Shaded area} = \pi R^2 - \pi r^2$

$= \pi \left(\frac{s\sqrt{2}}{2}\right)^2 - \pi \left(\frac{s}{2}\right)^2$

$= \pi \left(\frac{s^2}{2}\right) - \pi \left(\frac{s^2}{4}\right)$

$= \frac{\pi s^2}{2} - \frac{\pi s^2}{4}$

$= \frac{\pis^2}{4}$

Now, let's find the area enclosed by the inner circle. The area of a circle is given by $A = \pi r^2$, so the area enclosed by the inner circle is:

$\text{Area enclosed by inner circle} = \pi r^2$

$= \pi \left(\frac{s}{2}\right)^2$

$= \pi \left(\frac{s^2}{4}\right)$

$= \frac{\pi s^2}{4}$

Comparing the shaded area and the area enclosed by the inner circle, we see that they are equal:

$\text{Shaded area} = \frac{\pi s^2}{4} = \text{Area enclosed by inner circle}$

Therefore, we have proved that the shaded area between the two circles is the same as the area enclosed by the inner circle.