The figure below shows circle O inscribed inside square ABCD. Find the area of the shaded region to the nearest tenth.

To find the area of the shaded region, we need to subtract the area of the circle from the area of the square.

The diagonal of the square is equal to the diameter of the circle, which is 10 units. Since a square's diagonal is $\sqrt{2}$ times the length of a side, each side of the square is $10/\sqrt{2} = 10\sqrt{2}/2 = 5\sqrt{2}$ units long.

The area of the square is $(5\sqrt{2})^2 = 25 \cdot 2 = 50$ square units.

The area of the circle is $\pi r^2$, but since we are only given a radius of 5 units, we will need to approximate $\pi$ as 3.14.

So the area of the circle is $3.14 \cdot 5^2 = 3.14 \cdot 25 = 78.5$ square units.

Therefore, the area of the shaded region is $50 - 78.5 = -28.5$ square units.

However, the area of a shape cannot be negative, so there must be an error in the calculation or the diagram.

Please provide more information or check your calculations to correct the error.

To find the area of the shaded region, we need to subtract the area of the circle from the area of the square.

Let's start by finding the area of the square, which is a side multiplied by itself. Since the square is inscribed in the circle, the diameter of the circle is equal to the side length of the square.

Let's assume the side length of the square is 's'.

The formula for the area of the square is: Area_square = s^2

Now, let's find the area of the circle. The formula for the area of a circle is: Area_circle = πr^2

Since the diameter of the circle is equal to the side length of the square, the radius of the circle is half of the side length. Therefore, the radius (r) of the circle is 's/2'.

Now we can substitute the value of r in the formula for the area of the circle.

Area_circle = π(s/2)^2

Now let's calculate the area of the square, using the length of a side, s.

Area_square = s^2

Next, we can find the area of the shaded region by subtracting the area of the circle from the area of the square.

Area_shaded = Area_square - Area_circle

Now we can substitute the formulas we derived earlier to find the area of the shaded region.

Area_shaded = s^2 - π(s/2)^2

Let's calculate the area of the shaded region, rounding to the nearest tenth.

Would you like to provide the dimensions of the square?

To find the area of the shaded region, we need to subtract the area of the circle from the area of the square.

First, let's find the area of the square. The area of a square can be found by multiplying the length of one side by itself. Since all sides of the square are equal, we can choose any side length. Let's choose the length of one side of the square as "s".

Given that the circle is inscribed inside the square, the diagonal of the square is equal to the diameter of the circle. The diameter of the circle is twice the length of the radius. Let's call the radius of the circle "r".

Since the diagonal of the square is equal to the diameter of the circle, we have:

diagonal of square = 2 * radius of circle

By using the Pythagorean theorem, we can find the length of the diagonal of the square:

diagonal^2 = (side length)^2 + (side length)^2
diagonal^2 = 2(side length)^2
(s^2) = diagonal^2 / 2

Now, let's substitute the value of the radius into the equation:

r^2 = (diagonal^2) / 2

Simplifying further, we have:

s^2 = 2 * r^2
s = sqrt(2 * r^2) = sqrt(2) * r

Therefore, the side length of the square is sqrt(2) times the radius of the circle.

Now, let's find the area of the square:

Area of square = (side length)^2
Area of square = (sqrt(2) * r)^2
Area of square = 2 * r^2

Next, let's find the area of the circle. The area of a circle is given by the formula:

Area of circle = π * (radius^2)
Area of circle = π * r^2

Now, let's subtract the area of the circle from the area of the square to find the area of the shaded region:

Area of shaded region = Area of square - Area of circle
Area of shaded region = 2 * r^2 - π * r^2

Finally, we can calculate the area of the shaded region to the nearest tenth by plugging in the appropriate values for r and π.