We need to find the area of both the square and the circle and then subtract the two.  Inscribed means draw within a figure so as to touch in as many places as possible.  So the circle is drawn inside the square.  The opposite is circumscribed, meaning drawn outside.

A_square = s² = 36 cm² so the side is 6 cm

6 cm is also the diameter of the circle and thus the radius is 3 cm

A circle = πr² = 3.14 * 3² = 28.28 cm²

The resulting difference is 7.74 cm²

We can find the area of the shaded region by subtracting the area of the semicircles, which is much easier to find. Two semi-circles are equivalent to one full circle. Thus we can just use the area formula, where r = 6:

π(6²) → 36π

Now we must subtract the area of the semi-circles from the total area of the square. Since we know that the radius also covers half of a side, 6(2) = 12 is the full length of a side of the square. Squaring this, 12² = 144. Subtracting the area of the circles, we get our final terms,

= 144 – 36π

First find the area of both squares using the formula $\displaystyle A=s^2$.

For square A, s = 5.

$\displaystyle A_A=5^2=25$

For square B, s = 25.

$\displaystyle A_B=25^2=625$

The question is asking for the ratio of these two areas, which will tell us how many times bigger square B is. Divide the area of square B by the area of square A to find the answer.

$\displaystyle \frac{A_B}{A_A}=\frac{625}{25}=25$

To solve, we will need to find the area of the wall. We can do this by finding the areas of each section and adding them together. Break the area into a rectange and two triangles.

The area of the rectangle will be equal to the base times the height. The area of each triangle will be one half its given base times its height.

$\displaystyle A_{rec}=bh$

$\displaystyle A_{tri}=\frac{1}{2}bh$

For the rectangle, the base is 12 and the height is 4 (both given in the figure).

$\displaystyle A_{rec}=(12)(4)=48$

The triangle to the right has a given base of 6, but we need to solve for its height. The height will be equal to the difference between the total height (6) and the height of the rectangle (4).

$\displaystyle h_{tri}=6-4=2$

We now have the base and height of the triangle to the right, allowing us to calculate its area.

$\displaystyle A_{tri_r}=\frac{1}{2}(6)(2)=6$

Now we need to solve the triangle to the left. We solved for its height (2), but we still need to solve for its base. The total base of the rectangle is 12. Subtract the base of the right-side triangle (6) and the small segment at the top of the rectangle (3) from this total length to solve for the base of the left triangle.

$\displaystyle 12-6-3=3$

The left-side triangle has a base of 3 and a height of 2, allowing us to calculate its area.

$\displaystyle A_{tri_l}=\frac{1}{2}(3)(2)=3$

Add together the two triangles and the rectangle to find the total area.

$\displaystyle A=48+6+3=57$

We know that each bucket of paint will cover 5 square units, and we have 57 square units total. Divide to find how many buckets are required.

$\displaystyle \frac{57}{5}=11.4$

We will need 11 full buckets and part of a twelfth bucket to cover the wall, meaning that we will need 12 buckets total.

First, find the area of the circle.

$\displaystyle A=\pi(3\sqrt{2})^2$

$\displaystyle \small A=18\pi$

Next, find the length of 1 side of the square using the Pythagorean Theorem. Two radii from the center of the circle to adjacent corners of the square will create a right angle at the center of the circle. The radii will be the legs of the triangle and the side of the square will be the hypotenuse.

$\displaystyle \small (3\sqrt{2})^2 +(3\sqrt{2})^2 =c^2$

$\displaystyle \small 18+18=36=c^2$

$\displaystyle \small c=6$

Find the area of the square.

$\displaystyle \small A=c^2=6^2=36$

Subtract the area of the square from the area of the circle.

$\displaystyle \small 18\pi-36$

$\displaystyle \overline{BE }$ is both one side of Square $\displaystyle BCDE$ and the hypotenuse of $\displaystyle \Delta ABE$;  its hypotenuse can be calculated from the lengths of the legs using the Pythagorean Theorem:

$\displaystyle BE = \sqrt{(AB)^{2}+ \left ( AE\right )^{2}}$

$\displaystyle BE = \sqrt{x^{2}+ \left ( 2x\right )^{2}} = \sqrt{x^{2}+ 4x ^{2}} = \sqrt{5x ^{2}} = x\sqrt{5}$

Polygon $\displaystyle ABCDE$ is the composite of $\displaystyle \Delta ABE$ and Square $\displaystyle BCDE$, so its area is the sum of those of the two figures. 

The area of $\displaystyle \Delta ABE$ is half the product of its legs:

$\displaystyle A= \frac{1}{2} \cdot x \cdot 2x = x^{2}$

The area of  Square $\displaystyle BCDE$ is the square of the length of a side:

$\displaystyle A = \left ( x\sqrt{5} \right )^{2} = 5x^{2}$

Add the areas:

$\displaystyle x^{2}+ 5x^{2} = 6x^{2}$

This is the area of the polygon.

The area of  $\displaystyle ETAN$, the shaded region in question, is that of the rectangle minus those of $\displaystyle \Delta ERN$ and $\displaystyle \Delta ECT$. We look at both.

The answer is independent of the sidelengths of the rectangle, so to ease calculations - this will become more apparent later - we will arbitrarily assign to the rectangle the dimensions

$\displaystyle RN = CA = 6$

and, subsequently,

$\displaystyle RC = NA = 12$

and, since $\displaystyle T$ is the midpoint of $\displaystyle \overline{CA}$,

$\displaystyle CT = 3$.

The area of Rectangle $\displaystyle RCAN$ is equal to $\displaystyle RN \cdot RC = 6 \cdot 12 = 72$.

 Since $\displaystyle EC = 2 \cdot RE$, and $\displaystyle RE + EC = RC$,

$\displaystyle RE + 2 (RE)= 12$

$\displaystyle 3 (RE)= 12$

$\displaystyle RE = 4$ and $\displaystyle EC = 8$

The area of $\displaystyle \Delta ERN$ is equal to $\displaystyle \frac{1}{2}(RE)(RN)= \frac{1}{2} \cdot 4 \cdot 6 = 12$.

The area of $\displaystyle \Delta ECT$ is equal to $\displaystyle \frac{1}{2}(EC)(CT)= \frac{1}{2} \cdot 8 \cdot 3 = 12$

The area of the shaded region is therefore $\displaystyle 72-12-12 = 48$, which is

$\displaystyle \frac{48}{72} \times 100 \% = 66 \frac{2}{3} \%$ of the rectangle.

To determine the measure of the angles of a regular polygon use:

Angle = (n – 2) x 180° / n

Thus, (n – 2) x 180° / n = 140°

180° n - 360° = 140° n

40° n = 360°

n = 360° / 40° = 9

The formula for of interior angles based on a polygon with a number of side n is:

Each Interior  Angle = (n-2)*180/n

= (7-2)*180/7 = 128.57 degrees

This question is very misleading, because while each answer COULD be true, none of them MUST be true. Between angle A and C, onne of the angles could be very small (0.001 degrees) and the other one could be very large. For instance, if A = 89.9999 and C = 0.0001, AC = 0.009. On the other hand, the two angles could be very siimilar. If B = 90 and D = 90 then BD = 8100 and BD > AC. If we use these same values we disprove AD = BC as 8100 ≠ .009. Finally, if B is a very small value, then B/C will be very small and smaller than A/D.