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

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 .

For square A, s = 5.

For square B, s = 25.

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.

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.

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

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).

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

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.

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

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

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.

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.

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.

Find the area of the square.

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

We start by rearranging the equation into the form y = mx + b (where m is the slope and b is the y intercept); y = 4x – 22
Now we know the slope is 4 and so the equation we are looking for must have the m = 4 because the lines are parallel. We are also told that the equation must pass through the origin; this means that b = 0.

In 4x – y = 0 we can rearrange to get y = 4x. This fulfills both requirements.

The given equation is in standard form and needs to be converted to slope-intercept form which gives y = –2/5x + 6/5. The parallel line will have a slope of –2/5 (the same slope as the old line). The slope and the given point are substituted back into the slope-intercept form to yield y = –2/5x +5.

The slope of the given line is  and a parallel line would have the same slope, so we need to find a line through  with a slope of 2 by using the slope-intercept form of the equation for a line.  The resulting line is  which needs to be converted to the standard form to get .