Other Shapes - GED Math

The easiest way to solve this is to look at the exterior angles, each of which have measure . Since each exterior angle of a regular polygon with sides is , we solve for in the following equation:

The polygon has 36 sides.

A regular polygon with sides has interior angles of measure each. Substitute 72 for .

Two of the angles of the quadrilateral formed are angles of a regular hexagon, so each measures

.

The four angles of the quadrilateral are . Their sum is , so we can set up an equation and solve for :

Three of the angles of the pentagon formed are angles of a regular octagon, so each measures

.

The five angles of the pentagon are . Their sum is , so we can set up an equation and solve for :

When naming an angle after three points, the middle letter must be its vertex, or the point at which its sides meet - this is . The other two letters must refer to points on its two sides. Therefore, includes on one side, making one of its sides , and on the other, making the other side .

An alternative name for this angle must be one of two things:

It can be named only after its vertex - that is, - but only if there is no ambiguity as to which angle is being named. Since more than one angle in the diagram has vertex , is not a correct choice.

It can be named after three points. Again, the middle letter must be vertex , so we can throw out and .

The only possible choice is .

By angle addition,

.

, as an angle of an equilateral triangle, has measure .

, as an angle of a square, has measure .

Therefore,

.

By angle addition,

.

is an angle of a regular pentagon and has measure .

is one of two acute angles of isosceles right triangle , so .

First, we find .

By angle addition,

.

is an angle of a regular pentagon and has measure .

, as an angle of an equilateral triangle, has measure .

is equilateral, so ; Pentagon is regular, so . Therefore, , and by the Isosceles Triangle Theorem, .

The degree measures of three angles of a triangle total , so:

Let be an even angle. The next consecutive even values are .

Set up an equation such that all angles added equal to 180.

Divide by three on both sides.

The second largest angle is .

Substitute the value of in to the expression.

The answer is:

is a right triangle with legs ; its area is half the product of its legs, which is

is a right triangle with legs

and

;

its area is half the product of its legs, which is

The shaded region is the former triangle removed from the latter triangle; its area is the difference of the two: .

This region is therefore

of .

is a right triangle with legs ; its area is half the product of its legs, which is

is a right triangle with legs

and

;

its area is half the product of its legs, which is

The shaded region is the former triangle removed from the latter triangle; its area is the difference of the two: .

A regular hexagon can be divided into six triangles, each of which can be easily proved equilateral, as seen in the diagram below:

All segments shown are congruent, and, since the diameter shown in the original diagram is 4, each sidelength is half this, or 2.

Each equilateral triangle has area

.

There are six such triangles, so the total area of the hexagon is six times this, or .

A hexagon has six sides; a regular hexagon with perimeter 36 has sidelength

.

A regular hexagon can be divided into six triangles, each of which can be easily proved equilateral, as seen in the diagram below:

Each equilateral triangle has sidelength 6, so each has area

.

The total area of the hexagon is the area of six such triangles:

Write the area of a square.

Substitute the side into the formula.

The answer is:

Examine the bottom figure, in which the bottom two sides have been connected. Note that the figure is now a rectangle cut out of a rectangle, and, since the opposite sides of a rectangle have the same length, we can fill in some of the side lengths as shown:

The figure is a 60-by-40 rectangle cut from a 100-by-100 square, so, to get the area of the figure, subtract the area of the former from that of the latter. The area of a rectangle is equal to the product of its dimensions, so the areas of the rectangle and the square are, respectively,

and

,

making the area of the figure

.

Since the circle is inscribed in the square, the diameter of the circle is the same length as the length of a square.

Start by finding the area of the square.

For the given square,

Now, because the diameter of the circle is the same as the length of a side of the square, we now also know that the radius of the circle must be . Next recall how to find the area of a circle.

Plug in the found radius to find the area of the circle.

Now, the shaded area is the area left over when the area of the circle is subtracted from the area of the square. Thus, we can write the following equation to find the area of the shaded region.

Start by drawing out the square yard and the circular pool in a way that maximizes the area of the pool.

Notice that the diameter of the pool will be the same length as the side of the square.

Since the question asks about the area that is left over after the pool is built, we can find that area by subtracting the area of the pool from the area of the square.

Start by finding the area of the square.

Next, find the area of the circular pool.

Since the diameter of the pool is , the radius of the pool must be . Recall how to find the area of a circle:

Plug in the radius of the circle.

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

A regular octagon has eight sides of equal length, so multiply the length of one side by eight:

feet.

Divide by three to get the equivalent in yards:

yards.

A polygon with eight sides is called an octagon.