GED Math - 2-Dimensional Geometry

Find the measure of angle B if it is the supplement to angle A:

$m\angle A=115^{o}$

$m\angle B=65^{o}$

$m\angle B=21^{o}$

$m\angle B=205^{o}$

$m\angle B=5^{o}$

Explanation

If two angles are supplementary, that means the sum of their degrees of measure will add up to 180. In order to find the measure of angle B, subtract angle A from 180 like shown:

$m\angle B=180^{o}-115^{o}=65^{o}$

This gives us a final answer of 65 degrees for angle B.

A square and a circle share a center as shown by the figure below.

If the length of a side of the square is and is half the length of the radius of the circle, to the nearest hundredths place, find the area of the shaded region.

Explanation

In order to find the area of the shaded region, we will need to find the area of the circle then subtract the area of the square from it.

Start by finding the area of the square. Since we know the side length of the square, we can write the following:

$\text{Area of Square}=4^2=16$

Next, we find the area of the circle. Since the question states that the length of a side of the square is half the radius of the circle, the radius of the circle must be . Thus, we can find the area of the circle.

$\text{Area of Circle}=\pi r^2=\pi(8)^2=64\pi$

Now, subtract these two values to find the area of the shaded region.

$\text{Area of Shaded Region}=64\pi-16=185.06$

Find the area of a square with a side of .

Explanation

Write the formula for the area of a square.

Substitute the side.

The answer is:

A hexagon has a perimeter of 90in. Find the length of one side.

$15\text{in}$

$20\text{in}$

$18\text{in}$

$22\text{in}$

$9\text{in}$

Explanation

A hexagon has 6 equal sides. The formula to find perimeter of a hexagon is:

where a is the length of any side. Now, to find the length of one side, we will solve for a.

We know the perimeter of the hexagon is 90in. So, we will substitute and solve for a. We get

$90\text{in} = 6a$

$\frac{90\text{in}}{6} = \frac{6a}{6}$

$15\text{in} = a$

$a = 15\text{in}$

Therefore, the length of one side of the hexagon is 15in.

The image is not to scale.

If the sum of two angles results in a complementary angle, what is the measure of the unknown angle?

Capture2

$60^{\circ}$

$150^{\circ}$

$30^{\circ}$

$90^{\circ}$

$20^{\circ}$

Explanation

With the provided image, we are asked to solve for the measure of the unknown angle.

Capture2

First, we must understand some information before attempting to solve the problem. The problem provides the information that the two angles summed up result in a complimentary angle. This is another way to say that when we add the measures of the two angles, it will equal $90^{\circ}$ .

This becomes a problem where we solve for a missing variable now. We can call the unknown angle x. We would set this up in equation format accordingly:

Now, we can solve for x.

$x+30{\color{Red} -30}=90{\color{Red} -30}$

$x=60^{\circ}$

Therefore, the unknown angle is $60^{\circ}$ .

Two angles are complementary if they add up to:

$90^{\circ}$

$180^{\circ}$

$270^{\circ}$

$360^{\circ}$

$45^{\circ}$

Explanation

Two angles are complementary if they add up to $90^{\circ}$ .

If two angles are complementary, and one angle is $75^{\circ}$ , what is the value of the other angle?

$15^{\circ}$

$105^{\circ}$

$30^{\circ}$

$285^{\circ}$

$150^{\circ}$

Explanation

Two angles are complementary if they add up to $90^{\circ}$ . We can use the formula:

$x+y=90^{\circ}$

where x and y are the angles.

Now, we know one angle is $75^{\circ}$ . So, we will substitute and solve for the other angle. We get

$x+75^{\circ}=90^{\circ}$

$x+75^{\circ}-75^{\circ}=90^{\circ}-75^{\circ}$

$x+0^{\circ} = 15^{\circ}$

$x=15^{\circ}$

Find the the measure of angle B if it is complement of angle A:

$m\angle A=75^{o}$

$m\angle B=15^{o}$

$m\angle B=25^{o}$

$m\angle B=105^{o}$

$m\angle B=72^{o}$

Explanation

If two angles are complementary, that means the sum of their degrees of measure will add up to 90. In order to find the measure of angle B, subtract angle A from 90 like shown:

$m\angle B=90^{o}-75^{o}=15^{o}$