Angles - GED Math

Card 0 of 10

Question

What is the measure of each angle of a regular octagon?

Answer

The sum of the degree measures of the angles of a polygon with sides is . Since an octagon has eight sides, substitute to get:

$180(8-2 ) = 180 \times6 = 1,080$

Each angle of a regular polygon has equal measure, so divide this by 8 to get the measure of one angle:

$1,080 \div 8 = 135$ , the degree measure of one angle.

Compare your answer with the correct one above

Question

Give the number of sides of a regular polygon whose interior angles have measure $170^{\circ }$ .

Answer

The easiest way to solve this is to look at the exterior angles, each of which have measure $(180-170)^{\circ } = 10^{\circ }$ . Since each exterior angle of a regular polygon with sides is $\frac{360}{N}$ , we solve for in the following equation:

$\frac{360}{N} = 10$

$\frac{360}{N} \cdot \frac{N }{10} = 10 \cdot \frac{N }{10}$

The polygon has 36 sides.

Compare your answer with the correct one above

Question

Give the measure of each interior angle of a regular 72-sided polygon.

Answer

A regular polygon with sides has interior angles of measure $\frac{(n-2)180^{\circ }}{n}$ each. Substitute 72 for .

$\frac{(72-2)180^{\circ }}{72}= \frac{70 \times 180^{\circ }}{72} = 175^{\circ }$

Compare your answer with the correct one above

Question

Note: Figure NOT drawn to scale.

The above hexagon is regular. What is ?

Answer

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

$\frac{180^{\circ } (6-2)}{6}= 120^{\circ }$ .

The four angles of the quadrilateral are $120 ^{\circ }, 120^{\circ }, 45^{\circ }, x^{\circ }$ . Their sum is $360^{\circ }$ , so we can set up an equation and solve for :

Compare your answer with the correct one above

Question

The above octagon is regular. What is ?

Answer

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

$\frac{180^{\circ } (8-2)}{8}= 135^{\circ }$ .

The five angles of the pentagon are $135^{\circ }, 135^{\circ }, 135^{\circ }, 80^{\circ }, y^{\circ }$ . Their sum is $180^{\circ } (5-3) = 540^{\circ }$ , so we can set up an equation and solve for :

Compare your answer with the correct one above

Question

Refer to the above diagram.

Which of these is a valid alternative name for $\angle FEC$ ?

Answer

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, $\angle FEC$ includes on one side, making one of its sides $\overrightarrow{EF}$ , and on the other, making the other side $\overrightarrow{EC}$ .

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

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

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

The only possible choice is $\angle CEA$ .

Compare your answer with the correct one above

Question

Refer to the above figure. $\Delta OKR$ is equilateral, and Quadrilateral is a square.

Evaluate $m \angle ROP$ .

Answer

By angle addition,

$m \angle ROP = m \angle ROK + m \angle KOP$ .

$\angle ROK$ , as an angle of an equilateral triangle, has measure $60 ^{\circ }$ .

$\angle KOP$ , as an angle of a square, has measure $90 ^{\circ }$ .

Therefore,

$m \angle ROP = 60 ^{\circ } + 90^{\circ } = 150 ^{\circ }$ .

Compare your answer with the correct one above

Question

Refer to the above figure, which shows Square and regular Pentagon .

Evaluate $m \angle NOQ$ .

Answer

By angle addition,

$m \angle NOQ = m \angle NOK + m \angle KOQ$ .

$\angle NOK$ is an angle of a regular pentagon and has measure $\frac{180^{\circ}\times (5-2)}{5} = 108 ^{\circ}$ .

$\angle KOQ$ is one of two acute angles of isosceles right triangle $\Delta KOQ$ , so $m \angle KOQ = 45^{\circ }$ .

$m \angle NOQ = 108^{\circ } + 45^{\circ } = 153^{\circ }$

Compare your answer with the correct one above

Question

Note: Figure NOT drawn to scale.

Refer to the above figure. $\Delta OKR$ is equilateral and Pentagon is regular.

Evaluate $m \angle NRO$ .

Answer

First, we find $m \angle NOR$ .

By angle addition,

$m \angle NOR + m \angle ROK = m \angle NOK$ .

$\angle NOK$ is an angle of a regular pentagon and has measure $\frac{180^{\circ}\times (5-2)}{5} = 108 ^{\circ}$ .

$\angle ROK$ , as an angle of an equilateral triangle, has measure $60 ^{\circ }$ .

$m \angle NOR + 60 ^{\circ }= 108 ^{\circ }$

$m \angle NOR = 48 ^{\circ }$

$\Delta OKR$ is equilateral, so $\overline{OR} \cong \overline{OK}$ ; Pentagon is regular, so $\overline{OK} \cong \overline{NO}$ . Therefore, $\overline{OR} \cong \overline{ON}$ , and by the Isosceles Triangle Theorem, $m \angle NRO = m \angle RNO$ .

The degree measures of three angles of a triangle total $180 ^{\circ }$ , so:

$m \angle NRO + m \angle RNO + m \angle NOR = 180^{\circ }$

$m \angle NRO + m \angle NRO + 48^{\circ } = 180^{\circ }$

$2 m \angle NRO + 48^{\circ } = 180^{\circ }$

$2 m \angle NRO = 132^{\circ }$

$m \angle NRO = 66^{\circ }$

Compare your answer with the correct one above

Question

Three consecutive even angles add up to . What must be the value of the second largest angle?

Answer

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.

$\frac{3x}{3}=\frac{174}{3}$

The second largest angle is .

Substitute the value of in to the expression.

The answer is:

Compare your answer with the correct one above

Tap the card to reveal the answer