Geometry and Graphs - GED Math

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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.

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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.

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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 .

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

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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 :

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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 :

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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 $\angle CEA$ .

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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,

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

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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 $m \angle KOQ = 45^{\circ }$ .

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

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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:

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

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

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

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

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

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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:

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Question

In Rhombus , $m \angle R = 60 ^{\circ }$ . If $\overline{RO }$ is constructed, which of the following is true about $\Delta RHO$ ?

Answer

The figure referenced is below.

The sides of a rhombus are congruent by definition, so $\overline{RH} \cong \overline{HO}$ , making $\Delta RHO$ isosceles (and possibly equilateral).

Also, consecutive angles of a rhombus are supplementary, as they are with all parallelograms, so

$m \angle H = 180^{\circ} - m \angle HRM = 180^{\circ} - 60 ^{\circ} = 120^{\circ}$ .

$\angle H$ , having measure greater than $90^{\circ}$ , is obtuse, making $\Delta RHO$ an obtuse triangle. Also, the triangle is not equilateral, since such a triangle must have three $60 ^{\circ}$ angles.

The correct response is that $\Delta RHO$ is obtuse and isosceles, but not equilateral.

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Question

Given Quadrilateral , which of these statements would prove that it is a parallelogram?

I) $\angle A \cong \angle B$ and $\angle C \cong \angle D$

II) $\angle A \cong \angle C$ and $\angle B \cong \angle D$

III) $\angle A$ and $\angle B$ are supplementary and $\angle C$ and $\angle D$ are supplementary

Answer

Statement I asserts that two pairs of consecutive angles are congruent. This does not prove that the figure is a parallelogram. For example, an isosceles trapezoid has two pairs of congruent base angles, which are consecutive.

Statement II asserts that both pairs of opposite angles are congruent. By a theorem of geometry, this proves the quadrilateral to be a parallelogram.

Statement III asserts that two pairs of consecutive angles are supplementary. While all parallelograms have this characteristic, trapezoids do as well, so this does not prove the figure a parallelogram.

The correct response is Statement II only.

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Question

You are given Parallelogram with . Which of the following statements, along with what you are given, would be enough to prove that Parallelogram is a rectangle?

I) $m \angle A = 90 ^{\circ }$

II) $m \angle B = 90 ^{\circ }$

III) $m \angle C = 90 ^{\circ }$

Answer

A rectangle is defined as a parallelogram with four right, or $90 ^{\circ}$ , angles.

Since opposite angles of a paralellogram are congruent, if one angle measures $90 ^{\circ}$ , so does its opposite. Since consecutive angles of a paralellogram are supplementary - that is, their degree measures total $180 ^{\circ}$ - if one angle measures $90 ^{\circ}$ , then both of the neighboring angles measure $180 ^{\circ} - 90 ^{\circ} = 90 ^{\circ}$ .

In short, in a parallelogram, if one angle is right, all are right and the parallelogram is a rectangle. All three statements assert that one angle is right, so from any one, it follows that the figure is a rectangle. The correct response is Statements I, II, or III.

Note that the sidelengths are irrelevant.

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Question

If the rectangle has a width of 5 and a length of 10, what is the area of the rectangle?

Answer

Write the area for a rectangle.

Substitute the given dimensions.

The answer is:

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Question

In the figure below, find the measure of the largest angle.

Answer

Recall that in a quadrilateral, the interior angles must add up to $360^{\circ}$ .

Thus, we can solve for :

Now, to find the largest angle, plug in the value of into each expression for each angle.

The largest angle is $151^{\circ}$ .

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Question

Which of the following can be the measures of the three angles of an acute isosceles triangle?

Answer

For the triangle to be acute, all three angles must measure less than $90 ^{\circ }$ . We can eliminate $32 ^{\circ }, 32 ^{\circ }, 116^{\circ }$ and $45 ^{\circ }, 45 ^{\circ }, 90^{\circ }$ for this reason.

In an isosceles triangle, at least two angles are congruent, so we can eliminate $50 ^{\circ }, 60 ^{\circ }, 70 ^{\circ }$ .

The degree measures of the three angles of a triangle must total 180, so, since $80 ^{\circ }+ 80 ^{\circ }+ 40 ^{\circ } = 200^{\circ }$ , we can eliminate $80 ^{\circ }, 80 ^{\circ }, 40 ^{\circ }$ .

$36^{\circ }, 72 ^{\circ }, 72 ^{\circ }$ is correct.

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Question

Which of the following describes a triangle with sides of length 9 feet, 3 yards, and 90 inches?

Answer

One yard is equal to three feet, and one foot is equal to twelve inches. Therefore, 9 feet is equal to $9 \times 12 = 108$ inches, and 3 yards is equal to $3 \times 36 = 108$ inches. The triangle has sides of measure 90 inches, 108 inches, and 108 inches. Exactly two sides are of equal measure, so it is isosceles but not equilateral.

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Question

Note: Figure NOT drawn to scale.

Refer to the above triangle. Evaluate .

Answer

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

$4x \div 4 = 156 \div 4$

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Question

Note: Figure NOT drawn to scale.

Refer to the above figure. Evaluate .

Answer

The degree measures of the interior angles of a triangle total $180 ^{\circ }$ , so, if we let $x^{\circ }$ be the measure of the unmarked angle, then

Three angles with measures $67^{\circ}, y^{\circ}, y^{\circ}$ together form a straight angle, so

$2y \div 2 = 113 \div 2$

$y = 56 \frac{1}{2}$

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Question

Figure drawn to scale.

Refer to the above diagram.

Which of the following is a valid description of $\Delta EFC$ ?

Answer

One of the angles of $\Delta EFC$ - namely, $\angle EFC$ - can be seen to be an obtuse angle, as it is wider than a right angle. This makes $\Delta EFC$ , by definition, an obtuse triangle.

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