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HiSET: Math : HiSet: High School Equivalency Test: Math

Study concepts, example questions & explanations for HiSET: Math

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Example Questions

Example Question #2 : Angle Measure, Central Angles, And Inscribed Angles

Inscribed heptagon

The above figure shows a regular seven-sided polygon, or heptagon, inscribed inside a circle. is the common center of the figures.

Give the measure of $\angle DBO$ .

Possible Answers:

$63\frac{3}{7}^{\circ }$

$65 \frac{1}{7}^{\circ }$

$62 \frac{4}{7}^{\circ }$

$64 \frac{2}{7}^{\circ }$

$61 \frac{5}{7}^{\circ }$

Correct answer:

$64 \frac{2}{7}^{\circ }$

Explanation:

Consider the figure below, which adds some radii of the heptagon (and circle):

Inscribed heptagon

$\overline{OB}$ , as a radius of a regular polygon, bisects $\angle EBD$ . The measure of this angle can be calculated using the formula

$m = \frac{180 (n-2)}{n}$ ,

where n = 7 :

$m \angle EBD = \frac{180^{\circ } (7-2)}{7} = \frac{180^{\circ } (5)}{7} = \frac{900}{7}^{\circ } = 1 28 \frac{4}{7}^{\circ }$

Consequently,

$\angle DBO = \frac{1}{2} \cdot m \angle EBD = \frac{1}{2} \cdot 1 28 \frac{4}{7}^{\circ } = 64 \frac{2}{7}^{\circ }$ ,

the correct response.

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Example Question #1 : Properties Of Polygons And Circles

Inscribed heptagon

The above figure shows a regular seven-sided polygon, or heptagon, inscribed inside a circle. is the common center of the figures.

Give the measure of $\angle AOB$ .

Possible Answers:

$156 \frac{4}{7}^{\circ }$

$155 \frac{1}{7}^{\circ }$

$154 \frac{2}{7}^{\circ }$

$152 \frac{6}{7}^{\circ }$

$153 \frac{5}{7}^{\circ }$

Correct answer:

$154 \frac{2}{7}^{\circ }$

Explanation:

Examine the diagram below, which divides $\angle AOB$ into three congruent angles, one of which is $\angle 1$ :

Inscribed heptagon

The measure of a central angle of a regular -sided polygon which intercepts one side of the polygon is $\frac{360 ^{\circ }}{n}$ ; setting n = 7 , the measure of $\angle 1$ is

$m \angle 1 = \frac{360 ^{\circ }}{7} = 51 \frac{3}{7}^{\circ }$ . $\angle AOC$ has measure three times this; that is,

$m \angle AOC = 3 \times51 \frac{3}{7}^{\circ } = 154 \frac{2}{7}^{\circ }$

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Example Question #11 : Measurement And Geometry

Inscribed heptagon

The above figure shows a regular seven-sided polygon, or heptagon, inscribed inside a circle. is the common center of the figures.

Give the measure of $\angle AOC$ .

Possible Answers:

$105 \frac{1}{7}^{\circ }$

$101 \frac{3}{7}^{\circ }$

$103 \frac{5}{7}^{\circ }$

$102 \frac{6}{7}^{\circ }$

$104 \frac{2}{7}^{\circ }$

Correct answer:

$102 \frac{6}{7}^{\circ }$

Explanation:

Examine the diagram below, which divides $\angle AOC$ into two congruent angles, one of which is $\angle 1$ :

Inscribed heptagon

The measure of a central angle of a regular -sided polygon which intercepts one side of the polygon is $\frac{360 ^{\circ }}{n}$ ; setting n = 7 , the measure of $\angle 1$ is

$m \angle 1 = \frac{360 ^{\circ }}{7} = 51 \frac{3}{7}^{\circ }$ . $\angle AOC$ has measure twice this; that is,

$m \angle AOC = 2 \times51 \frac{3}{7}^{\circ } = 102 \frac{6}{7}^{\circ }$

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Example Question #6 : Angle Measure, Central Angles, And Inscribed Angles

Inscribed nonagon

The above figure shows a regular ten-sided polygon, or decagon, inscribed inside a circle. is the common center of the figures.

Give the measure of $\angle AOB$ .

Possible Answers:

$75^{\circ }$

$72^{\circ }$

$80^{\circ }$

$84^{\circ }$

$76^{\circ }$

Correct answer:

$72^{\circ }$

Explanation:

Examine the diagram below, which divides $\angle AOB$ into two congruent angles, one of which is $\angle 1$ :

Inscribed nonagon

The measure of a central angle of a regular -sided polygon which intercepts one side of the polygon is $\frac{360 ^{\circ }}{n}$ ; setting n = 10 , the measure of $\angle 1$ is

$m \angle 1 = \frac{360 ^{\circ }}{10} =36^{\circ }$ . $\angle AOB$ has measure twice this; that is,

$m \angle AOB = 2 \times 36 ^{\circ } = 72 ^{\circ }$ .

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Example Question #7 : Angle Measure, Central Angles, And Inscribed Angles

Inscribed nonagon

The above figure shows a regular ten-sided polygon, or decagon, inscribed inside a circle. is the common center of the figures.

Give the measure of $\angle OBC$ .

Possible Answers:

$32^{\circ }$

$42^{\circ }$

$45^{\circ }$

$40^{\circ }$

$36 ^{\circ }$

Correct answer:

$36 ^{\circ }$

Explanation:

Consider the triangle $\bigtriangleup OBC$ . Since $\overline{OB}$ and $\overline{OC}$ are radii, they are congruent, and by the Isosceles Triangle Theorem, $m \angle OBC = m \angle OCB$ .

Now, examine the figure below, which divides $\angle BOC$ into three congruent angles, one of which is $\angle 1$ :

Inscribed nonagon

The measure of a central angle of a regular -sided polygon which intercepts one side of the polygon is $\frac{360 ^{\circ }}{n}$ ; setting n = 10 , the measure of $\angle 1$ is

$m \angle 1 = \frac{360 ^{\circ }}{10} =36^{\circ }$ . $\angle BOC$ has measure three times this; that is,

$m \angle BOC = 3 \times 36 ^{\circ } = 108 ^{\circ }$ .

The measures of the interior angles of a triangle total $108^{\circ }$ , so

$m \angle OCB + m \angle OBC+m \angle BOC = 180$

Substituting 108 for $\angle BOC$ and $m \angle OBC$ for $m \angle OCB$ :

$m \angle OBC+ m \angle OBC+108 = 180$

$2 m \angle OBC+108 = 180$

$2 m \angle OBC+108- 108 = 180- 108$

$2 m \angle OBC=72$

$2 m \angle OBC \div 2 =72 \div 2$

$m \angle OBC= 36$

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Example Question #1 : Angle Measure, Central Angles, And Inscribed Angles

Inscribed nonagon

The above figure shows a regular ten-sided polygon, or decagon, inscribed inside a circle. is the common center of the figures.

Give the measure of $\angle CBE$ .

Possible Answers:

$30^{\circ }$

$32^{\circ }$

$42^{\circ }$

$36^{\circ }$

$40^{\circ }$

Correct answer:

$36^{\circ }$

Explanation:

Through symmetry, it can be seen that Quadrilateral BCDE is a trapezoid, such that $\overline{BC} || \overline{DE }$ . By the Same-Side Interior Angle Theorem, $\angle CBE$ and $\angle DEB$ are supplementary - that is,

$m \angle CBE +\angle DEB = 180$ .

The measure of $\angle DEB$ can be calculated using the formula

$m = \frac{180 (n-2)}{n}$ ,

where n = 10 :

$m \angle DEB= \frac{180^{\circ } (10-2)}{10} = \frac{180^{\circ } (8)}{10} = \frac{1,440}{10}^{\circ } =144 ^{\circ }$

Substituting:

$m \angle CBE +144= 180$

$m \angle CBE +144 - 144 = 180 - 144$

$m \angle CBE = 36$

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Example Question #1 : Properties Of Polygons And Circles

If two angles are supplementary and one angle measures $95^{\circ}$ , what is the measurement of the second angle?

Possible Answers:

$80^{\circ}$

$90^{\circ}$

$95^{\circ}$

$85^{\circ}$

Correct answer:

$85^{\circ}$

Explanation:

Step 1: Define supplementary angles. Supplementary angles are two angles whose sum is $180^{\circ}$ .

Step 2: Find the other angle by subtracting the given angle from the maximum sum of the two angles.

So, $180^{\circ}-95^{\circ}=85^\circ$

The missing angle (or second angle) is $85^{\circ}$

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Example Question #2 : Properties Of Polygons And Circles

$\angle 1$ and $\angle 2$ are complementary angles.

$\angle 1$ and $\angle 3$ are supplementary angles.

$\angle 2 \cong \angle 4$

$m \angle 3= 112^{\circ }$

Evaluate $m \angle 4$ .

Possible Answers:

$112^{\circ }$

$90^{\circ }$

$158^{\circ }$

$68^{\circ }$

$22^{\circ }$

Correct answer:

$22^{\circ }$

Explanation:

$\angle 1$ and $\angle 3$ are supplementary angles, so, by definition,

$m \angle 1 + m \angle 3 = 180^{\circ }$

$m \angle 3= 112^{\circ }$ , so substitute and solve for $m \angle 1$ :

$m \angle 1 + 112^{\circ } = 180^{\circ }$

$m \angle 1 + 112^{\circ } -112^{\circ } = 180^{\circ }-112^{\circ }$

$m \angle 1 = 68 ^{\circ }$

$\angle 1$ and $\angle 2$ are complementary angles, so, by definition,

$m \angle 1 + m \angle 2 = 90^{\circ }$

Substitute and solve for $m \angle 2$ :

$68 ^{\circ } + m \angle 2 = 90^{\circ }$

$68 ^{\circ } + m \angle 2 - 68 ^{\circ } = 90^{\circ }-68 ^{\circ }$

$m \angle 2 =22^{\circ }$

$\angle 2 \cong \angle 4$ - that is, the angles have the same measure. Therefore,

$m \angle 4 =22^{\circ }$ .

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Example Question #71 : Hi Set: High School Equivalency Test: Math

$\angle 1$ and $\angle 2$ are a pair of vertical angles.

$\angle 3$ and $\angle 4$ are a linear pair.

$\angle 2$ and $\angle 3$ are the two acute angles of a right triangle.

Which of the following must be true?

Possible Answers:

$m \angle 1 = m \angle 4$

$m \angle 4 + 90^{\circ }= m \angle 1$

$m \angle 1 + 90^{\circ }= m \angle 4$

$m \angle 1 +m \angle 4 = 180^{\circ }$

$m \angle 1 +m \angle 4 = 90^{\circ }$

Correct answer:

$m \angle 1 + 90^{\circ }= m \angle 4$

Explanation:

$\angle 1$ and $\angle 2$ are a pair of vertical angles; it follows that

$m \angle 1 = m \angle 2$

$\angle 3$ and $\angle 4$ are a linear pair; it follows that they are supplementary - that is,

$m \angle 3 + m \angle 4 = 180^{\circ }$ .

$\angle 2$ and $\angle 3$ are the two acute angles of a right triangle; it follows that they are complementary - that is,

$m \angle 2 + m \angle 3 = 90^{\circ }$ .

Therefore, we have the three statements

$m \angle 1 = m \angle 2$

$m \angle 2 + m \angle 3 = 90^{\circ }$

$m \angle 3 + m \angle 4 = 180^{\circ }$

From the second statement, we can subtract $m \angle 2$ from both sides to get

$m \angle 2 + m \angle 3 - m \angle 2 = 90^{\circ } - m \angle 2$

$m \angle 3 = 90^{\circ } - m \angle 2$

Substitute this expression for $m \angle 3$ in the third expression to get

$90^{\circ } - m \angle 2 + m \angle 4 = 180^{\circ }$

Substitute $m \angle 1$ for $m \angle 2$ :

$90^{\circ } - m \angle 1 + m \angle 4 = 180^{\circ }$

Add $m \angle 1 - 90^{\circ }$ to both sides:

$90^{\circ } - m \angle 1 + m \angle 4 +m \angle 1 - 90^{\circ }= 180^{\circ }+m \angle 1 - 90^{\circ }$

$m \angle 4 = 90^{\circ }+m \angle 1$ ,

or, rearranged,

$m \angle 1 + 90^{\circ }= m \angle 4$ .

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Example Question #72 : Hi Set: High School Equivalency Test: Math

A five sided irregular polygon has sides of the following lengths:

$1,\ 4,\ 7,\ 8,\ \textup{and } 13$

Find its perimeter.

Possible Answers:

Correct answer:

Explanation:

Perimeters can be calculated using the following formula.

P=s_1+s_2+s_3...

In this formula, the variable, , represents a side of the polygon.

Substitute and solve.

P=1+4+7+8+13

P=33

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HiSET: Math : HiSet: High School Equivalency Test: Math

Study concepts, example questions & explanations for HiSET: Math

All HiSET: Math Resources

Example Questions

Example Question #2 : Angle Measure, Central Angles, And Inscribed Angles

Example Question #1 : Properties Of Polygons And Circles

Example Question #11 : Measurement And Geometry

Example Question #6 : Angle Measure, Central Angles, And Inscribed Angles

Example Question #7 : Angle Measure, Central Angles, And Inscribed Angles

Example Question #1 : Angle Measure, Central Angles, And Inscribed Angles

Example Question #1 : Properties Of Polygons And Circles

Example Question #2 : Properties Of Polygons And Circles

Example Question #71 : Hi Set: High School Equivalency Test: Math

Example Question #72 : Hi Set: High School Equivalency Test: Math

All HiSET: Math Resources

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