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HiSET: Math : Properties of polygons and circles

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Example Questions

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 #1 : Perimeter And Circumference

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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Example Question #81 : Hi Set: High School Equivalency Test: Math

The area of a square is . In terms of , give the perimeter of the square.

Possible Answers:

$\sqrt{2A}$

$2\sqrt{A}$

$4A^{2}$

$4 \sqrt{A }$

$2A^{2}$

Correct answer:

$4 \sqrt{A }$

Explanation:

The length of one side of a square is equal to the square root of its area, so, if the area of a square is , the common sidelength is $\sqrt{A}$ . Since a square comprises four sides of equal length, the perimeter is equal to four times this length, or $4\sqrt{A}$ .

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Example Question #1 : Perimeter And Circumference

The perimeter of a regular octagon is $16\sqrt{6}$ . Give the length of one side.

Possible Answers:

$4\sqrt{6}$

$2\sqrt{6}$

$8\sqrt{6}$

Correct answer:

$2\sqrt{6}$

Explanation:

A regular octagon has eight sides of equal length. Its perimeter is equal to the sum of the lengths of its sides, so the length of one side can be computed by dividing the perimeter by 8, as follows:

$l = \frac{P}{8}= \frac{16\sqrt{6}}{8} = \frac{16}{8} \sqrt{6}= 2\sqrt{6}$ ,

the correct response.

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Example Question #1 : Perimeter And Circumference

Give the perimeter of a regular octagon in yards if the length of each side is feet.

Possible Answers:

24x yards

$\frac{3}{8}x$ yards

$\frac{8}{3}x$ yards

3x+8 yards

$\frac{1}{24}x$ yards

Correct answer:

$\frac{8}{3}x$ yards

Explanation:

The perimeter of a regular octagon - the sum of the lengths of its (eight congruent) sides - is eight times the common sidelength, so the perimeter of the octagon is feet. One yard is equivalent to three feet, so divide this by conversion factor 3 to get $\frac{8x}{3}$ yards - the correct response.

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Example Question #1 : Perimeter And Circumference

Hexagon

Hexagon ABCDEF is regular. Give the perimeter of Trapezoid ABEF .

Possible Answers:

$9+ 3\sqrt{3}$

$9+ 3\sqrt{2}$

Correct answer:

Explanation:

A regular hexagon can be divided into six equilateral triangles, as follows:

Hexagon

The perimeter of Trapezoid ABEF can be seen to be 3+3+(3+3)+3 = 15 .

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Example Question #1 : Arc Length And Area Of A Sector

Circle area

The depicted circle has a radius of 3 cm. The arc length between the two points shown on the circle is $\pi$ cm. Find the area of the enclosed sector (highlighted in green).

Possible Answers:

$27\; cm^2$

$\frac{3\pi}{2}\; cm^2$

$\frac{1}{6}\; cm^2$

$\pi\; cm^2$

$\frac{\pi}{6}\; cm^2$

Correct answer:

$\frac{3\pi}{2}\; cm^2$

Explanation:

First, find the area and circumference of the circle using the radius and the following formulae for circles:

$Circumference = 2\pi r$

$Area = \pi r^2$

Substituting in 3 for yields:

$Circumference = 6\pi \ cm$

$Area = 9\pi \ cm^2$

Next, find what fraction of the total circumference is between the two points on the circle (the arc length).

$\frac{Arc \ Length }{Circumference } = \frac{\pi}{6\pi} = \frac{1}{6}$

Finally, use this fraction to calculate the area of the enclosed sector. Note that this area is proportional to the above fraction. In other words:

$\frac{Arc \ Length }{Circumference } = \frac{Sector \ Area}{Total \ Area}$

So the Sector Area is one sixth of the total area.

$\frac{1 }{6} = \frac{Sector \ Area}{9\pi}$

Cross multiply:

$Sector \ Area\times6 = 9\pi$

Divide both sides by 6, then simplify to get the final answer:

$Sector \ Area = \frac{9\pi}{6} = \frac{3\pi}{2}\; cm^2$

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Example Question #1 : Arc Length And Area Of A Sector

Intercepted 2

is the center of the above circle, and AO = 6 . Evaluate the length of $\overarc{AB}$ .

Possible Answers:

$\frac{15}{2} \pi$

$\frac{9}{4} \pi$

$6 \pi$

$\frac{15}{4} \pi$

$\frac{9}{2} \pi$

Correct answer:

$\frac{9}{2} \pi$

Explanation:

The radius of the circle is given to be r= AO = 6 . The total circumference of the circle is $2 \pi$ times this, or

$C = 2 \pi r = 2 \pi (6)= 12\pi$ .

The length of $\overarc{AB}$ is equal to

$l = \frac{m \overarc {AB}}{360} \cdot C$

Thus, it is first necessary to find the degree measure of $\overarc{AB}$ . If the sides of an angle with vertex outside the circle are both tangent to the circle, the angle formed is half the difference of the measures of the arcs. Therefore,

$\frac{m \overarc{ACB} - m \overarc{AB} }{2} = m \angle AVB$

Letting $x= m \overarc{AB}$ , since the total arc measure of a circle is 360 degrees,

$m \overarc{ACB} = 360 - x$

We are also given that

$m \angle AVB = 45$

Making substitutions, and solving for :

$\frac{(360- x )- x }{2} =45$

$\frac{360- 2 x }{2} =45$

Multiply both sides by 2:

$\frac{360- 2 x }{2} \cdot 2 =45 \cdot 2$

360- 2 x = 90

Subtract 360 from both sides:

360- 2 x - 360 = 90 - 360

-2x = -270

Divide both sides by :

$-2x \div (-2) = -270 \div (-2)$

x = 135 ,

the degree measure of $\overarc{AB}$ .

Thus, the length of $\overarc{AB}$ is

$l = \frac{m \overarc {AB}}{360} \cdot C$

$= \frac{225}{360} \cdot (12 \pi)$

$= \frac{3}{8} (12\pi)$

$= \frac{36}{8} \pi$

$= \frac{9}{2} \pi$

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Example Question #3 : Arc Length And Area Of A Sector

Intercepted 2

is the center of the above circle, and AO = 8 . Evaluate the area of the shaded sector.

Possible Answers:

The information given is insufficient to answer the question.

$40 \pi$

$24 \pi$

$12 \pi$

$20 \pi$

Correct answer:

$24 \pi$

Explanation:

The radius of the circle is given to be r= AO = 8 . The total area of the circle can be found using the area formula:

$A = \pi r^{2}$

$A = \pi (8^{2})$

$A =64 \pi$

The area of the sector is equal to

$a = \frac{m \overarc {AB}}{360} \cdot A$

$\frac{m \overarc{ACB} - m \overarc{AB} }{2} = m \angle AVB$

Letting $x= m \overarc{AB}$ , since the total arc measure of a circle is 360 degrees,

$m \overarc{ACB} = 360 - x$

We are also given that

$m \angle AVB = 45$

Making substitutions, and solving for :

$\frac{(360- x )- x }{2} =45$

$\frac{360- 2 x }{2} =45$

Multiply both sides by 2:

$\frac{360- 2 x }{2} \cdot 2 =45 \cdot 2$

360- 2 x = 90

Subtract 360 from both sides:

360- 2 x - 360 = 90 - 360

-2x = -270

Divide both sides by :

$-2x \div (-2) = -270 \div (-2)$

x = 135 ,

the degree measure of $\overarc{AB}$ .

Thus, the area of the shaded sector is

$a = \frac{m \overarc {AB}}{360} \cdot A$

$= \frac{135}{360} \cdot (64 \pi)$

$= \frac{3}{8} \cdot 64 \pi$

$= \frac{192}{8} \pi$

$=24 \pi$

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Example Question #4 : Arc Length And Area Of A Sector

Intercepted 2

is the center of the above circle, and AO = 6 . Evaluate the area of the shaded sector.

Possible Answers:

$\frac{27}{2} \pi$

$\frac{45}{2} \pi$

The information given is insufficient to answer the question.

$\frac{45}{4} \pi$

$\frac{27}{4} \pi$

Correct answer:

$\frac{45}{2} \pi$

Explanation:

The radius of the circle is given to be r= AO = 6 . The total area of the circle can be found using the area formula:

$A = \pi r^{2}$

$A = \pi (6^{2})$

$A =36\pi$

The area of the sector is equal to

$a = \frac{m \overarc {ACB}}{360} \cdot A$

Thus, it is first necessary to find the degree measure of $\overarc{ACB}$ . If the sides of an angle with vertex outside the circle are both tangent to the circle, the angle formed is half the difference of the measures of the arcs. Therefore,

$\frac{m \overarc{ACB} - m \overarc{AB} }{2} = m \angle AVB$

Letting $x= m \overarc{ACB}$ , since the total arc measure of a circle is 360 degrees,

$m \overarc{AB} = 360 - x$

We are also given that

$m \angle AVB = 45$

Making substitutions, and solving for :

$\frac{x- (360- x ) }{2} =45$

$\frac{ 2 x -360 }{2} =45$

Multiply both sides by 2:

$\frac{2 x -360}{2} \cdot 2 =45 \cdot 2$

2 x -360 = 90

Add 360 to both sides:

2 x -360+ 360 = 90 + 360

2 x= 450

Divide both sides by 2:

$2 x \div 2 = 450 \div 2$

x = 225 ,

the degree measure of $\overarc{ACB}$ .

Therefore, the area of the shaded sector is

$a = \frac{m \overarc {ACB}}{360} \cdot A$

$= \frac{225}{360} \cdot 36 \pi$

$= \frac{5}{8 } \cdot 36 \pi$

$= \frac{180}{8 } \pi$

$= \frac{45}{2} \pi$

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HiSET: Math : Properties of polygons and circles

Study concepts, example questions & explanations for HiSET: Math

All HiSET: Math Resources

Example Questions

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

Example Question #1 : Perimeter And Circumference

Example Question #81 : Hi Set: High School Equivalency Test: Math

Example Question #1 : Perimeter And Circumference

Example Question #1 : Perimeter And Circumference

Example Question #1 : Perimeter And Circumference

Example Question #1 : Arc Length And Area Of A Sector

Example Question #1 : Arc Length And Area Of A Sector

Example Question #3 : Arc Length And Area Of A Sector

Example Question #4 : Arc Length And Area Of A Sector

All HiSET: Math Resources

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