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HiSET: Math : Properties of polygons and circles

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Example Questions

Example Question #21 : Properties Of Polygons And Circles

Intercepted 2

is the center of the above circle, and AO = 8 . Evaluate the length of $\overarc{ACB}$ .

Possible Answers:

$3 \pi$

$6 \pi$

$10 \pi$

$5 \pi$

$12 \pi$

Correct answer:

$10 \pi$

Explanation:

The radius of the circle is given to be r= AO = 8 .

The total circumference of the circle is $2 \pi$ times this, or

$C = 2 \pi r = 2 \pi (8)= 16\pi$ .

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

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

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 $\overarc{ACB}$ .

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

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

$= \frac{225}{360} \cdot 16 \pi$

$= \frac{5}{8} \cdot 16 \pi$

$= \frac{80}{8} \pi$

$= 10 \pi$

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Example Question #84 : Hi Set: High School Equivalency Test: Math

Intercepted

Refer to the above figure. Give the ratio of the area of Sector 2 to that of Sector 1.

Possible Answers:

4:1

5:1

6:1

25:1

16:1

Correct answer:

5:1

Explanation:

The ratio of the area of the larger Sector 2 to that of smaller Sector 1 is equal to the ratio of their respective arc measures - that is,

$m \overarc{ACB} : m \overarc{AD}$ .

Therefore, it is sufficient to find these arc measures.

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 $\overarc{ACB}$ .

It follows that

$m \overarc{AB} = 360 - 225 = 135$

By the Arc Addition Principle,

$m \overarc{AB} = m \overarc{AD} + m \overarc{DB}$

Since $\angle DOB$ , the central angle which intercepts $\overarc{DB}$ , is a right angle, $m \overarc{DB} = 90$ . By substitution,

$135 = m \overarc{AD} + 90$ ,

and

$135 - 90 = m \overarc{AD} + 90 - 90$

$45= m \overarc{AD}$

The ratio $m \overarc{ACB} : m \overarc{AD}$ is equal to

$\frac{225}{45} = \frac{225 \div 45}{45 \div 45 } = \frac{5}{1}$ ,

a 5 to 1 ratio.

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Example Question #1 : Area

Find the area of a square with the following side length:

s=5

Possible Answers:

Correct answer:

Explanation:

We can find the area of a circle using the following formula:

A=s^2

In this equation the variable, , represents the length of a single side.

Substitute and solve.

A=5^2

A=25

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Example Question #1 : Area

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

Possible Answers:

$\frac{1}{4}P^{2}$

$\frac{1}{16} \sqrt{P}$

$\frac{1}{4} \sqrt{P}$

$\frac{1}{16}P^{2}$

Correct answer:

$\frac{1}{16}P^{2}$

Explanation:

Since a square comprises four segments of the same length, the length of one side is equal to one fourth of the perimeter of the square, which is $\frac{1}{4}P$ . The area of the square is equal to the square of this sidelength, or

$\left (\frac{1}{4}P \right )^{2} = \left (\frac{1}{4} \right )^{2} P^{2} = \frac{1}{16}P^{2}$ .

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Example Question #1 : Area

The volume of a sphere is equal to $108 \pi$ . Give the surface area of the sphere.

Possible Answers:

$36 \pi \sqrt[3]{3}$

$12 \pi \sqrt[3]{9}$

$36 \pi \sqrt[3]{9}$

None of the other choices gives the correct response.

$12 \pi \sqrt[3]{3}$

Correct answer:

$36 \pi \sqrt[3]{9}$

Explanation:

The volume of a sphere can be calculated using the formula

$V= \frac{4}{3} \pi r^{3}$

Solving for :

Set $V = 108 \pi$ . Multiply both sides by $\frac{3}{4}$ :

$\frac{3}{4} \cdot \frac{4}{3} \pi r^{3} = \frac{3}{4} \cdot 108 \pi$

$\pi r^{3} = 81\pi$

Divide by $\pi$ :

$\frac{\pi r^{3} }{\pi}= \frac{81\pi}{\pi}$

$r^{3} = 81$

Take the cube root of both sides:

$r = \sqrt[3]{81} = \sqrt[3]{27(3)} = \sqrt[3]{27} \cdot \sqrt[3]{3} = 3 \sqrt[3]{3}$

Now substitute for in the surface area formula:

$A = 4 \pi r^{2}$

$A = 4 \pi (3 \sqrt[3]{3})^{2}= 4 \pi (3 )^{2}( \sqrt[3]{3})^{2} = 4 \pi (9) ( \sqrt[3]{3^{2}}) = 36 \pi \sqrt[3]{9}$ ,

the correct response.

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Example Question #1 : Area

Express the area of a square plot of land 60 feet in sidelength in square yards.

Possible Answers:

600 square yards

400 square yards

200 square yards

3,600 square yards

Correct answer:

400 square yards

Explanation:

One yard is equal to three feet, so convert 60 feet to yards by dividing by conversion factor 3:

$60 \textup{ ft} \div 3 \textup{ ft / yd}= 20 \textup{ yd}$

Square this sidelength to get the area of the plot:

$20 \textup{ yd} \times 20 \textup{ yd} = 400 \textup{ yd} ^{2}$ ,

the correct response.

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Example Question #1 : Area

A square has perimeter 12y + 4 . Give its area in terms of .

Possible Answers:

$9y ^{2}+ 12y+ 4$

$9y ^{2}+ 24y+ 16$

$9y ^{2}+ 6y+ 1$

$y ^{2}+ 2y+ 1$

$y ^{2}+ 8y+ 16$

Correct answer:

$9y ^{2}+ 6y+ 1$

Explanation:

Divide the perimeter to get the length of one side of the square.

P = 12y + 4

$s= \frac{P}{4}$

$s= \frac{12y+4}{4}$

Divide each term by 4:

$s= \frac{12y}{4}+ \frac{4}{4} = 3y+ 1$

Square this sidelength to get the area of the square. The binomial can be squared by using the square of a binomial pattern:

$A= s^{2}$

$=( 3y+ 1)^{2}$

$=( 3y )^{2}+ 2 (3y)(1)+ 1^{2}$

$= 3^{2}y ^{2}+ 2 (3 )(1)y+ 1^{2}$

$=9y ^{2}+ 6y+ 1$

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Example Question #96 : Hi Set: High School Equivalency Test: Math

A cube has surface area 6. Give the surface area of the sphere that is inscribed inside it.

Possible Answers:

$\pi$

$\frac{1}{3} \pi$

$\frac{1}{2} \pi$

$\frac{1}{6} \pi$

$\frac{2}{3} \pi$

Correct answer:

$\pi$

Explanation:

A cube with surface area 6 has six faces,each with area 1. As a result, each edge of the cube has length the square root of this, which is 1.

This is the diameter of the sphere inscribed in the cube, so the radius of the sphere is half this, or $r= \frac{1}{2}$ . Substitute this for in the formula for the surface area of a sphere:

$A = 4\pi r^{2} = 4\pi \left ( \frac{1}{2} \right )^{2} = 4\pi \left ( \frac{1}{4} \right ) = \pi$ ,

the correct choice.

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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 #21 : Properties Of Polygons And Circles

Example Question #84 : Hi Set: High School Equivalency Test: Math

Example Question #1 : Area

Example Question #1 : Area

Example Question #1 : Area

Example Question #1 : Area

Example Question #1 : Area

Example Question #96 : Hi Set: High School Equivalency Test: Math

All HiSET: Math Resources

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