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HiSET: Math : Measurement and Geometry

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Example Questions

Example Question #7 : Cones

A cone has as its base a circle whose radius is 20% greater that of a base of the cylinder; its height is twice that of the cylinder. Which of the following is true of the volume of the cone?

Possible Answers:

The volume of the cone is 4% less than that of the cylinder.

The volume of the cone is 52% less than that of the cylinder.

The volume of the cone is equal than that of the cylinder.

The volume of the cone is 52% greater than that of the cylinder.

The volume of the cone is 4% greater than that of the cylinder.

Correct answer:

The volume of the cone is 4% less than that of the cylinder.

Explanation:

The volume of a cylinder with a base of radius and with height is

$V = \pi r^{2}h$

The cone has height twice that of the cylinder, which is . Its radius is 20% greater than, or 120% of, that of the cylinder, which is equal to 1.2r .

The volume of a cone with a base of radius and height is

$V = \frac{1}{3}\pi r'^{2}h'$

Set r'=1.2r and h' = 2h :

$V = \frac{1}{3}\pi \left ( 1.2r\right )^{2}(2h)$

$= \frac{1}{3}\pi \left ( 1.2^{2} \right )r^{2} (2h)$

$= \frac{1}{3} \left ( 1.44 \right ) (2) \cdot \pi r^{2} h$

$=0.96 \cdot \pi r^{2} h$

Substitute for $\pi r^{2}h$ :

V'= 0.96 V

This means that the volume of the cone is $0.96= 0.96 \times 100\% = 96\%$ that of the cylinder—or $(100 - 96) \% = 4 \%$ less.

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Example Question #11 : Cones

About the y-axis, rotate the triangle formed with the x-axis, the y-axis, and the line of the equation

2x + y = 8 .

Give the volume of the solid of revolution formed.

Possible Answers:

$\frac{128}{3} \pi$

None of the other choices gives the correct response.

$96 \pi$

$\frac{256}{3} \pi$

$48 \pi$

Correct answer:

$\frac{128}{3} \pi$

Explanation:

The vertices of the triangle are the points of intersection of the three lines. The -axis and the -axis meet at the origin (0,0) .

The point of intersection of the -axis and the graph of 2x + y = 8 - the -intercept of the latter - can be found by substituting 0 for :

2x + y = 8

2x + 0= 8

2x = 8

Divide both sides by 2 to isolate :

$2x \div 2 = 8 \div 2$

x = 4

The point of intersection is at (4,0) .

Similarly, The point of intersection of the -axis and the graph of 2x + y = 8 - the -intercept of the latter - can be found by substituting 0 for :

2x + y = 8

2 (0) + y = 8

0 + y= 8

y = 8

The point of intersection is at (0,8) .

The three vertices of the triangle are at the origin, (4,0) , and (0,8) . When this triangle is rotated about the -axis, the resulting solid of revolution is a cone whose base has radius r= 4 , and which has height h= 8 . Substitute these values into the formula for the volume of a cone:

$V = \frac{1}{3} \pi r^{2}h$

$= \frac{1}{3} \pi (4)^{2}(8)$

$= \frac{1}{3}\cdot 16 \cdot 8 \cdot \pi$

$= \frac{128}{3} \pi$

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Example Question #181 : Hi Set: High School Equivalency Test: Math

A right cone has slant height 20; its base has radius 10. Which of the following gives its volume to the nearest whole number?

(Round to the nearest whole number).

Possible Answers:

3,142

942

1,885

1,814

2,094

Correct answer:

1,814

Explanation:

The volume of a cone, given radius and height , can be calculated using the formula

$V = \frac{1}{3} \pi r^{2}h$ .

We are given that r = 10 , but we are not given the value of . We are given slant height l = 20 , and since the cone is a right cone, its radius, height, and slant height can be related using the Pythagorean relation

$r^{2}+ h^{2}= l^{2}$ .

Substituting 10 for and 20 for , we can find :

$10^{2}+ h^{2}= 20^{2}$

$100+ h^{2}=400$

$100+ h^{2} - 100 =400 - 100$

$h^{2} =300$

$h = \sqrt{300 }$

Now substitute in the volume formula:

$V = \frac{1}{3} \pi r^{2}h$

$= \frac{1}{3} \pi (10^{2}) \sqrt{300}$

$\approx \frac{1}{3} \cdot 3.1416 \cdot 100 \cdot 17.3$

$\approx 1813.8$

To the nearest whole, this is 1,814.

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Example Question #13 : Cones

The volume of a cone is $12.56\;\textup{cm}^3$ . Which of the following most closely approximates the radius of the base of the cone if its height is $1\;\textup{cm}$ ?

Possible Answers:

$\frac{3}{2}\;\textup{cm}$

$\sqrt{2}\;\textup{cm}$

$\sqrt{12}\;\textup{cm}$

$3\;\textup{cm}$

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

Correct answer:

$\sqrt{12}\;\textup{cm}$

Explanation:

The volume of a cone is given by the formula

$V= \frac{1}{3}\pi r^2h$

We are given that h=1 . Thus, the volume of our cone is given by

$V= \frac{1}{3}\pi r^2$ .

We are given that the volume of our cone is 12.56 .

Thus,

$12.56=\frac{1}{3}\pi r^2$

$\frac{12.56 \times3}{\pi} = r^2$

and

12 = r^2 .

Thus,

$\sqrt{12}=r$ .

so the correct answer is $\sqrt{12}\;cm$ .

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Example Question #1 : Spheres

The surface area of a sphere is equal to $100 \pi$ . Give the volume of the sphere.

Possible Answers:

$\frac{500}{3}\pi$

$\frac{1,000}{3}\pi$

$\frac{250 }{3}\pi$

$250 \pi$

$500 \pi$

Correct answer:

$\frac{500}{3}\pi$

Explanation:

The surface area of a sphere can be calculated using the formula

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

Solving for :

Set $A = 100 \pi$ and divide both sides by $4 \pi$ :

$4 \pi r^{2} = 100 \pi$

$\frac{4 \pi r^{2}}{4 \pi} = \frac{100 \pi}{4 \pi}$

$r^{2} =25$

Take the square root of both sides:

$r = \sqrt{25} = 5$

Set r=5 in the volume formula:

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

$V= \frac{4}{3} \pi \left (5 \right )^{3} = \frac{4}{3} \pi \left (125 \right ) = \frac{500}{3}\pi$ ,

the correct response.

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Example Question #181 : Hi Set: High School Equivalency Test: Math

A sphere has surface area $64 \pi$ . Give its volume.

Possible Answers:

$64 \pi$

$128 \pi$

$\frac{64}{3}\pi$

$256 \pi$

$\frac{256}{3}\pi$

Correct answer:

$\frac{256}{3}\pi$

Explanation:

The surface area of a sphere, given its radius , is equal to

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

Substitute $64\pi$ for and solve for :

$4 \pi r^{2} = 64 \pi$

Divide by $4 \pi$ :

$\frac{4 \pi r^{2}}{4 \pi} = \frac{64 \pi}{4 \pi}$

$r^{2} = 16$

Extract the square root:

r= 4

Substitute for in the volume formula:

$V = \frac{4}{3} \pi r^{3}= \frac{4}{3} \pi \cdot 4^{3} =\frac{4}{3} \pi \cdot 64 = \frac{256}{3}\pi$ ,

the correct response.

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Example Question #1 : Apply Concepts Of Density In Modeling Situations

A farmer has to make a square pen to hold chickens. If each chicken has to have $1 \;\textup{m}^2$ of area to roam and there are chickens total, what is the length of the amount of fencing required to pen in the chickens?

Possible Answers:

$20\;\textup{m}$

$75 \;\textup{m}$

$83\;\textup{m}$

$50 \;\textup{m}$

$25 \;\textup{m}$

Correct answer:

$20\;\textup{m}$

Explanation:

A square has area formula

A = l^2

The total area required for the chickens will be

$25 \;\textup{m}^2$

since each chicken requires $1\;\textup{m}^2$ of space and there are chickens.

Thus, we have

25 = l^2

for the length of our chicken fence.

$\sqrt{25}=\sqrt{l^2}^$

l = 5

Since there are four sides of a square and each side has a length of 5, the total length of fence required is

$5\times4 = 20\;\textup{m}$ .

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Example Question #1 : Apply Concepts Of Density In Modeling Situations

Census

Source: United States Census Bureau

The average population density of a geographic area is defined to be the average number of residents per square mile.

Above is a table with the land areas and populations of five states.Which state among the five has the greatest population density?

Possible Answers:

Kentucky

Mississippi

Arkansas

Tennessee

Alabama

Correct answer:

Tennessee

Explanation:

For each state, divide the population by the land area. We can round each figure to the nearest whole for simplicity's sake.

Alabama:

$4,860,000 \div 50,645 \approx 96$ persons per square mile.

Arkansas:

$2,980,000 \div 52,035 \approx 57$ persons per square mile.

Kentucky:

$4,430,000 \div 39,486 \approx 112$ persons per square mile.

Mississippi:

$2,990,000 \div 46,923 \approx 64$ persons per square mile.

Tennessee:

$6,600,000 \div 41,235 \approx 160$ persons per square mile.

Tennessee has the greatest population density among the five states.

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Example Question #3 : Apply Concepts Of Density In Modeling Situations

Census

Source: United States Census Bureau

The average population density of a geographic area is defined to be the average number of residents per square mile.

Above is a table with the population densities and the land areas of five states. Of the five, which state has the greatest population?

Possible Answers:

South Dakota

North Dakota

Vermont

Wyoming

Alaska

Correct answer:

South Dakota

Explanation:

Multiply the population density of each state by its corresponding area to get an estimate of the population (round to the nearest thousand for simplicity's sake):

Alaska:

$1.3 \times 570,641 \approx 742,000$

North Dakota:

$11.0 \times 69,001 \approx 759,000$

South Dakota:

$11.1 \times 75,811 \approx 842,000$

Vermont:

$67.9 \times 9,217 \approx 626,000$

Wyoming:

$6.0 \times 97,093 \approx 583,000$

Of the five states, South Dakota is the most populous.

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Example Question #1 : Apply Concepts Of Density In Modeling Situations

County 1

The average population density of a geographic area is defined to be the average number of residents per square mile.

Above is the map of a county whose population is about 120,000. Which of the following is the best estimate of the average population density?

Possible Answers:

150 persons per square mile

50 persons per square mile

100 persons per square mile

200 persons per square mile

250 persons per square mile

Correct answer:

100 persons per square mile

Explanation:

The county is in the shape of a trapezoid with bases of length B = 44 and b= 33 , and with height h = 30 . Its area in square miles can be found by substituting in the formula for the area of a trapezoid:

$A = \frac{1}{2} (B+b)h$

$A = \frac{1}{2} (44+33) (30)$

$A = \frac{1}{2} (77) (30)$

A = 1,155 square miles

Divide the population by this area to obtain an estimate of the population density:

$120,000\div 1,155 \approx 44.8 \approx 103.9$ persons per square mile.

Of the given choices, 100 persons per square mile comes closest.

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HiSET: Math : Measurement and Geometry

Study concepts, example questions & explanations for HiSET: Math

All HiSET: Math Resources

Example Questions

Example Question #7 : Cones

Example Question #11 : Cones

Example Question #181 : Hi Set: High School Equivalency Test: Math

Example Question #13 : Cones

Example Question #1 : Spheres

Example Question #181 : Hi Set: High School Equivalency Test: Math

Example Question #1 : Apply Concepts Of Density In Modeling Situations

Example Question #1 : Apply Concepts Of Density In Modeling Situations

Example Question #3 : Apply Concepts Of Density In Modeling Situations

Example Question #1 : Apply Concepts Of Density In Modeling Situations

All HiSET: Math Resources

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