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HiSET: Math : Cones

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Example Question #1 : Cones

About the -axis, rotate the triangle with its vertices at (5, 0) , (0, 6 ) , and the origin. What is the volume of the solid of revolution formed?

Possible Answers:

$50 \pi$

$60 \pi$

$55 \pi$

None of the other choices gives the correct response.

$66 \pi$

Correct answer:

$50 \pi$

Explanation:

When this triangle is rotated about the -axis, the resulting solid of revolution is a cone whose base has radius r= 5 , and which has height h= 6 . Substitute these values into the formula for the volume of a cone:

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

$= \frac{1}{3} \pi (5^{2}) (6)$

$= 25 \cdot 6 \cdot \frac{1}{3} \pi$

$= 50 \pi$

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Example Question #1 : Use Volume Formulas To Solve Problems

About the x-axis, rotate the triangle with its vertices at (5, 0) , (0, 6 ) , and the origin. What is the volume of the solid of revolution formed?

Possible Answers:

$55 \pi$

$60 \pi$

$50 \pi$

$66 \pi$

None of the other choices gives the correct response.

Correct answer:

$60 \pi$

Explanation:

When this triangle is rotated about the -axis, the resulting solid of revolution is a cone whose base has radius r= 6 , and which has height h= 5 . Substitute these values into the formula for the volume of a cone:

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

$= \frac{1}{3} \pi (6)^{2}(5)$

$= 36 \cdot 5 \cdot \frac{1}{3} \pi$

$=60 \pi$

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Example Question #1 : Use Volume Formulas To Solve Problems

A right square pyramid has height 10 and a base of perimeter 36.

Inscribe a right cone inside this pyramid. What is its volume?

Possible Answers:

$\frac{405}{2} \pi$

None of the other choices gives the correct response.

$30 \pi$

$90 \pi$

$\frac{135}{2} \pi$

Correct answer:

$\frac{135}{2} \pi$

Explanation:

The length of one side of the square is one fourth of its perimeter, or $36 \div 4 = 9$ . The cone inscribed inside this pyramid has the same height. Its base is the circle inscribed inside the square. This circle will have as its diameter the length of one side of the square, or 9, and, as its radius, half this, or $9 \div 2 = \frac{9}{2}$ .

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

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

Set $r = \frac{9}{2}$ and h = 10 :

$V = \frac{1}{3} \pi \left ( \frac{9}{2} \right )^{2} 10$

$= \frac{1}{3} \left ( \frac{81}{4} \right ) 10 \pi$

$= \frac{135}{2} \pi$

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Example Question #1 : Cones

A right square pyramid has height 10 and a base of area 36.

Inscribe a right cone inside this pyramid. What is its volume?

Possible Answers:

$\frac{405}{2} \pi$

$90 \pi$

$\frac{135}{2} \pi$

None of the other choices gives the correct response.

$30 \pi$

Correct answer:

$30 \pi$

Explanation:

The length of one side of the square is the square root of the area, or $\sqrt{36} = 6$ . The cone inscribed inside this pyramid will have as its base the circle inscribed inside the square. This circle will have as its diameter the length of one side of the square, or 6, and, as its radius, half this, or 3.

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

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

Set r = 3 and h = 10 :

$V = \frac{1}{3} \pi \left (3 \right )^{2} 10$

$= \frac{1}{3} \left ( 9 \right ) 10 \pi$

$=30 \pi$

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Example Question #1 : Cones

A cylinder has volume 120. A cone with base the same size as a base of the cylinder has the same height. Give the volume of the cone.

Possible Answers:

Insufficient information is given to answer the question.

Correct answer:

Explanation:

A cone with the same height as a given cylinder and a base the same radius as those of that cylinder has as its volume one-third that of the cylinder. That makes the volume of the cone one-third of 120, or

$V = \frac{1}{3}(120) = 40$

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Example Question #171 : Hi Set: High School Equivalency Test: Math

A right cone has height 10 and slant height 20. Which of the following correctly gives its volume? (Round to the nearest whole number).

Possible Answers:

942

1,885

1,814

3,142

2,094

Correct answer:

3,142

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 h = 10 , but we are not given the value of . We are given that 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 :

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

$r^{2}+ 100=400$

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

$r^{2} =300$ , which is what we need in the formula.

Now substitute in the volume formula:

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

$\approx \frac{1}{3}(3.1416 )(300)(10)$

$\approx 3141.6$

This rounds to 3,142

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Example Question #12 : Use Volume Formulas To Solve Problems

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

Possible Answers:

The volume of the cone is 60% less than that of the cylinder.

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

The volume of the cone is 20% greater than that of the cylinder.

The volume of the cone is 20% less than that of the cylinder.

The volume of the cone is 60% greater than that of the cylinder.

Correct answer:

The volume of the cone is 60% greater 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 radius twice that of the cylinder, which is . Its height is 20% greater than, or 120% of, that of the cylinder, which is equal to 1.2h .

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

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

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

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

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

$= \frac{1}{3}\pi (4 r^{2} )\left (1.2h \right )$

$= \left (\frac{1}{3} \cdot 4 \cdot 1.2 \right ) \pi r^{2} h$

$= 1.6 \pi r^{2} h$

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

V' =1.6V .

This means that the volume of the cone is $1.6 = 1.6 \times 100\% = 160\%$ times that of the cylinder—or 60% greater.

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Example Question #1 : Cones

About the -axis, rotate the triangle whose sides are along the -axis, the -axis, and the line of the equation

2x + y = 8 .

Give the volume of the solid of revolution formed.

Possible Answers:

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

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

None of the other choices gives the correct response.

$96 \pi$

$48 \pi$

Correct answer:

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

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= 8 , and which has height h= 4 . Substitute these values into the formula for the volume of a cone:

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

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

$= \frac{1}{3}\cdot 64 \cdot 4 \cdot \pi$

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

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Example Question #1 : Cones

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

Possible Answers:

942

3,142

2,094

1,814

1,885

Correct answer:

2,094

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 and h= 20 , so we can substitute and calculate:
$V = \frac{1}{3} \pi (10^{2} )(20)$

$\approx \frac{1}{3} \cdot 3.1416 \cdot 100 \cdot 20$

$\approx 2094.4$

To the nearest whole, this is 2,094.

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Example Question #2 : 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% greater than that of the cylinder.

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

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

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

The volume of the cone is 52% 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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HiSET: Math : Cones

Study concepts, example questions & explanations for HiSET: Math

All HiSET: Math Resources

Example Questions

Example Question #1 : Cones

Example Question #1 : Use Volume Formulas To Solve Problems

Example Question #1 : Use Volume Formulas To Solve Problems

Example Question #1 : Cones

Example Question #1 : Cones

Example Question #171 : Hi Set: High School Equivalency Test: Math

Example Question #12 : Use Volume Formulas To Solve Problems

Example Question #1 : Cones

Example Question #1 : Cones

Example Question #2 : Cones

All HiSET: Math Resources

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