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HiSET: Math : Use volume formulas to solve problems

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125 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #1 : Cylinders

What is the volume of a cylinder with a diameter of cm and a height of 15.5 cm?

Possible Answers:

$300\pi$

$181.25\pi$

$155\pi$

$387.5 \pi$

Correct answer:

$387.5 \pi$

Explanation:

Step 1: Find the diameter.

If we are given the diameter, the length of the radius is one-half the diameter.

So, the radius is $\dfrac {1}{2}\times 10=5$

Step 2: Recall the volume formula...

Volume formula of cylinder is $\pi r^2 h$ .

$V=\pi (5)^2(15.5)\rightarrow \pi(25)(15.5)\implies 387.5 \pi$

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Example Question #1 : Pyramids

What is the area of the base of the pyramid with volume $12 \;\textup{m}^3$ and height $1\;\textup{m}$ ?

Possible Answers:

$32\;\textup{m}^2$

$42\;\textup{m}^2$

$19\;\textup{m}^2$

$36\;\textup{m}^2$

$23\;\textup{m}^2$

Correct answer:

$36\;\textup{m}^2$

Explanation:

The formula for the volume of a pyramid is

$V = \frac{lwh}{3}$

The height of the pyramid is $1\;\textup{m}$ , so

h=1 .

The volume of the pyramid is $12 \;\textup{m}^3$ .

Thus,

$12=\frac{lw}{3}$

$lw=12\times3=36$ .

Note, the area of the base of the pyramid is

A = lw .

Thus,

A = lw = 36 .

Hence,

$A = 36 \;\textup{m}^2$

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Example Question #1 : Use Volume Formulas To Solve Problems

A pyramid with a square base is inscribed inside a right cone with radius 24 and height 10.

Give the volume of the pyramid.

Possible Answers:

$7,680\sqrt{2}$

7,680

$3,840\sqrt{2}$

3,840

None of the other choices gives the correct response.

Correct answer:

3,840

Explanation:

The volume of a pyramid with height and a base of area can be determined using the formula

$V = \frac{1}{3}Bh$ .

The height of the inscribed pyramid is equal to that of the cone, so we can set h= 10 . The base of the pyramid is a square inscribed inside a circle, so the length of each diagonal of the square is equal to the diameter of the circle. See the figure below, which shows the bases of the pyramid and the cone.

Circle in square

The circle has radius 24, so its diameter - and the lengths of the diagonals - is twice this, or 48.

The area of a square - which is also a rhombus - is equal to half the product of the lengths of its diagonals, so

$B = \frac{1}{2}d_{1}d_{2} = \frac{1}{2} (48) (48) = 1,152$

Substituting for and , we get

$V = \frac{1}{3} (1,152)(10) = 3,840$ .

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Example Question #1 : Use Volume Formulas To Solve Problems

The base of a right pyramid with height 6 is a regular hexagon with sides of length 6.

Give its volume.

Possible Answers:

$108 \sqrt{2}$

$54 \sqrt{3}$

$108 \sqrt{3}$

$54 \sqrt{2}$

108

Correct answer:

$108 \sqrt{3}$

Explanation:

The regular hexagonal base can be divided by its diameters into six equilateral triangles, as seen below:

Hexagon 1

Each of the triangles has as its sidelength that of the hexagon. If we let this common sidelength be , each of the triangles has area

$A = \frac{s^{2}\sqrt{3}}{4}$ ;

the total area of the base is six times this.

Substituting 6 for , the area of each triangle is

$A = \frac{6^{2}\sqrt{3}}{4} = \frac{36\sqrt{3}}{4} = 9 \sqrt{3}$

The total area of the base is six times this, or

$B = 6 A =6 \cdot 9 \sqrt{3} = 54 \sqrt{3}$

The volume of a pyramid with height and a base of area can be determined using the formula

$V = \frac{1}{3}Bh$ .

Set $B = 54 \sqrt{3}$ and h = 6 ;

$V = \frac{1}{3} (54\sqrt{3}) (6) = 108 \sqrt{3}$

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Example Question #1 : Use Volume Formulas To Solve Problems

A right pyramid and a right rectangular prism both have square bases. The base of the pyramid has sides that are 20% longer than those of the bases of the prism; the height of the pyramid is 20% greater than that of the prism.

Which of the following is closest to being correct?

Possible Answers:

The volume of the pyramid is 42.4% less than that of the prism.

The volume of the pyramid is 82.9% less than that of the prism.

The volume of the pyramid is 61.6% less than that of the prism.

The volume of the pyramid is 33.3% less than that of the prism.

The volume of the pyramid is 74.4% less than that of the prism.

Correct answer:

The volume of the pyramid is 42.4% less than that of the prism.

Explanation:

The volume of a right prism with height and bases of area can be determined using the formula

V = Bh .

Since its base is a square, if we let be the length of one side, then $B = s^{2}$ , and

$V =s^{2}h$

The volume of a right pyramid with height and a base of area can be determined using the formula

$V '=\frac{1}{3} B'h'$ .

Since its base is also a square, if we let be the length of one side, then $B' = s'^{2}$ , and

$V '=\frac{1}{3} s'^{2}h'$ .

The height of the pyramid is 20% greater than the height of the prism - this is 120% of , so h'= 1.2h . Similarly, the length of a side of the base of the pyramid is 20% greater than that of a base of the prism, so s' = 1.2s . Substitute in the pyramid volume formula:

$V '=\frac{1}{3} (1.2s )^{2} \cdot 1.2h$

$=\frac{1}{3} \cdot 1.2^{2} \cdot s ^{2} \cdot 1.2h$

$=\frac{1}{3} \cdot 1.44 \cdot 1.2 \cdot s ^{2} h$

$=0.576 s ^{2} h$

We can substitute , the volume of the prism, for $s^{2}h$ . This yields

V'=0.576 V

The volume of the pyramid is equal to 57.6 % of that of the prism, or, equivalently, $(100 - 57.6) \% = 42.4 \%$ less.

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Example Question #161 : Hi Set: High School Equivalency Test: Math

A triangular pyramid in coordinate space has its vertices at the origin, (c,0,0) , (0,c,0) , and (0,0,c) . In terms of , give its volume.

Possible Answers:

$\frac{1}{9}c^{3}$

$\frac{1}{6}c^{3}$

$\frac{1}{4}c^{3}$

$\frac{1}{8}c^{3}$

$\frac{1}{12}c^{3}$

Correct answer:

$\frac{1}{6}c^{3}$

Explanation:

The pyramid in question can be seen in the diagram below:

Pyramid 1

This pyramid can be seen as having as its base the triangle on the -plane with vertices at the origin, (c,0,0) , and (0,c,0) ; this is a right triangle with two legs of length , so its area is half their product, or $B = \frac{1}{2}c^{2}$ .

The altitude (perpendicular to the base) is the segment from the origin to (0,0,c) , which has length (the height of the pyramid) h = c .

Setting $B = \frac{1}{2}c^{2}$ and h = c in the formula for the volume of a pyramid:

$V = \frac{1}{3} Bh$

$V = \frac{1}{3}\cdot \frac{1}{2}c^{2} \cdot c= \frac{1}{6}c^{3}$ , the correct response.

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Example Question #111 : Measurement And Geometry

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:

$60 \pi$

$50 \pi$

None of the other choices gives the correct response.

$66 \pi$

$55 \pi$

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:

$50 \pi$

$55 \pi$

$66 \pi$

$60 \pi$

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 #2 : 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:

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

None of the other choices gives the correct response.

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

$30 \pi$

$90 \pi$

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 #111 : Measurement And Geometry

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:

None of the other choices gives the correct response.

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

$90 \pi$

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

$30 \pi$

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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HiSET: Math : Use volume formulas to solve problems

Study concepts, example questions & explanations for HiSET: Math

All HiSET: Math Resources

Example Questions

Example Question #1 : Cylinders

Example Question #1 : Pyramids

Example Question #1 : Use Volume Formulas To Solve Problems

Example Question #1 : Use Volume Formulas To Solve Problems

Example Question #1 : Use Volume Formulas To Solve Problems

Example Question #161 : Hi Set: High School Equivalency Test: Math

Example Question #111 : Measurement And Geometry

Example Question #1 : Use Volume Formulas To Solve Problems

Example Question #2 : Use Volume Formulas To Solve Problems

Example Question #111 : Measurement And Geometry

All HiSET: Math Resources

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