Use volume formulas to solve problems - HiSET

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

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

This pyramid can be seen as having as its base the triangle on the -plane with vertices at the origin, , and ; 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 , which has length (the height of the pyramid) .

Setting $B = $$\frac{1}{2}$c^{2}$$ and 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.

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.

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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$

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

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The volume of a cone is given by the formula

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

We are given that . 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 .

Thus,

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

so

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

and

.

Thus,

$$\sqrt{12}$=r$ .

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

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

.

Give the volume of the solid of revolution formed.

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

The point of intersection of the -axis and the graph of - the -intercept of the latter - can be found by substituting 0 for :

Divide both sides by 2 to isolate :

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

The point of intersection is at .

Similarly, The point of intersection of the -axis and the graph of - the -intercept of the latter - can be found by substituting 0 for :

The point of intersection is at .

The three vertices of the triangle are at the origin, , and . When this triangle is rotated about the -axis, the resulting solid of revolution is a cone whose base has radius , and which has height . 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$

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

.

Give the volume of the solid of revolution formed.

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

The point of intersection of the -axis and the graph of —the -intercept of the latter—can be found by substituting 0 for :

Divide both sides by 2 to isolate :

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

The point of intersection is at .

Similarly, the point of intersection of the -axis and the graph of —the -intercept of the latter—can be found by substituting 0 for :

The point of intersection is at .

The three vertices of the triangle are at the origin, , and . When this triangle is rotated about the -axis, the resulting solid of revolution is a cone whose base has radius , and which has height . 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$

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

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When this triangle is rotated about the -axis, the resulting solid of revolution is a cone whose base has radius , and which has height . 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$

$= 50 \pi$

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

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When this triangle is rotated about the -axis, the resulting solid of revolution is a cone whose base has radius , and which has height . 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$

$=60 \pi$

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).

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

.

Substituting 10 for and 20 for , we can find :

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

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

$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.

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).

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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 and , 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.

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

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

.

Substituting 10 for and 20 for , we can find :

, 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

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?

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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 .

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

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

Set and :

$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$ :

.

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

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?

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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 .

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

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

Set and :

$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$ :

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

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

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

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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 :

$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$

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

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

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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 and :

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

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

$=30 \pi$

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

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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$

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

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The formula for the volume of a pyramid is

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

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

.

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

Thus,

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

so

$lw=12\times3=36$ .

Note, the area of the base of the pyramid is

.

Thus,

.

Hence,

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

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?

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

.

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 . Similarly, the length of a side of the base of the pyramid is 20% greater than that of a base of the prism, so . 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 . This yields

The volume of the pyramid is equal to 57.6 % of that of the prism, or, equivalently, less.

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

Give its volume.

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

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 ;

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

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

Give the volume of the pyramid.

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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 . 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.

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$ .

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

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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} = $\frac{100 \pi}{4 \pi}$

Take the square root of both sides:

$r = $\sqrt{25}$ = 5$

Set 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.