Use volume formulas to solve problems

Help Questions

HiSET › Use volume formulas to solve problems

Questions 1 - 10

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

$387.5 \pi$

$155\pi$

$300\pi$

$181.25\pi$

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 $\pi r^2 h$ .

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

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?

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 74.4% 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.

Explanation

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 $s^{2}h$ . This yields

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

Which of the following is closest to being correct?

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 74.4% 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.

Explanation

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

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

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

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

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

$=0.576 s ^{2} h$

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

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

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

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

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

$64 \pi$

$128 \pi$

$256 \pi$

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:

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.

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

$60 \pi$

$50 \pi$

$55 \pi$

$66 \pi$

None of the other choices gives the correct response.

Explanation

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

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

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

$64 \pi$

$128 \pi$

$256 \pi$

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:

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.

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

$60 \pi$

$50 \pi$

$55 \pi$

$66 \pi$

None of the other choices gives the correct response.

Explanation

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

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

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

$=60 \pi$

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

$387.5 \pi$

$155\pi$

$300\pi$

$181.25\pi$

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 $\pi r^2 h$ .

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

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

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

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

$30 \pi$

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

$90 \pi$

None of the other choices gives the correct response.

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 $9 \div 2 = \frac{9}{2}$ .