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Prisms and Cylinders: Core practice

10 practice problems for this lesson. Work on paper, use hints when you need them, and check the answer or the full solution when you are ready.

Difficulty: Core (core-course level)

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Problem 1 of 10

Problem 1 A constant section

Every section of a prism parallel to its bases has area $11.5$ square cm. The perpendicular distance between its base planes is $6.4$ cm. Find its volume and explain the units.
1. Hint 1
  Think about how a prism repeats one base area through its height.
2. Hint 2
  Multiply the section area by the perpendicular distance.
Answer

$73.6$ cubic cm.

Full solution

Each base-parallel section is a translated copy of the base, so the base area is 11.5 square cm.

Multiply by the perpendicular height.
$\begin{aligned}V&=11.5(6.4)\\V&=73.6\text{ cm}^3.\end{aligned}$

Square centimeters times centimeters give cubic centimeters, as required for volume.

Answer

$73.6$ cubic cm.

Key idea

A constant base-parallel section area supplies the base area in the prism volume product.
Problem 2 Smaller measuring cubes

A rectangular prism is filled without gaps by cubes with edge length $0.5$ cm. There are seven cubes along its length, four along its width, and three along its height.

Find the prism volume in cubic centimeters. Explain why its cube count and its volume number differ.
1. Hint 1
  Identify the volume of one measuring cube before totaling the cubes.
2. Hint 2
  There are three factors of one half in each small cube volume.
Answer

$10.5$ cubic cm; there are 84 cubes, each occupying $0.125$ cubic cm.

Full solution

The array contains
$7(4)(3)=84$
small cubes.

One cube has volume
$0.5^3=0.125\text{ cm}^3.$

Therefore
$\begin{aligned}V&=84(0.125)\\V&=10.5\text{ cm}^3.\end{aligned}$

The count is 84, but these are not one-cubic-centimeter cubes.
As a length check, the whole prism is 3.5 cm by 2 cm by 1.5 cm, whose product is also 10.5 cubic cm.

Answer

$10.5$ cubic cm; there are 84 cubes, each occupying $0.125$ cubic cm.

Key idea

A cube count becomes a volume only after multiplying by the volume of each measuring cube.
Problem 3 A height hidden in the lean

An oblique prism has base area $45$ square cm. Its joining edge is $13$ cm long. In a side view perpendicular to the base planes, that edge, the perpendicular height, and a $5$ cm horizontal offset form a right triangle.

Find the perpendicular height and the prism volume.

A perpendicular side view isolates the height; this is not a base shape.

Text description of this figure

A right triangle in a perpendicular side view has slanted hypotenuse 13 cm, horizontal leg 5 cm, and vertical leg h; the prism base area is 45 square centimeters.
1. Hint 1
  Distinguish the distance between base planes from the slanted joining edge.
2. Hint 2
  Use the right triangle to recover the perpendicular distance, then find the volume.
Answer

Height $12$ cm; volume $540$ cubic cm.

Full solution

The 13 cm joining edge is the hypotenuse of the supplied right triangle.

Subtract the offset square.
$\begin{aligned}h^2&=13^2-5^2\\h^2&=144\\h&=12\text{ cm}.\end{aligned}$

Every base-parallel section has area 45 square cm.
$\begin{aligned}V&=45(12)\\V&=540\text{ cm}^3.\end{aligned}$

The positive height checks the right triangle because its square plus 25 is 169.

Answer

Height $12$ cm; volume $540$ cubic cm.

Key idea

Recover the perpendicular height before using the base-area volume product.
Problem 4 Different sections, equal amounts

A right prism has base area $26$ square cm and perpendicular height $6.5$ cm. An oblique circular cylinder has the same base area and perpendicular height.

Explain why their volumes are equal, checking Cavalieri's hypotheses across the full height. Then give the common volume.
1. Hint 1
  Compare cuts parallel to the bases at corresponding heights.
2. Hint 2
  Each prism section repeats its polygonal base; each cylinder section repeats its disk base.
Answer

Both volumes are $169$ cubic cm. Every corresponding base-parallel section has area $26$ square cm over the common $6.5$ cm height.

Full solution

Place the lower base planes at the same level.
Equal perpendicular heights put both upper bases in one parallel plane.

At every height between these planes, the prism section is a translated base polygon of area 26 square cm.
The oblique cylinder section is a translated base disk of area 26 square cm.

The section shapes differ, but all corresponding areas agree.
Cavalieri therefore gives equal volumes.

The common volume is
$\begin{aligned}V&=26(6.5)\\V&=169\text{ cm}^3.\end{aligned}$

Answer

Both volumes are $169$ cubic cm. Every corresponding base-parallel section has area $26$ square cm over the common $6.5$ cm height.

Key idea

Cavalieri compares section areas at every height, rather than requiring congruent section shapes.
Problem 5 The two ends and the side strip

A closed right prism has base area $46$ square cm and base perimeter $32$ cm. Its total surface area is $348$ square cm.

Find its perpendicular height and its volume.
1. Hint 1
  Separate the two end faces from the lateral rectangles.
2. Hint 2
  After removing the two base areas, divide the remaining area by the base perimeter.
Answer

Height $8$ cm; volume $368$ cubic cm.

Full solution

Both bases together occupy
$2B=92\text{ cm}^2.$

Subtract them to isolate the side strip.
$\begin{aligned}L&=348-92\\L&=256\text{ cm}^2.\end{aligned}$

Because the prism is right, the side rectangles share its height.
$\begin{aligned}32h&=256\\h&=8\text{ cm}.\end{aligned}$

Now find volume.
$\begin{aligned}V&=46(8)\\V&=368\text{ cm}^3.\end{aligned}$

The surface check adds 256 and 92 to recover 348 square cm.

Answer

Height $8$ cm; volume $368$ cubic cm.

Key idea

A closed right prism surface area separates into two bases and a perimeter-by-height side strip.
Problem 6 Choose the end disks

A right circular cylinder has radius $2.5$ cm and height $6.2$ cm. Ignore thickness and seams. Give exact areas in terms of pi for each specified boundary.

Case A includes the curved side only, with neither end disk.

Case B includes the curved side and one end disk.

Case C includes the curved side and both end disks.

The cases, rather than the outline alone, specify the included end disks.

Text description of this figure

A right circular cylinder with radius 2.5 cm and height 6.2 cm. Circular boundaries locate the possible end disks; the cases specify which disks are included.
1. Hint 1
  Identify the pieces common to all three boundaries.
2. Hint 2
  Find the curved-side area and one disk area separately, then include the required number of disks.
Answer

$\begin{aligned}A&:31\pi\text{ cm}^2\\B&:37.25\pi\text{ cm}^2\\C&:43.5\pi\text{ cm}^2.\end{aligned}$

Full solution

The common curved-side area is
$\begin{aligned}L&=2\pi(2.5)(6.2)\\L&=31\pi\text{ cm}^2.\end{aligned}$

One disk has area
$B=6.25\pi\text{ cm}^2.$

Case A includes no disk, so its area is $31\pi$ square cm.

Case B includes one disk.
$31\pi+6.25\pi=37.25\pi\text{ cm}^2.$

Case C includes two disks.
$31\pi+12.5\pi=43.5\pi\text{ cm}^2.$

Answer

$\begin{aligned}A&:31\pi\text{ cm}^2\\B&:37.25\pi\text{ cm}^2\\C&:43.5\pi\text{ cm}^2.\end{aligned}$

Key idea

The words describing the boundary decide how many base areas to add.
Problem 7 Capacity and sheet material

An ideal open rectangular tank has inside base dimensions $11$ dm by $8$ dm and capacity $1{,}320$ liters. It has a bottom and four upright sides, with no lid. Use $1$ liter as one cubic decimeter.

Find its inside height and the area of sheet material needed. Ignore thickness, seams, and waste.

Inside dimensions determine capacity; the open rim contributes no lid.

Text description of this figure

An open rectangular tank has inside base 11 dm by 8 dm and unknown perpendicular height h; the top rim is open and hidden lower edges are dashed.
1. Hint 1
  Use capacity to determine the missing inside dimension.
2. Hint 2
  List the bottom and four sides before totaling the material area.
Answer

Height $15$ dm; sheet area $658$ square dm.

Full solution

The inside base area is
$B=88\text{ dm}^2.$
Capacity is 1320 cubic dm.

Divide volume by base area.
$\begin{aligned}h&=1320/88\\h&=15\text{ dm}.\end{aligned}$

The four side rectangles total
$\begin{aligned}L&=2(11+8)(15)\\L&=570\text{ dm}^2.\end{aligned}$

Add one bottom, since the top is open.
$\begin{aligned}S&=570+88\\S&=658\text{ dm}^2.\end{aligned}$

The recovered dimensions give 1320 cubic dm.
A lid would require another 88 square dm, which the stated boundary excludes.

Answer

Height $15$ dm; sheet area $658$ square dm.

Key idea

Capacity determines the inside space, while the listed boundary pieces determine material area.
Problem 8 A ratio of different measurements

First consider a right circular cylinder with radius $6$ cm and height $5$ cm. Find its volume divided by its curved-side area, including the unit of the quotient.

Then take any right circular cylinder with positive radius r and height h. Derive its volume divided by its curved-side area in terms of r. Explain why the height cancels.
1. Hint 1
  Write both measurements before taking their quotient.
2. Hint 2
  The same positive height and one radius factor occur in numerator and denominator.
Answer

Numerical quotient $3$ cm. General quotient $V/L=r/2$ , a length.

Full solution

For the numerical cylinder,
$\begin{aligned}V&=180\pi\text{ cm}^3\\L&=60\pi\text{ cm}^2.\end{aligned}$

Their quotient is
$V/L=3\text{ cm}.$
Cubic units divided by square units leave a length.

For arbitrary positive r and h,
$\begin{aligned}\frac VL&=\frac{\pi r^2h}{2\pi rh}\\\frac VL&=\frac r2.\end{aligned}$

The positive common height, pi, and one radius factor cancel.
Increasing only the height multiplies volume and curved-side area by the same factor, which leaves this quotient unchanged.

Answer

Numerical quotient $3$ cm. General quotient $V/L=r/2$ , a length.

Key idea

The volume-to-curved-side-area quotient recovers half the radius of a right circular cylinder.
Problem 9 How much section evidence?

Two ordinary solids lie between the planes at heights 0 cm and 6 cm. At heights 0 cm, 2 cm, 4 cm, and 6 cm, each solid has section area 9 square cm in the plane parallel to those boundary planes.

Does this evidence establish equal volumes by Cavalieri? State exactly what additional equality must be known. No volume calculation is required.
1. Hint 1
  Read the quantifier in Cavalieri's principle.
2. Hint 2
  Compare the finite list of tested heights with all heights between the planes.
Answer

No. Section areas must be equal at every corresponding height throughout the interval from 0 cm to 6 cm.

Full solution

The boundary planes already provide a common height.
The four measurements provide equal section areas only at four selected levels.

Nothing states that the sections agree between those levels.
Cavalieri requires equal section areas in every parallel plane throughout the full height.

The volumes might be equal, but the supplied evidence does not establish that conclusion by Cavalieri.

Answer

No. Section areas must be equal at every corresponding height throughout the interval from 0 cm to 6 cm.

Key idea

A finite collection of matching samples does not supply an every-height hypothesis.
Problem 10 Recover a cylinder and its boundary

A right circular cylinder has volume $243\pi$ cubic cm and curved-side area $54\pi$ square cm. All dimensions are positive.

Find its radius and height.

Find the boundary area if the curved side and exactly one base are included.

Find the extra area required to close the other end. Check the recovered dimensions in both supplied measurements.
1. Hint 1
  Both measurements contain the height; compare them to isolate the radius.
2. Hint 2
  Divide volume by curved-side area, then substitute the recovered radius into either original relation.
3. Hint 3
  The specified boundary has one disk; closing the other end adds one more disk.
Answer

Radius $9$ cm; height $3$ cm. One-base boundary $135\pi$ square cm; extra closing disk $81\pi$ square cm.

Full solution

Divide the two measurements.
$\begin{aligned}\frac{243\pi}{54\pi}&=\frac r2\\\frac92&=\frac r2\\r&=9\text{ cm}.\end{aligned}$

Use the curved-side area to find height.
$\begin{aligned}54\pi&=2\pi(9)h\\h&=3\text{ cm}.\end{aligned}$

One base has area
$B=81\pi\text{ cm}^2.$

The specified one-base boundary has area
$S=135\pi\text{ cm}^2.$
Closing the other end requires an additional $81\pi$ square cm.

Check volume and side area separately.
$\begin{aligned}\pi(9)^2(3)&=243\pi\text{ cm}^3\\2\pi(9)(3)&=54\pi\text{ cm}^2.\end{aligned}$
Both agree with the givens.

Answer

Radius $9$ cm; height $3$ cm. One-base boundary $135\pi$ square cm; extra closing disk $81\pi$ square cm.

Key idea

Coupled volume and lateral-area data determine dimensions before the boundary pieces are added.

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