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Volume and Surface Area: 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 The cube sculpture

The figure shows a sculpture made from cubes with edges 1 cm long. It is one cube deep, and each column is filled from the bottom to the top with no gaps. Find its volume.

A sculpture built from 1 cm cubes, one cube deep.

Text description of this figure

A sculpture built from unit cubes, drawn in an oblique view from the front and slightly above, so the front of every cube, the top of each column and the exposed right sides can be seen. It is one cube deep and has five columns standing side by side in a straight row, all starting from the same bottom level. From left to right the columns are 1 cube, 3 cubes, 2 cubes, 1 cube and 2 cubes tall. Lines show the boundary between every pair of neighboring cubes. The bottom front edge of the leftmost cube is labeled 1 cm.
1. Hint 1
  Each cube contributes one cubic cm to the volume.
2. Hint 2
  Count the cubes in each vertical column, then combine the counts.
Answer

9 cubic cm.

Full solution

Reading from left to right, the columns contain 1, 3, 2, 1, and 2 unit cubes.
Each cube occupies one cubic cm.
$\begin{aligned}V&=1+3+2+1+2\\V&=9\end{aligned}$
The sculpture has volume 9 cubic cm.

As a check, its bottom level contains five cubes, its middle level contains three, and its top level contains one, giving the same total.

Answer

9 cubic cm.

Key idea

The volume of a solid made from unit cubes is the total number of cubes it contains.
Problem 1: Mark complete
Problem 2 The folded carton

The figure is the complete net of a rectangular carton. It folds without overlap into a closed box. Find the volume enclosed by the box.

The complete net of a rectangular carton; dashed lines are folds.

Text description of this figure

A net of six rectangles drawn to one scale, with solid outer cut edges and dashed fold lines wherever two rectangles share an edge. In the center is a rectangle wider than it is tall: its lower edge is labeled 5 cm and its left edge is labeled 3 cm. A narrow rectangle as wide as the central one is attached to its top edge, and another to its bottom edge; the outer left edge of the upper narrow rectangle is labeled 2 cm. A small rectangle, as tall as the central one and as wide as the narrow rectangles are tall, is attached to each of its left and right edges. A sixth rectangle, the same size as the central one, is attached above the upper narrow rectangle. No other lengths are labeled.
1. Hint 1
  Adjacent faces in the net show the three edge lengths of the folded box.
2. Hint 2
  The central face becomes the base; its adjacent side face gives the height.
Answer

30 cubic cm.

Full solution

The central face is 5 cm by 3 cm.
The adjacent faces rise 2 cm when folded, so the box has dimensions 5 cm, 3 cm, and 2 cm.

The base area $B$ , in square cm, is
$\begin{aligned}B&=5(3)\\B&=15\end{aligned}$
So one 1 cm layer over the base holds 15 cubic cm, and the 2 cm height stacks two layers.
$\begin{aligned}V&=15(2)\\V&=30\end{aligned}$
The volume is 30 cubic cm.

Answer

30 cubic cm.

Key idea

A net can reveal the three dimensions needed to measure the space inside a folded box.
Problem 2: Mark complete
Problem 3 The unrolled sleeve

A cylinder is 7 cm tall. A removable sleeve covers its curved side exactly, with no overlap. When the sleeve is unrolled flat, it is a rectangle with area $70\pi$ square cm. Find the cylinder's radius.
1. Hint 1
  The rectangle's edge that wrapped around the cylinder is as long as the distance around a circular end.
2. Hint 2
  The rectangle's other edge ran up the side, so it is as long as the cylinder is tall. Divide the area by that length.
3. Hint 3
  Set $2\pi r$ equal to the length of the wrapped edge and solve for $r$ .
Answer

5 cm.

Full solution

The rectangle's edges that ran up the side of the cylinder are 7 cm long, the cylinder's height.
Call the length of the edge that wrapped around the cylinder $w$ cm.
$\begin{aligned}7w&=70\pi\\w&=10\pi\end{aligned}$

The wrapped edge is as long as the circumference of a circular end, $2\pi r$ .
Since $\pi$ is positive, dividing by $2\pi$ is valid.
$\begin{aligned}2\pi r&=10\pi\\r&=5\end{aligned}$
The radius is 5 cm.

As a check, a circle of radius 5 cm has circumference $10\pi$ cm, and a rectangle $10\pi$ cm by 7 cm has area $70\pi$ square cm.

Answer

5 cm.

Key idea

A cylinder's unrolled side is a rectangle whose height is the cylinder's height and whose other edge is the circumference of an end.
Problem 3: Mark complete
Problem 4 The packing arrangement

A tray holds 72 cubes, each with edges 1 cm long. All of them are rearranged into a rectangular prism with 4 complete layers and 3 equal rows in each layer. How many cubes are in each row, and what are the prism's dimensions?
1. Hint 1
  Every layer contains the same number of cubes, and every row within a layer has the same length.
2. Hint 2
  Divide the total cube count by the number of layers, then by the rows per layer.
3. Hint 3
  A cube edge is 1 cm, so the three cube counts give the three lengths in cm.
Answer

6 cubes per row; the prism is 6 cm by 3 cm by 4 cm (the three dimensions in any order).

Full solution

There are 4 equal layers, so the number of cubes in each layer is
$72\div4=18$
Each layer has 3 equal rows, so the number of cubes in one row is
$18\div3=6$

The rows are 6 cm long, the three rows span 3 cm, and the four layers span 4 cm.
Multiply to check the volume in cubic cm.
$\begin{aligned}V&=6(3)(4)\\V&=72\end{aligned}$
This counts all 72 cubes exactly once.

Answer

6 cubes per row; the prism is 6 cm by 3 cm by 4 cm (the three dimensions in any order).

Key idea

Dividing a cube count into equal layers and rows reverses the volume product.
Problem 4: Mark complete
Problem 5 The solid display stand

The figure shows a prism used as a solid display stand. Its matching ends are trapezoids, and its joining edges are perpendicular to those ends. Find its volume.

A solid display stand shaped as a prism with trapezoid ends.

Text description of this figure

A right prism drawn in an oblique view, so its front end, its top face and its slanted right face can be seen, and its three hidden edges are drawn dashed. The front end is a trapezoid. Its lower side is horizontal and labeled 7 cm, and its upper side is horizontal, centered above the lower side and labeled 3 cm; matching chevron marks show that these two sides are parallel. A dashed segment runs straight down from the upper left corner of the trapezoid to the lower side, meets it at a right angle marked with a small square, and is labeled 4 cm. From every corner of the front trapezoid an edge runs back, all in the same direction, to a matching trapezoid at the back. The lower right one of these edges is labeled 5 cm.
1. Hint 1
  A prism stacks matching copies of its end shape through its length.
2. Hint 2
  Find the trapezoid area from its parallel sides and perpendicular height.
3. Hint 3
  Multiply that area by the distance between the trapezoid ends.
Answer

100 cubic cm.

Full solution

The trapezoid ends have parallel sides 3 cm and 7 cm, separated by a perpendicular height of 4 cm.
Their area $B$ , in square cm, is
$\begin{aligned}B&=\frac{1}{2}(3+7)(4)\\B&=20\end{aligned}$

The prism extends 5 cm between its matching ends.
Multiply the base area by that distance.
$\begin{aligned}V&=20(5)\\V&=100\end{aligned}$
The volume is 100 cubic cm.

Answer

100 cubic cm.

Key idea

A prism volume uses the area of its end shape and the distance through which that shape is repeated.
Problem 5: Mark complete
Problem 6 The banded container

A closed cylinder has radius 4 cm and height 9 cm. A band covers a 5 cm tall strip of the curved side all the way around, without overlap. The band touches neither circular end. Find the cylinder's volume and the area of the cylinder's surface not covered by the band, including both circular ends. Give exact answers in terms of $\pi$ .
1. Hint 1
  The inside is measured by stacking the circular base through the height; the exposed outside consists of the ends and uncovered side portions.
2. Hint 2
  The uncovered side heights add to the cylinder height minus the band height.
3. Hint 3
  Unroll the uncovered side portions into rectangles with width equal to the cylinder circumference.
Answer

Volume $144\pi$ cubic cm; uncovered surface area $64\pi$ square cm.

Full solution

The circular base has area $\pi(4)^2=16\pi$ square cm.
Multiply by the height of 9 cm for volume.
$\begin{aligned}V&=(16\pi)(9)\\V&=144\pi\end{aligned}$
The volume is $144\pi$ cubic cm.

Both circular ends are uncovered.
Their combined area $E$ , in square cm, is
$\begin{aligned}E&=2\pi(4)^2\\E&=32\pi\end{aligned}$

The uncovered side heights total $9-5=4$ cm.
Each unrolled portion has width equal to the circumference, $2\pi(4)=8\pi$ cm.
Their combined area $L$ , in square cm, is
$\begin{aligned}L&=(8\pi)(4)\\L&=32\pi\end{aligned}$

Add the uncovered end and side areas.
$\begin{aligned}S&=32\pi+32\pi\\S&=64\pi\end{aligned}$
The uncovered area is $64\pi$ square cm.

As a check, the full surface area is $32\pi+8\pi(9)=104\pi$ square cm, and the band covers $8\pi(5)=40\pi$ square cm.
Subtracting leaves $104\pi-40\pi=64\pi$ square cm, as before.

Answer

Volume $144\pi$ cubic cm; uncovered surface area $64\pi$ square cm.

Key idea

A covering removes only the surface area it actually touches from the exposed area.
Problem 6: Mark complete
Problem 7 The folded prism

The figure shows the complete net of a closed triangular prism. Find its surface area.

The complete net of a closed triangular prism; dashed lines are folds.

Text description of this figure

A net drawn to one scale, with solid outer cut edges and dashed fold lines wherever two faces share an edge. Three rectangles of the same height stand side by side in a row; from left to right they are labeled 3 cm, 4 cm and 5 cm wide, and the left edge of the row is labeled 7 cm. A right triangle is attached to the top edge of the middle rectangle, pointing upward, and an identical right triangle is attached to its bottom edge, pointing downward. In each triangle the right angle, marked with a small square, is at the left end of the attached edge. Each triangle's free vertical side is labeled 3 cm and its slanted side is labeled 5 cm.
1. Hint 1
  The net contains every outer surface exactly once.
2. Hint 2
  Find the areas of the three rectangles and the two right triangles separately.
3. Hint 3
  Add all five areas, using the perpendicular triangle sides to find each triangle area.
Answer

96 square cm.

Full solution

The rectangles have dimensions 3 cm by 7 cm, 4 cm by 7 cm, and 5 cm by 7 cm.
Their areas in square cm are
$\begin{aligned}A_1&=3(7)\\A_1&=21\\A_2&=4(7)\\A_2&=28\\A_3&=5(7)\\A_3&=35\end{aligned}$

Each triangle is attached along the top or bottom edge of the 4 cm rectangle, so that side is 4 cm, and the right angle sits between it and the 3 cm side.
So each triangular end is a right triangle with perpendicular sides 3 cm and 4 cm, and its area in square cm is
$\begin{aligned}A_t&=\frac{1}{2}(3)(4)\\A_t&=6\end{aligned}$
Add the three rectangles and the two triangles.
$\begin{aligned}S&=21+28+35+2(6)\\S&=96\end{aligned}$
The surface area is 96 square cm.

Answer

96 square cm.

Key idea

The surface area of a prism is the sum of the areas of all the faces in its net.
Problem 7: Mark complete
Problem 8 Turning the box

A box filled with unit cubes is 6 cm long, 4 cm wide, and 2 cm high. It is turned to stand 6 cm high, resting on a 4 cm by 2 cm face. Mira says that its volume stays the same even though it now has more horizontal layers. Is she correct? Explain by counting the cubes in one layer in each position.
1. Hint 1
  Think about what volume counts, and compare that count before and after turning.
2. Hint 2
  Compare the number of cubes per layer and the number of layers in each position.
Answer

Yes. Originally: 24 cubes per layer and 2 layers. Turned: 8 cubes per layer and 6 layers. Both volumes are 48 cubic cm.

Full solution

In the original position, one layer contains $6(4)=24$ cubes and there are 2 layers.
$\begin{aligned}V_1&=24(2)\\V_1&=48\end{aligned}$
The volume is 48 cubic cm.

After turning, one layer contains $4(2)=8$ cubes and there are 6 layers.
$\begin{aligned}V_2&=8(6)\\V_2&=48\end{aligned}$
The volume remains 48 cubic cm.
Mira is correct: more layers are balanced by fewer cubes in each layer.

Answer

Yes. Originally: 24 cubes per layer and 2 layers. Turned: 8 cubes per layer and 6 layers. Both volumes are 48 cubic cm.

Key idea

Changing which face is underneath can change the number of layers and the cubes in each, but not the total number of cubes.
Problem 8: Mark complete
Problem 9 The joined blocks

Two cubes each have edges 2 cm long. They are joined along one entire square face to form a rectangular prism, with no gap or overlap. Noel says the surface area of the joined solid is the sum of the two original cube surface areas. Is Noel correct? Find the surface area of the joined solid and explain.
1. Hint 1
  Surface area counts the exposed outer faces of the joined solid.
2. Hint 2
  Look at the two faces that meet when the cubes are joined.
3. Hint 3
  Start from the surface areas of the separate cubes and remove the areas that become internal.
Answer

No; the joined solid has surface area 40 square cm.

Full solution

Each cube has six square faces, each of area $2^2=4$ square cm.
Their separate surface areas, in square cm, total
$\begin{aligned}S_0&=2(6)(4)\\S_0&=48\end{aligned}$

The two touching faces become internal, so remove both of their areas.
$\begin{aligned}S&=48-2(4)\\S&=40\end{aligned}$
The joined solid has surface area 40 square cm.
Noel included surfaces that are no longer exposed.

As a check, the joined prism is 4 cm by 2 cm by 2 cm.
Its three pairs of faces contribute, in square cm,
$\begin{aligned}S&=2(8+8+4)\\S&=40\end{aligned}$
This agrees with the face count.

Answer

No; the joined solid has surface area 40 square cm.

Key idea

Joining solids hides the touching faces, so their areas no longer contribute to the outer surface area.
Problem 9: Mark complete
Problem 10 Half the radius, twice the height

A cylinder has radius 4 cm and height 5 cm. A second cylinder has half the radius and twice the height. Tess says the two cylinders have the same volume, because one measurement is halved and the other is doubled. Is she correct? Find both volumes in terms of $\pi$ and explain.
1. Hint 1
  Look at how each measurement enters the volume: which one is squared?
2. Hint 2
  The second cylinder has radius 2 cm and height 10 cm.
3. Hint 3
  Find each base area, then multiply it by that cylinder's height.
Answer

No. The first volume is $80\pi$ cubic cm and the second is $40\pi$ cubic cm, half as much.

Full solution

The first base is a circle of radius 4 cm, with area $\pi(4)^2=16\pi$ square cm.
Stack it through the height of 5 cm.
$\begin{aligned}V_1&=(16\pi)(5)\\V_1&=80\pi\end{aligned}$
The first volume is $80\pi$ cubic cm.

The second cylinder has radius 2 cm and height 10 cm, so its base area is $\pi(2)^2=4\pi$ square cm.
$\begin{aligned}V_2&=(4\pi)(10)\\V_2&=40\pi\end{aligned}$
The second volume is $40\pi$ cubic cm, half the first.

Tess is not correct.
The radius is squared in the base area, so halving it divides the base area by 4, from $16\pi$ to $4\pi$ square cm.
Doubling the height only multiplies the volume by 2, so the volume ends up half as large.

Answer

No. The first volume is $80\pi$ cubic cm and the second is $40\pi$ cubic cm, half as much.

Key idea

In $V=\pi r^2 h$ the radius is squared, so halving the radius divides the volume by 4, while doubling the height only multiplies it by 2.
Problem 10: Mark complete

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