Volume and Surface Area: Practice

First find the base area, then divide the volume by it.

$\text{base area} = \pi r^2 = 3.14 \times 16 = 50.24 \text{ cm}^2$

$h = \frac{V}{\text{base area}} = \frac{150.72}{50.24} = 3 \text{ cm}$

Divide the box volume by the small cube's volume.

$V_{\text{box}} = 8 \times 6 \times 4 = 192 \text{ cm}^3, \qquad V_{\text{cube}} = 2^3 = 8 \text{ cm}^3$

$\frac{192}{8} = 24$

Each dimension also splits evenly into $2$ cm pieces ($4 \times 3 \times 2 = 24$), confirming that $24$ small cubes fit.

Find one face of each kind, then double the total.

$lw = 12 \times 5 = 60, \quad lh = 12 \times 4 = 48, \quad wh = 5 \times 4 = 20$

$S = 2(60 + 48 + 20) = 2 \times 128 = 256 \text{ cm}^2$

The value $240$ would be the volume $12 \times 5 \times 4$, a cubic measure, not the surface area.

With equal heights, the larger base area wins. Each volume is base area times height.

$V_{\text{cyl}} = 28.26 \times 10 = 282.6 \text{ cm}^3, \qquad V_{\text{box}} = 30 \times 10 = 300 \text{ cm}^3$

The box is larger by

$300 - 282.6 = 17.4 \text{ cm}^3$

Work back to the edge from the surface area $6s^2$.

$6s^2 = 96 \;\Rightarrow\; s^2 = 16 \;\Rightarrow\; s = 4 \text{ cm}$

Then the volume is $s^3$.

$V = 4^3 = 64 \text{ cm}^3$

Find the triangular base area first.

$B = \frac{1}{2} \times 0.6 \times 0.4 = 0.12 \text{ m}^2$

Then multiply by the length.

$V = 0.12 \times 3 = 0.36 \text{ m}^3$

Add the two circular ends and the rolled-out side.

$2\pi r^2 = 2 \times 3.14 \times 25 = 157 \text{ cm}^2$

$2\pi r h = 2 \times 3.14 \times 5 \times 4 = 125.6 \text{ cm}^2$

$S = 157 + 125.6 = 282.6 \text{ cm}^2$

If the base side is $s$ and the height is also $s$, the box is a cube of volume $s^3$.

$s^3 = 125 \;\Rightarrow\; s = 5 \text{ cm}$

because $5 \times 5 \times 5 = 125$.

Volume is $\pi r^2 h$, and the radius is squared. Doubling the radius multiplies $r^2$ by $4$.

$\frac{\pi (2r)^2 h}{\pi r^2 h} = \frac{4 r^2}{r^2} = 4$

So the second cylinder holds $4$ times as much.

Split the L-shaped front face into two rectangles: the tall part is $6$ cm by $2$ cm and the foot is $8$ cm by $2$ cm. Each forms a box $4$ cm deep, so find each volume and add.

$V_1 = 6 \times 2 \times 4 = 48 \text{ cm}^3, \qquad V_2 = 8 \times 2 \times 4 = 64 \text{ cm}^3$

$V = 48 + 64 = 112 \text{ cm}^3$

Equal volume does not force equal surface area. Compute each.

$S_A = 2(4 \cdot 4 + 4 \cdot 3 + 4 \cdot 3) = 2(16 + 12 + 12) = 80 \text{ cm}^2$

$S_B = 2(8 \cdot 3 + 8 \cdot 2 + 3 \cdot 2) = 2(24 + 16 + 6) = 92 \text{ cm}^2$

Box B has the larger surface area, because the more stretched-out a box is, the more skin it needs for the same volume.

A closed can needs both ends plus the curved side: $S = 2\pi r^2 + 2\pi r h$.

$2\pi r^2 = 2 \times 3.14 \times 16 = 100.48 \text{ cm}^2$

$2\pi r h = 2 \times 3.14 \times 4 \times 6 = 150.72 \text{ cm}^2$

$S = 100.48 + 150.72 = 251.2 \text{ cm}^2$