Area, Surface Area, and Volume: practice solutions, Grade 10 – download the PDF

1 Area of a triangle ★★★

\(A = \dfrac{1}{2}bh = \dfrac{1}{2} \cdot 12 \cdot 7 = 42\).

The area is 42 cm².

2 A trapezoid garden ★★★

\(A = \dfrac{1}{2}(9 + 15)\cdot 6 = 12 \cdot 6 = 72\).

The garden covers 72 m².

3 A shoebox ★★★

Volume: \(V = 12 \cdot 7 \cdot 5 = 420\).

Surface area: \(\text{SA} = 2(12\cdot7 + 12\cdot5 + 7\cdot5) = 2(84 + 60 + 35) = 358\).

The box holds 420 in³ and needs 358 in² of cardboard.

4 A soup can ★★★

\(V = \pi r^2 h = \pi \cdot 16 \cdot 10 = 160\pi\).

That is about 502.65 cm³.

5 A party hat ★★★

Slant height: \(\ell = \sqrt{9^2 + 12^2} = \sqrt{225} = 15\) cm.

Volume: \(V = \dfrac{1}{3}\pi \cdot 81 \cdot 12 = 324\pi \approx 1017.88\) cm³.

Lateral area: \(\pi r \ell = \pi \cdot 9 \cdot 15 = 135\pi \approx 424.12\) cm².

6 A globe ★★★

\(V = \dfrac{4}{3}\pi \cdot 125 = \dfrac{500\pi}{3} \approx 523.60\) in³.

\(\text{SA} = 4\pi \cdot 25 = 100\pi \approx 314.16\) in².

7 Name the cross section ★★★

1. A circle, congruent to the bases.
2. A rectangle whose width is the diameter and whose height is the height of the cylinder.
3. A triangle (isosceles), with the base diameter as its base.
4. A square, smaller than the base.

8 Doubling the radius ★★★

False. For \(r = 2\): \(V = \dfrac{4}{3}\pi \cdot 8 = \dfrac{32\pi}{3}\). For \(r = 4\): \(V = \dfrac{4}{3}\pi \cdot 64 = \dfrac{256\pi}{3}\).

The ratio is \(\dfrac{256}{32} = 8\), so the volume is multiplied by \(2^3 = 8\). The surface area is multiplied by \(2^2 = 4\).

9 A patio with a curved end ★★★

The semicircle has diameter 12 ft, so its radius is 6 ft.

Rectangle: \(20 \cdot 12 = 240\). Semicircle: \(\dfrac{1}{2}\pi \cdot 6^2 = 18\pi \approx 56.55\).

Total: \(240 + 18\pi \approx \) 296.55 ft².

10 A hexagonal tile ★★★

Perimeter: \(P = 6 \cdot 6 = 36\) cm.

\(A = \dfrac{1}{2}aP = \dfrac{1}{2} \cdot 3\sqrt{3} \cdot 36 = 54\sqrt{3}\).

That is about 93.53 cm².

11 A tent shaped like a prism ★★★

Base area: \(B = \dfrac{1}{2} \cdot 6 \cdot 8 = 24\) cm².

Volume: \(V = Bh = 24 \cdot 15 = 360\) cm³.

Perimeter of the base: \(6 + 8 + 10 = 24\) cm. Lateral area: \(24 \cdot 15 = 360\). Then \(\text{SA} = 2 \cdot 24 + 360 = 408\) cm².

12 A glass pyramid ★★★

Volume: \(V = \dfrac{1}{3} \cdot 100 \cdot 12 = 400\) m³.

The slant height of a face joins the apex to the midpoint of a base edge: \(\ell = \sqrt{12^2 + 5^2} = 13\) m.

Lateral area: \(\dfrac{1}{2} \cdot 40 \cdot 13 = 260\). With the base: \(\text{SA} = 100 + 260 = 360\) m².

13 An ice cream cone ★★★

Height: \(h = \sqrt{10^2 - 6^2} = \sqrt{64} = 8\) cm.

Volume: \(\dfrac{1}{3}\pi \cdot 36 \cdot 8 = 96\pi \approx 301.59\) cm³.

Lateral area: \(\pi \cdot 6 \cdot 10 = 60\pi \approx 188.50\) cm².

14 Slicing a cone ★★★

The section is a circle similar to the base with scale factor \(k = \dfrac{12}{20} = 0.6\).

Its radius is \(8 \cdot 0.6 = 4.8\) cm, so \(A = \pi \cdot 4.8^2 = 23.04\pi \approx 72.38\) cm².

15 Slicing a sphere ★★★

The radius \(\rho\) satisfies \(\rho^2 = 13^2 - 5^2 = 169 - 25 = 144\), so \(\rho = 12\) cm.

The area is \(\pi \cdot 144 = 144\pi \approx 452.39\) cm².

16 A water tank ★★★

The radius is 3 ft. \(V = \pi \cdot 9 \cdot 5 = 45\pi \approx 141.37\) ft³.

Gallons: \(141.37 \cdot 7.48 \approx 1057\).

The tank holds about 1,057 gallons.

17 A steel rod ★★★

\(V = \pi \cdot 4 \cdot 10 = 40\pi \approx 125.66\) cm³.

\(m = \rho V = 7.85 \cdot 40\pi \approx 986.5\).

The mass is about 986.5 g.

18 A hollow pipe ★★★

The material is the large cylinder minus the hollow cylinder.

\(V = \pi \cdot 5^2 \cdot 20 - \pi \cdot 4^2 \cdot 20 = 500\pi - 320\pi = 180\pi\).

That is about 565.49 in³.

19 A stack of coins ★★★

The stack height is \(40 \cdot 0.2 = 8\) cm. Every horizontal slice of either stack is a disk of radius 1.2 cm, so the cross sections have equal area at every level.

By Cavalieri’s principle the volumes are equal: \(V = \pi \cdot 1.2^2 \cdot 8 = 11.52\pi \approx 36.19\) cm³.

20 An ice cream treat ★★★

Cone: \(\dfrac{1}{3}\pi \cdot 9 \cdot 9 = 27\pi\).

Hemisphere: \(\dfrac{1}{2}\cdot\dfrac{4}{3}\pi \cdot 27 = 18\pi\).

Total: \(45\pi \approx\) 141.37 cm³.

21 Scaling similar cones ★★★

The scale factor is \(k = \dfrac{3}{2}\).

Volume scales by \(k^3 = \dfrac{27}{8}\): \(40 \cdot \dfrac{27}{8} = 135\) cm³.

Surface area scales by \(k^2 = \dfrac{9}{4}\): \(50 \cdot \dfrac{9}{4} = 112.5\) cm².

22 A grain silo ★★★

Cylinder: \(\pi \cdot 36 \cdot 20 = 720\pi\). Hemisphere: \(\dfrac{2}{3}\pi \cdot 216 = 144\pi\). Total: \(864\pi \approx 2714.34\) ft³.

Wheat volume: \(0.8 \cdot 2714.34 \approx 2171.47\) ft³. Mass: \(2171.47 \cdot 48 \approx 104{,}230\).

The wheat weighs about 104,230 lb.

23 Designing a can ★★★

Set \(\pi r^2 \cdot 12 = 355\), so \(r^2 = \dfrac{355}{12\pi} \approx 9.417\).

Then \(r \approx 3.07\). The radius is about 3.07 cm.