Every worked solution shows the formula, the substitution and the final answer in bold, with exact answers in terms of \(\pi\) and rounded values. Compare each step with your own work to find where a mistake started.
Problem 1: Circumference from the radius
\(C=2\pi r=2\pi(7)=14\pi\). Then \(14\pi\approx 43.98\).
The circumference is \(14\pi\approx 43.98\text{ cm}\).
Problem 2: Circumference from the diameter
\(C=\pi d=\pi(18)=18\pi\approx 56.55\).
The circumference is \(18\pi\approx 56.55\text{ in}\).
Problem 3: Radius from the circumference
\(C=2\pi r\), so \(25.12=2(3.14)r=6.28r\). Divide both sides by 6.28: \(r=4\).
Check: \(2(3.14)(4)=25.12\).
The radius is \(4\text{ m}\).
Problem 4: Area of a circle
\(A=\pi r^{2}=\pi(9)^{2}=81\pi\approx 254.47\).
The area is \(81\pi\approx 254.47\text{ ft}^{2}\).
Problem 5: Area from the diameter
The radius is \(14: 2=7\text{ in}\). Then \(A=\pi(7)^{2}=49\pi\approx 153.94\).
The area is \(49\pi\approx 153.94\text{ in}^{2}\).
Problem 6: Radius from the area
\(\pi r^{2}=144\pi\), so \(r^{2}=144\) and \(r=12\) (a length is positive).
Then \(C=2\pi(12)=24\pi\approx 75.40\).
The radius is \(12\text{ cm}\) and the circumference is \(24\pi\text{ cm}\).
Problem 7: Sectors into a parallelogram
(a) \(C=2\pi(10)=20\pi\text{ cm}\).
(b) The base is half the circumference: \(\frac{20\pi}{2}=10\pi\text{ cm}\).
(c) The height is the radius, \(10\text{ cm}\). Area \(=10\pi\times 10=100\pi\approx 314.16\text{ cm}^{2}\). This matches \(\pi r^{2}=\pi(10)^{2}\).
Answers: \(20\pi\text{ cm}\); \(10\pi\text{ cm}\); \(100\pi\text{ cm}^{2}\).
Problem 8: Volume of a cylinder
\(V=\pi r^{2}h=\pi(4)^{2}(10)=160\pi\approx 502.65\).
The volume is \(160\pi\approx 502.65\text{ cm}^{3}\).
Problem 9: Height of a cylinder
\(225\pi=\pi(5)^{2}h=25\pi h\). Divide both sides by \(25\pi\): \(h=9\).
Check: \(\pi(25)(9)=225\pi\).
The height is \(9\text{ in}\).
Problem 10: Equal stacks of coins
(a) Height \(=6\times 0.3=1.8\text{ cm}\) for each stack.
(b) Volume \(=\) area \(\times \) height \(=5\times 1.8=9\text{ cm}^{3}\) for each stack.
(c) At every level, both stacks have a cross section with area \(5\text{ cm}^{2}\), and both are \(1.8\text{ cm}\) tall. By Cavalieri's principle the volumes are equal.
Answers: \(1.8\text{ cm}\); \(9\text{ cm}^{3}\); same cross-sectional areas and heights.
Problem 11: Volume of a pyramid
The base area is \(B=6\times 6=36\text{ m}^{2}\). Then \(V=\frac{1}{3}Bh=\frac{1}{3}(36)(9)=108\).
The volume is \(108\text{ m}^{3}\).
Problem 12: Rectangular pyramid
\(B=8\times 5=40\text{ ft}^{2}\). Then \(V=\frac{1}{3}(40)(12)=160\).
The volume is \(160\text{ ft}^{3}\).
Problem 13: Volume of a cone
\(V=\frac{1}{3}\pi r^{2}h=\frac{1}{3}\pi(9)(7)=21\pi\approx 65.97\).
The volume is \(21\pi\approx 65.97\text{ in}^{3}\).
Problem 14: Height of a cone
\(100\pi=\frac{1}{3}\pi(5)^{2}h=\frac{25\pi}{3}h\). Multiply by 3: \(300\pi=25\pi h\). Divide by \(25\pi\): \(h=12\).
The height is \(12\text{ cm}\).
Problem 15: Volume of a sphere
\(V=\frac{4}{3}\pi r^{3}=\frac{4}{3}\pi(216)=288\pi\approx 904.78\).
The volume is \(288\pi\approx 904.78\text{ cm}^{3}\).
Problem 16: Volume of a hemisphere
A hemisphere is half a sphere: \(V=\frac{1}{2}\cdot\frac{4}{3}\pi(5)^{3}=\frac{2}{3}\pi(125)=\frac{250\pi}{3}\approx 261.80\).
The bowl holds \(\frac{250\pi}{3}\approx 261.80\text{ in}^{3}\).
Problem 17: Cone compared with cylinder
Cylinder: \(\pi(4)^{2}(9)=144\pi\approx 452.39\text{ cm}^{3}\).
Cone: \(\frac{1}{3}\pi(16)(9)=48\pi\approx 150.80\text{ cm}^{3}\).
\(\frac{48\pi}{144\pi}=\frac{1}{3}\).
Answer: cylinder \(144\pi\text{ cm}^{3}\), cone \(48\pi\text{ cm}^{3}\), cone \(=\frac{1}{3}\) of cylinder.
Problem 18: Grain bin
(a) \(V=\pi(6)^{2}(20)=720\pi\approx 2{,}261.95\text{ ft}^{3}\).
(b) \(2{,}261.95: 50\approx 45.24\), so 45 loads leave some grain behind. The bin needs 46 loads.
Problem 19: Scoop on a sugar cone
Scoop: \(\frac{4}{3}\pi(1.5)^{3}=\frac{4}{3}\pi(3.375)=4.5\pi\approx 14.14\text{ in}^{3}\).
Cone: \(\frac{1}{3}\pi(1.5)^{2}(5)=\frac{1}{3}\pi(11.25)=3.75\pi\approx 11.78\text{ in}^{3}\).
Difference: \(4.5\pi-3.75\pi=0.75\pi\approx 2.36\).
The melted ice cream is larger than the cone, so it overflows by about \(2.36\text{ in}^{3}\).
Problem 20: Depth of water in a tank
The radius is \(5\text{ ft}\). Then \(1000=\pi(5)^{2}h=25\pi h\), so \(h=\frac{1000}{25\pi}=\frac{40}{\pi}\approx 12.73\).
The depth is about \(12.73\text{ ft}\).
Problem 21: Melting candles
One candle: \(\frac{1}{3}(3\times 3)(5)=15\text{ in}^{3}\). Four candles: \(60\text{ in}^{3}\).
Mold: \(60=\pi(2)^{2}h=4\pi h\), so \(h=\frac{15}{\pi}\approx 4.77\).
The wax is about \(4.77\text{ in}\) tall.
Problem 22: Cone cup poured into a cylinder
Cone: \(\frac{1}{3}\pi(16)(12)=64\pi\text{ cm}^{3}\).
Cylinder: \(64\pi=\pi(16)h\), so \(h=4\).
The water is \(4\text{ cm}\) deep, one third of the cone's height.
Problem 23: Ball in a box
Ball: \(r=3\), \(V=\frac{4}{3}\pi(27)=36\pi\approx 113.10\text{ cm}^{3}\). Box: \(6^{3}=216\text{ cm}^{3}\).
(a) \(\frac{36\pi}{216}=\frac{\pi}{6}\approx 0.5236\), which is about \(52.4\%\).
(b) \(216-36\pi\approx 102.90\text{ cm}^{3}\).
Answers: about \(52.4\%\); about \(102.90\text{ cm}^{3}\).
See every chapter on the Geometry math page.