Each worked solution names the shape or formula, shows the substitution and ends with the final answer in bold. Use the answers to check units, and compare your reasoning for the design problems step by step.
Problem 1: Slicing a cube
A cut parallel to a face of a cube is a square the same size as the face.
The area is \(6\times 6=36\text{ cm}^{2}\).
Answer: a square with area \(36\text{ cm}^{2}\).
Problem 2: Rectangle from a cylinder
The cross section is a rectangle. Its width is the diameter, \(2\times 4=8\text{ cm}\), and its height is \(10\text{ cm}\).
Area \(=8\times 10=80\text{ cm}^{2}\).
Answer: a rectangle, \(8\text{ cm}\) by \(10\text{ cm}\), area \(80\text{ cm}^{2}\).
Problem 3: Circle from a cylinder
The cross section is a circle with the same radius, \(5\text{ cm}\).
Area \(=\pi(5)^{2}=25\pi\approx 78.54\text{ cm}^{2}\). Circumference \(=2\pi(5)=10\pi\approx 31.42\text{ cm}\).
Answer: a circle; \(25\pi\text{ cm}^{2}\); \(10\pi\text{ cm}\).
Problem 4: Cutting a cone
(a) A circle. (b) An isosceles triangle with base \(2\times 3=6\text{ in}\) and height \(4\text{ in}\).
(c) Area \(=\frac{1}{2}(6)(4)=12\text{ in}^{2}\).
Answers: circle; triangle; \(12\text{ in}^{2}\).
Problem 5: Slicing a sphere
The radius of the sphere, the distance \(5\), and the slice radius form a right triangle with hypotenuse 13.
\(x^{2}+5^{2}=13^{2}\), so \(x^{2}=144\) and \(x=12\text{ in}\).
Area \(=\pi(12)^{2}=144\pi\approx 452.39\text{ in}^{2}\).
Answer: radius \(12\text{ in}\); area \(144\pi\text{ in}^{2}\).
Problem 6: Slicing a pyramid
Halfway up, the cross section is a smaller square similar to the base with scale factor \(\frac{1}{2}\).
Side \(=\frac{1}{2}(10)=5\text{ cm}\). Area \(=5^{2}=25\text{ cm}^{2}\), which is \(\frac{1}{4}\) of the base area \(100\text{ cm}^{2}\).
Answer: a square with side \(5\text{ cm}\) and area \(25\text{ cm}^{2}\).
Problem 7: Slices of a prism
A cut parallel to a face is a rectangle congruent to that face.
The \(6\) by \(4\) faces give \(24\text{ cm}^{2}\). The \(6\) by \(3\) faces give \(18\text{ cm}^{2}\). The \(4\) by \(3\) faces give \(12\text{ cm}^{2}\).
Answer: \(24\text{ cm}^{2}\), \(18\text{ cm}^{2}\), \(12\text{ cm}^{2}\).
Problem 8: Rotating a rectangle
The solid is a cylinder with radius \(3\text{ cm}\) and height \(7\text{ cm}\).
\(V=\pi(3)^{2}(7)=63\pi\approx 197.92\text{ cm}^{3}\).
Answer: cylinder, \(r=3\text{ cm}\), \(h=7\text{ cm}\), \(V=63\pi\text{ cm}^{3}\).
Problem 9: Rotating a right triangle
The solid is a cone with radius \(6\text{ in}\) and height \(8\text{ in}\).
Slant height \(=\sqrt{6^{2}+8^{2}}=\sqrt{100}=10\text{ in}\).
\(V=\frac{1}{3}\pi(6)^{2}(8)=96\pi\approx 301.59\text{ in}^{3}\).
Answer: cone; slant height \(10\text{ in}\); \(V=96\pi\text{ in}^{3}\).
Problem 10: Rotating about the other leg
Now the radius is \(8\text{ in}\) and the height is \(6\text{ in}\): \(V=\frac{1}{3}\pi(8)^{2}(6)=128\pi\approx 402.12\text{ in}^{3}\).
Difference: \(128\pi-96\pi=32\pi\approx 100.53\text{ in}^{3}\).
Answer: \(128\pi\text{ in}^{3}\); \(32\pi\text{ in}^{3}\) more.
Problem 11: Rotating a semicircle
The solid is a sphere with radius \(5\text{ cm}\).
\(V=\frac{4}{3}\pi(5)^{3}=\frac{500\pi}{3}\approx 523.60\text{ cm}^{3}\).
Answer: sphere; \(\frac{500\pi}{3}\approx 523.60\text{ cm}^{3}\).
Problem 12: Rotating a triangle about its altitude
The solid is a cone. The base of the triangle becomes the circle, so \(r=8: 2=4\text{ in}\), and \(h=5\text{ in}\).
\(V=\frac{1}{3}\pi(4)^{2}(5)=\frac{80\pi}{3}\approx 83.78\text{ in}^{3}\).
Answer: cone; \(\frac{80\pi}{3}\approx 83.78\text{ in}^{3}\).
Problem 13: Modeling a tree trunk
The radius is \(1\text{ ft}\). \(V=\pi(1)^{2}(30)=30\pi\approx 94.25\text{ ft}^{3}\).
Answer: about \(94.25\text{ ft}^{3}\).
Problem 14: Population density
Density \(=45{,}000: 30=1{,}500\).
Answer: 1,500 people per square mile.
Problem 15: Density of a county
Area \(=12\times 9=108\text{ mi}^{2}\).
Density \(=54{,}000: 108=500\).
Answer: \(108\text{ mi}^{2}\); 500 people per square mile.
Problem 16: Heating a room
Volume \(=12\times 10\times 8=960\text{ ft}^{3}\).
BTUs \(=960\times 6=5{,}760\).
Answer: 5,760 BTUs.
Problem 17: Density of a metal block
Volume \(=5\times 4\times 3=60\text{ cm}^{3}\).
Density \(=162: 60=2.7\).
Answer: \(2.7\text{ g/cm}^{3}\).
Problem 18: Comparing towns
Town A: \(9{,}600: 12=800\) people per \(\text{mi}^{2}\). Town B: \(7{,}350: 7=1{,}050\) people per \(\text{mi}^{2}\).
\(1{,}050-800=250\).
Answer: Town B, by 250 people per square mile.
Problem 19: Fencing a pen
The length and width add to \(48: 2=24\text{ ft}\). Test pairs: \(8\times 16=128\), \(10\times 14=140\), \(11\times 13=143\), \(12\times 12=144\), and the areas then decrease.
The best pen is \(12\text{ ft}\) by \(12\text{ ft}\) with area \(144\text{ ft}^{2}\). Compared with \(128\text{ ft}^{2}\), it adds \(16\text{ ft}^{2}\).
Answer: \(12\text{ ft}\) by \(12\text{ ft}\); \(16\text{ ft}^{2}\) more.
Problem 20: Two soup cans
Volumes: A: \(\pi(9)(8)=72\pi\). B: \(\pi(16)(4.5)=72\pi\).
Surface area \(=2\pi r^{2}+2\pi rh\). A: \(18\pi+48\pi=66\pi\approx 207.35\text{ cm}^{2}\). B: \(32\pi+36\pi=68\pi\approx 213.63\text{ cm}^{2}\).
Answer: Can A needs less metal (\(66\pi\text{ cm}^{2}\) versus \(68\pi\text{ cm}^{2}\)).
Problem 21: Sizing a water tank
Volume needed \(=1000: 7.5\approx 133.33\text{ ft}^{3}\).
\(\pi(3)^{2}h\ge 133.33\), so \(h\ge\frac{133.33}{9\pi}\approx 4.72\).
Rounding up to the next tenth so the tank holds enough: \(4.8\text{ ft}\).
Problem 22: Choosing a shipping box
Volumes: \(10\times 8\times 6=480\text{ in}^{3}\) and \(12\times 8\times 5=480\text{ in}^{3}\).
Surface areas: Box 1: \(2(80+60+48)=376\text{ in}^{2}\). Box 2: \(2(96+60+40)=392\text{ in}^{2}\).
Costs: \(376\times 0.02=\$7.52\) and \(392\times 0.02=\$7.84\).
Answer: Box 1 is cheaper by \(\$0.32\).
Problem 23: Banner grid
Margins use \(2\times 3=6\text{ in}\) and gutters use \(2\times 3=6\text{ in}\), so \(36-6-6=24\text{ in}\) remain for the columns.
Each column is \(24: 3=8\text{ in}\).
Answer: \(8\text{ in}\).
Problem 24: Crowded field
Area \(=120\times 80=9{,}600\text{ yd}^{2}\).
(a) \(14{,}400: 9{,}600=1.5\) people per \(\text{yd}^{2}\).
(b) Maximum \(=2\times 9{,}600=19{,}200\) people, so \(19{,}200-14{,}400=4{,}800\) more.
Answers: 1.5 people per square yard; 4,800 more people.
See every chapter on the Geometry math page.