Each solution shows the radius, the formula, and the arithmetic with \(\pi=3.14\) unless the problem asks for an answer in terms of \(\pi\). Compare every step with your own work, check units, and notice whether you squared only the radius in area problems.
Problem 1: Radius and diameter
The radius is half the diameter: \(14: 2=7\).
The diameter is twice the radius: \(9\times 2=18\).
Radius \(7\text{ cm}\); diameter \(18\text{ in}\)
Problem 2: Diameter and radius pairs
The radius is half the diameter, and the diameter is twice the radius.
- \(22 : 2 = 11\): \(11\text{ cm}\)
- \(15 \times 2 = 30\): \(30\text{ in}\)
- \(3.5 : 2 = 1.75\): \(1.75\text{ ft}\)
Problem 3: Circumference from the diameter
\(C=\pi d=3.14\times 10=31.4\).
\(31.4\text{ in}\)
Problem 4: Labeled diameter
\(C=\pi d=3.14\times 25\)
\(3.14\times 25=78.5\)
\(78.5\text{ cm}\)
Problem 5: Circumference from the radius
\(C=2\pi r=2\times 3.14\times 7=6.28\times 7=43.96\).
\(43.96\text{ ft}\)
Problem 6: Fence around a garden
The fence is the circumference: \(C=2\pi r=2\times 3.14\times 12\).
\(2\times 3.14=6.28\) and \(6.28\times 12=75.36\).
\(75.36\text{ m}\)
Problem 7: Area from the radius
\(A=\pi r^{2}=3.14\times 6^{2}=3.14\times 36=113.04\).
\(113.04\text{ m}^{2}\)
Problem 8: Labeled radius
\(A=\pi r^{2}=3.14\times 9^{2}=3.14\times 81\)
\(3.14\times 81=254.34\)
\(254.34\text{ in}^{2}\)
Problem 9: Area from the diameter
The radius is \(20: 2=10\).
\(A=3.14\times 10^{2}=3.14\times 100=314\).
\(314\text{ cm}^{2}\)
Problem 10: Circular patio
The radius is \(16: 2=8\).
\(A=3.14\times 8^{2}=3.14\times 64=200.96\).
\(200.96\text{ ft}^{2}\)
Problem 11: Answers in terms of pi
\(C=2\pi r=2\times 8\times \pi=16\pi\).
\(A=\pi r^{2}=\pi\times 64=64\pi\).
\(C=16\pi\text{ cm}\) and \(A=64\pi\text{ cm}^{2}\)
Problem 12: Exact answers with pi
\(C=2\pi r=2\times 11\times \pi=22\pi\).
\(A=\pi r^{2}=\pi\times 121=121\pi\).
\(C=22\pi\text{ m}\) and \(A=121\pi\text{ m}^{2}\)
Problem 13: Work backward from circumference
\(C=\pi d\), so \(d=C: \pi=62.8: 3.14=20\).
The radius is \(20: 2=10\).
Diameter \(20\text{ cm}\); radius \(10\text{ cm}\)
Problem 14: Work backward from area
\(A=\pi r^{2}\), so \(r^{2}=706.5: 3.14=225\).
\(15\times 15=225\), so \(r=15\).
Diameter: \(2\times 15=30\). Circumference: \(C=3.14\times 30=94.2\).
Check: \(3.14\times 15^{2}=3.14\times 225=706.5\).
Radius \(15\text{ cm}\); diameter \(30\text{ cm}\); circumference \(94.2\text{ cm}\)
Problem 15: Bicycle wheel
One turn rolls one circumference: \(C=3.14\times 26=81.64\text{ in}\).
Ten turns: \(81.64\times 10=816.4\).
\(816.4\text{ in}\)
Problem 16: Lace trim
The edge is the circumference: \(C=2\pi r=2\times 3.14\times 36=226.08\text{ in}\).
Convert to feet: \(226.08: 12=18.84\text{ ft}\).
Cost: \(18.84\times 1.50=28.26\).
\(226.08\text{ in}\), or \(18.84\text{ ft}\); the lace costs \(\$28.26\).
Problem 17: Sprinkler coverage
Square: \(24\times 24=576\).
Circle: \(3.14\times 12^{2}=3.14\times 144=452.16\).
Not watered: \(576-452.16=123.84\).
\(123.84\text{ ft}^{2}\)
Problem 18: Window with a semicircle
Rectangle: \(6\times 4=24\).
The semicircle has diameter \(6\), so \(r=3\). Half a circle: \(\frac{1}{2}\times 3.14\times 3^{2}=\frac{1}{2}\times 28.26=14.13\).
Total: \(24+14.13=38.13\).
\(38.13\text{ ft}^{2}\)
Problem 19: Gravel path
The outer circle has radius \(6+2=8\text{ ft}\). Its area is \(3.14\times 8^{2}=3.14\times 64=200.96\).
The pond has area \(3.14\times 6^{2}=3.14\times 36=113.04\).
The path is the difference: \(200.96-113.04=87.92\).
\(87.92\text{ ft}^{2}\)
Problem 20: Quarter-circle flower bed
(a) The radius is \(10\). Area: \(\frac{1}{4}\times 3.14\times 10^{2}=\frac{1}{4}\times 314=78.5\text{ ft}^{2}\).
(b) Curved edge: \(\frac{1}{4}\times 2\times 3.14\times 10=15.7\). Add the two straight edges: \(15.7+10+10=35.7\).
(a) \(78.5\text{ ft}^{2}\); (b) \(35.7\text{ ft}\)
See every chapter on the Grade 7 math page.