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Last updated: October 22, 2026 · Written by the Math Made Easy math teachers

Standards: G-GMD.A.1, G-GMD.A.3 · Level: Grade 10 Geometry
Problem 1 Area of a Circle from Wedges
(a) The base is half the circumference: ½ · 2π(8) = 8π. The height is the radius, 8.
(b) Area = base · height = 8π · 8 = 64π.
(a) base 8π in, height 8 in (b) 64π in²
Problem 2 Volume of a Cylinder
V = πr²h = π(5²)(12) = 300π.
300π ≈ 942.5.
300π in³ ≈ 942.5 in³
Problem 3 Cylinder from a Diameter
The radius is 14 ÷ 2 = 7. V = π(7²)(9) = 441π ≈ 1385.44.
≈ 1385.4 cm³
Problem 4 Volume of a Cone
V = 13π(6²)(10) = 13π · 360 = 120π.
120π ≈ 376.99.
120π m³ ≈ 377.0 m³
Problem 5 Volume of a Square Pyramid
Base area = 9 · 9 = 81. V = 13 · 81 · 14 = 27 · 14 = 378.
378 m³
Problem 6 Volume of a Sphere
V = 43π(9³) = 43π · 729 = 972π.
972π ≈ 3053.63.
972π cm³ ≈ 3053.6 cm³
Problem 7 Cavalieri and an Oblique Cylinder
(a) Both have the same height, and every horizontal slice of each is a circle of radius 4, so every pair of slices has equal area. By Cavalieri’s principle the volumes are equal. The slant side does not matter.
(b) V = π(4²)(10) = 160π ≈ 502.7.
160π ft³ ≈ 502.7 ft³
Problem 8 One Third of a Cube
(a) 6³ = 216.
(b) 216 ÷ 3 = 72. Formula: 13 · 36 · 6 = 72.
(a) 216 in³ (b) 72 in³
Problem 9 Height of a Cone
13π(6²)h = 96π, so 12h = 96 and h = 8.
8 in
Problem 10 Radius of a Cylinder
πr²(6) = 486π, so r² = 81 and r = 9.
9 m
Problem 11 Radius of a Sphere
43πr³ = 288π, so r³ = 288 · 34 = 216 and r = 6.
6 ft
Problem 12 Hemisphere Bowl
Hemisphere volume = ½ · 43π(10³) = 2000π3 ≈ 2094.4 cm³.
Liters: 2094.4 ÷ 1,000 ≈ 2.09.
≈ 2094.4 cm³ ≈ 2.1 L
Problem 13 Rectangular Pyramid
Base area = 8 · 15 = 120. V = 13 · 120 · 11 = 40 · 11 = 440.
440 ft³
Problem 14 Cone with a Slant Height
The height, radius and slant height form a right triangle: h = √(17² − 8²) = √225 = 15.
V = 13π(8²)(15) = 320π ≈ 1005.31.
h = 15 cm; V = 320π cm³ ≈ 1005.3 cm³
Problem 15 Grain Silo
Cylinder: π(6²)(24) = 864π. Cone: 13π(6²)(5) = 60π.
Total = 924π ≈ 2902.83.
924π ft³ ≈ 2903 ft³
Problem 16 Ice Cream Cone
Cone: 13π(3²)(11) = 33π. Hemisphere: 23π(3³) = 18π.
Total = 51π ≈ 160.22.
51π cm³ ≈ 160.2 cm³
Problem 17 Water Tank
Capacity = π(3²)(4.5) = 40.5π ≈ 127.23 m³.
80%: 0.8 · 127.23 ≈ 101.79 m³ = 101,788 L.
≈ 102,000 L
Problem 18 Melting a Sphere
Sphere volume = 43π(6³) = 288π.
Cylinder: π(4²)h = 16πh = 288π, so h = 18.
18 cm
Problem 19 Pyramid Monument
Monument: 13 · 32² · 21 = 13 · 1,024 · 21 = 7,168 m³.
Each dimension is divided by 8, so the volume is divided by 8³ = 512: 7,168 ÷ 512 = 14.
Monument 7,168 m³; model 14 m³
Problem 20 Sand Timer
Cone volume = 13π(3²)(6) = 18π.
Cylinder: π(3²)h = 9πh = 18π, so h = 2.
(Equal bases mean the cone fills one third of a cylinder of its own height, 6 ÷ 3 = 2.)
2 cm
