← Back to the Cross Sections And Geometric Modeling worksheet
Last updated: October 22, 2026 · Written by the Math Made Easy math teachers
Standards: G-GMD.B.4, G-MG.A.1, G-MG.A.2, G-MG.A.3 · Level: Grade 10 Geometry
Problem 1 Name the Cross Section
(a) circle. (b) rectangle. (c) circle (smaller than the base). (d) triangle (isosceles). (e) circle (a great circle with the same radius as the sphere).
(a) circle (b) rectangle (c) circle (d) triangle (e) circle
Problem 2 Slices of a Rectangular Box
A slice parallel to a face of a rectangular box is congruent to that face.
(a) 8 · 5 = 40. (b) 5 · 3 = 15.
(a) 40 in² (b) 15 in²
Problem 3 Slicing a Pyramid at Half Height
The cross section is similar to the base with scale factor ½. Its sides are 6 ft and 4 ft.
Area = 6 · 4 = 24, which is ¼ of the base area, 96.
6 ft by 4 ft; area 24 ft²
Problem 4 Slicing a Cone
The small cone is similar to the large one with scale factor 915 = 35.
Radius = 10 · 35 = 6. Area = π(6²) = 36π ≈ 113.1.
Radius 6 cm; area 36π cm² ≈ 113.1 cm²
Problem 5 Slicing a Sphere
The sphere radius is the hypotenuse of a right triangle whose legs are the distance 8 and the cross-section radius x.
x = √(17² − 8²) = √225 = 15. Area = π(15²) = 225π ≈ 706.9.
Radius 15 cm; area 225π cm² ≈ 706.9 cm²
Problem 6 Rectangle Rotated into a Cylinder
The solid is a cylinder with radius 4 and height 9.
V = π(4²)(9) = 144π ≈ 452.4.
Cylinder; 144π ft³ ≈ 452.4 ft³
Problem 7 Triangle Rotated Two Ways
(a) Radius 6, height 8: V = 13π(6²)(8) = 96π ≈ 301.6.
(b) Radius 8, height 6: V = 13π(8²)(6) = 128π ≈ 402.1.
Both solids are cones, but the one with the larger radius holds more.
(a) 96π ≈ 301.6 (b) 128π ≈ 402.1
Problem 8 Half Disk Rotated into a Sphere
The solid is a sphere of radius 5. V = 43π(5³) = 500π3 ≈ 523.6.
Sphere; ≈ 523.6 cm³
Problem 9 Isosceles Triangle About Its Altitude
The altitude splits the triangle into two right triangles. The solid is a cone with radius 5 (half the base) and height 12.
V = 13π(5²)(12) = 100π ≈ 314.16.
Cone; ≈ 314.2 in³
Problem 10 Population Density
Density = 842,600 ÷ 197 ≈ 4277.16.
≈ 4277 people per square mile
Problem 11 Mass from Density
Volume = π(3²)(10) = 90π ≈ 282.74 cm³.
Mass = 2.7 · 282.74 ≈ 763.41 g.
≈ 763 g
Problem 12 Heating a Room
(a) Volume = 18 · 14 · 8 = 2,016 ft³. BTUs = 2,016 · 22 = 44,352.
(b) 40,000 is less than 44,352, so the heater is not enough.
(a) 44,352 BTUs (b) No
Problem 13 Does the Soup Can Match Its Label?
The radius is 3.3 cm. V = π(3.3²)(10.2) ≈ 348.96 cm³ ≈ 349 mL.
This is within about 1 mL (0.3%) of 350, so the claim is reasonable.
≈ 349 mL; reasonable
Problem 14 Volume of a Tree Trunk
Radius = 942π ≈ 14.96 in.
Height = 30 ft = 360 in. V = π(14.96)²(360) ≈ 253,133 in³.
Convert: ÷ 1,728 ≈ 146.49 ft³.
Radius ≈ 15 in; volume ≈ 146.5 ft³
Problem 15 Design a Can
V = πr²(2r) = 2πr³ = 500, so r³ = 250π ≈ 79.58.
r ≈ 4.3 cm and h = 2r ≈ 8.6 cm.
r ≈ 4.3 cm; h ≈ 8.6 cm
Problem 16 Which Town Is Denser?
Town A: 18,400 ÷ 7.2 ≈ 2555.56. Town B: 31,500 ÷ 13.5 ≈ 2333.33.
Difference ≈ 222.22.
Town A by about 222 people per square mile
Problem 17 Tank Design
Volume needed = 12,000 ÷ 7.48 ≈ 1604.28 ft³.
π(5²)h = 1604.28, so h = 1604.28 ÷ (25π) ≈ 20.43.
≈ 20.4 ft
Problem 18 Cans in a Box
Box: 4 · 6.4 = 25.6 by 3 · 6.4 = 19.2 by 11, volume = 5406.72 cm³.
Cans: 12 · π(3.2²)(11) ≈ 4246.43 cm³.
Percent = 4246.43 ÷ 5406.72 ≈ 78.54%.
≈ 78.5%
Problem 19 Water in a Conical Cup
The scale factor is 812 = 23. The surface radius is 4.5 · 23 = 3 cm.
Water volume = 13π(3²)(8) = 24π ≈ 75.4 cm³.
Cup volume = 13π(4.5²)(12) = 81π. Fraction = 24π ÷ 81π = 827 ≈ 29.6%.
Radius 3 cm; ≈ 75.4 cm³; 8/27 of the cup (about 29.6%)
Problem 20 Shipping Tube Box
Box = 4.5 · 4.5 · 36 = 729 in³. Tube = π(2.25²)(36) ≈ 572.56 in³.
Empty = 729 − 572.56 = 156.44 in³, which is 21.46% of the box.
Box 729 in³; tube ≈ 572.6 in³; empty ≈ 21.5%