Geometry Geometric Modeling Applications Worksheet Solutions (PDF)

Geometry math worksheet solutions: Geometric Modeling Applications – Maths PDF

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This modeling practice solutions for Geometry covers geometric modeling applications (Common Core G-MG.A.1, G-MG.A.2, G-MG.A.3), with a free PDF to print.

Use this answer key to check the 20 problems on geometric modeling. Every solution shows the formula or theorem used, each step of the work, and the final answer in bold. If an answer differs from yours, compare the steps line by line to find where the work changed.

Geometry · Common Core State Standards: G-MG.A.1, G-MG.A.2, G-MG.A.3

Modeling practice solutions: exercises 1 to 10

Exercise 1 – Can as a cylinder

V is about 393.4 cm³.

Radius $3.3$ cm.

$V = \pi \cdot 3.3^{2} \cdot 11.5 = 393.4$

393.4 cubic centimeters

Exercise 2 – A log

Radius $12$ in $= 1$ ft. Length 20 ft.

$V = \pi \cdot 1^{2} \cdot 20 = 20\pi = 62.8$

62.8 cubic feet

Exercise 3 – Density of aluminum

V = 400 cm³.

Volume: $10 \cdot 8 \cdot 5 = 400$ cm³.

$\frac{1{,}080}{400} = 2.7$

2.7 g per cm³

Exercise 4 – Mass of a lead cylinder

$V = \pi \cdot 2^{2} \cdot 10 = 40\pi = 125.7$ cm³.

$11.34 \cdot 125.66 = 1,425$

About 1,425 grams

Exercise 5 – Population density

$\frac{31{,}250}{12.5} = 2{,}500$

2,500 people per square mile

Exercise 6 – Population from density

$4{,}200 \cdot 6.5 = 27{,}300$

27,300 people

Exercise 7 – Painting a room

Walls: $2(12 + 10) \cdot 8 = 352$ sq ft.

$\frac{352}{350} = 1.01$, so 2 gallons are needed.

Cost: $2 \cdot \$32 = \$64$.

352 sq ft; 2 gallons; $64

Exercise 8 – Weight of a beam

Cross section: $\frac{4}{12} \cdot \frac{6}{12} = \frac{1}{6}$ sq ft.

Volume: $\frac{1}{6} \cdot 12 = 2$ cu ft.

Weight: $2 \cdot 35 = 70$ pounds.

2 cu ft; 70 pounds

Exercise 9 – Gallons in a tank

$V = \pi \cdot 16 \cdot 10 = 160\pi = 502.7$ cu ft.

$502.7 \cdot 7.48 = 3,760$

About 3,760 gallons

Exercise 10 – Concrete slab

Thickness $\frac{4}{12} = \frac{1}{3}$ ft.

Volume: $20 \cdot 12 \cdot \frac{1}{3} = 80$ cu ft $= \frac{80}{27} = 2.96$ cu yd.

Cost: $2.96 \cdot \$145 = \$430$

2.96 cu yd; about $430

Exercises 11 and up

Exercise 11 – Fence design

Best of the three designs: 30 ft by 60 ft.

Constraint: $2w + l = 120$, so $l = 120 – 2w$.

w = 25: l = 70, area 1,750. w = 30: l = 60, area 1,800. w = 35: l = 50, area 1,750.

Width 30 ft by length 60 ft, area 1,800 sq ft

Exercise 12 – Minimum height of a tank

$\pi \cdot 5^{2} \cdot h \ge 1{,}000$

$h \ge \frac{1{,}000}{25\pi} = 12.7$

At least 12.7 feet

Exercise 13 – Material for a can

$2\pi r^{2} + 2\pi rh = 2\pi(3.3)^{2} + 2\pi(3.3)(11.5)$

$= 68.4 + 238.4 = 307$

About 307 square centimeters

Exercise 14 – Painting a tank

$2\pi \cdot 4 \cdot 10 = 80\pi = 251.3$ sq ft.

$\frac{251.3}{250} = 1.005$, so buy 2 gallons.

251.3 sq ft; 2 gallons

Exercise 15 – Attic volume

$V = \frac{1}{3} \cdot (30 \cdot 20) \cdot 8 = 1{,}600$

1,600 cubic feet

Exercise 16 – Basketball

V is about 448.9 in³.

Radius 4.75 in.

$V = \frac{4}{3}\pi \cdot 4.75^{3} = 448.9$

448.9 cubic inches

Exercise 17 – Grass seed

Lot with the house footprint.

Lot: $60 \cdot 90 = 5{,}400$. House: $40 \cdot 30 = 1{,}200$.

Seeded area: $5{,}400 – 1{,}200 = 4{,}200$ sq ft.

$\frac{4{,}200}{400} = 10.5$

10.5 pounds

Exercise 18 – Honeycomb cell

$A = \frac{3\sqrt{3}}{2} \cdot 3^{2} = \frac{27\sqrt{3}}{2} = 23.4$

23.4 square millimeters

Exercise 19 – Tennis ball can

Can: $\pi r^{2} \cdot 6r = 6\pi r^{3}$.

Balls: $3 \cdot \frac{4}{3}\pi r^{3} = 4\pi r^{3}$.

Empty: $6\pi r^{3} – 4\pi r^{3} = 2\pi r^{3}$, which is $\frac{2}{6} = \frac{1}{3}$ of the can.

One third

Exercise 20 – Best pizza value

10-inch: $\pi \cdot 5^{2} = 78.5$ sq in, so $\frac{78.5}{8} = 9.8$ sq in per dollar.

14-inch: $\pi \cdot 7^{2} = 153.9$ sq in, so $\frac{153.9}{14} = 11.0$ sq in per dollar.

The 14-inch pizza is the better value

Key facts behind this modeling practice solutions

Key facts

A model represents a real object with a simple shape, for example a can as a cylinder.
Density = amount / area or volume (people per square mile, grams per cubic centimeter).
Mass = density x volume.
Design problems have constraints such as fixed fencing, minimum volume or a cost limit.
State answers with units and check they are reasonable.

How to use this modeling practice solutions

Cover each solution, solve the exercise again, then uncover it and compare every line. If your answer differs, find the first step that changed and redo the exercise from there.

Download the modeling practice solutions as a PDF with the button below and keep it with your notes.

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