Word Problems Involving Geometric Shapes Practice Problems (page 2)
Geometry Word Problems Practice Problems
Set 1: Introduction to Geometry Word Problems - Rectangles and Squares
To review word problems involving geometric shapes, go to Word Problems Involving Geometric Shapes Help
- The diameter of a square is 60 feet. What is the length of its sides?
- A rectangle has one side 14 cm longer than the other. Its diameter is 34 cm. What are its dimensions?
- The length of a rectangle is 7 inches more than its width. The diagonal is 17 inches. What are its dimensions?
- The width of a rectangle is three-fourths its length. The diagonal is 10 inches. What are its dimensions?
- The diameter of a rectangular classroom is 34 feet. The room’s length is 14 feet longer than its width. How wide and long is the classroom?
When there is more than one solution to an equation and one of them is not valid, only the valid solution will be given.
Let x represent the length of each side (in feet). The diagonal is 60 feet, so D = 60. The formula D 2 = L 2 + W 2 becomes 60 2 = x 2 + x 2.
The length of the square’s sides is feet, or approximately 42.4 feet.
The length is 14 cm more than the width, so L = W + 14. The diameter is 34 cm, so D = 34. The formula D 2 = L 2 + W 2 becomes 34 2 = ( W + 14) 2 + W 2 .
The width is 16 cm and the length is 16 + 14 = 30 cm.
The length is 7 inches more than the width, so L = W + 7. The diagonal is 17 inches. The formula D 2 = L 2 + W 2 becomes
The rectangle’s width is 8 inches and its length is 8 + 7 = 15 inches.
The width is three-fourths its length, so . The diagonal is 10 inches, so the formula D 2 = L 2 + W 2 becomes .
The rectangle’s length is 8 inches and its width is inches.
The classroom’s length is 14 feet more than its width, so L = W + 14. The diameter is 34 feet. The formula D 2 = L 2 + W 2 becomes 34 2 = ( W + 14) 2 + W 2.
The classroom is 16 feet wide and 16 + 14 = 30 feet long.
Set 2: Triangles and Miscellaneous Shapes
To review quadratic applications, go to Word Problems Involving Geometric Shapes Help
- The area of a triangle is 12 in 2. The length of its base is two-thirds its height. What are the base and height?
- The area of a triangle is 20 cm 2. The height is 3 cm more than its base. What are the base and height?
- The sum of the base and height of a triangle is 14 inches. The area is 20 in 2. What are the base and height?
- The hypotenuse of a right triangle is 85 cm long. One leg is 71 cm longer than the other. What are the lengths of its legs?
- The manufacturer of a food can wants to increase the capacity of one of its cans. The can is 5 inches tall and its diameter is 6 inches. The manufacturer wants to increase the can’s capacity by 50% and wants the can’s height to remain 5 inches. How much does the diameter need to increase?
- A pizza restaurant advertises that its large pizza is 20% larger than the competition’s large pizza. The restaurant’s large pizza is 16 inches in diameter. What is the diameter of the competition’s large pizza?
The area formula for a triangle is . The area is 12. The length of its base is two-thirds its height, so . The formula becomes .
The height of the triangle is 6 inches. Its base is inches.
The formula for the area of a triangle is . The area is 20. The height is 3 cm more than the base, so H = B + 3. The formula becomes .
The triangle’s base is 5 cm and its height is 5 + 3 = 8 cm.
The formula for the area of a triangle is . The area is 20. B + H = 14 so H = 14 – B . The formula becomes .
There are two triangles that satisfy the conditions. If the base is 10 inches, the height is 14 – 10 = 4 inches. If the base is 4 inches, the height is 14 – 4 = 10 inches.
By the Pythagorean theorem, a 2 + b 2 = c . The hypotenuse is c , so c = 85. One leg is 71 longer than the other so a = b + 71 ( b = a + 71 also works). The Pythagorean theorem becomes 85 2 = ( b + 71) 2 + b 2.
The shorter leg is 13 cm and the longer leg is 13 + 71 = 84 cm.
Because the can’s diameter is 6, the radius is 3. Let x represent the increase in the can’s radius. The radius of the new can is 3 + x. The volume of the current can is V = π r 2 h = π(3) 2 5 = 45π. To increase the volume by 50% means to add half of 45π to itself; the new volume would be . The volume formula for the new can becomes .
(The other solution is negative.)
The manufacturer should increase the can’s radius by about 0.674 inches. Because the diameter is twice the radius, the manufacturer should increase the can’s diameter by about 2(0.674) = 1.348 inches.
A pizza’s shape is circular so we need the area formula for a circle which is A = π r 2. The radius is half the diameter, so the restaurant’s large pizza has a radius of 8 inches. The area of the restaurant’s large pizza is π(8) 2 = 64π ≈ 201. The restaurant’s large pizza is 20% larger than the competition’s large pizza. Let A represent the area of the competition’s large pizza. Then 201 is 20% more than A :
The competition’s radius is approximately 7.3 inches, so its diameter is approximately 2(7.3) = 14.6 inches.
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