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Word Problems Involving Geometric Shapes Help

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By — McGraw-Hill Professional
Updated on Sep 27, 2011

Introduction to Geometry Word Problems

To solve word problems involving geometric shapes, write down the formula or formulas referred to in the problem. For example, after reading “The perimeter of a rectangular room . . .” write P = 2 L + 2 W . Then fill in the information given about the formula. For example, after reading “The perimeter of the room is 50 feet . . .” write P = 50 and 50 = 2 L + 2 W . “Its width is two-thirds its length.” Write Geometric Problems and 50 = 2 L + 2 W becomes Geometric Problems .

Geometry Word Problems - Rectangles and Squares

The formulas you will need in this section are listed below.

Rectangle Formulas

Geometric Problems

  • Area A = LW
  • Perimeter (the length around its sides) P = 2 L + 2 W
  • Diagonal D 2 = L 2 + W 2

Examples

Example 1:

A square has a diameter of 50 cm. What is the length of each side?

Let x represent the length of each side.

Geometric Problems Examples

The diagonal formula for a rectangle is D 2 = L 2 + W 2 . In this example, D = 50, L = x , and W = x. D 2 = L 2 + W 2 becomes x 2 + x 2 = 50 2 .

Geometric Problems Examples

Example 2:

A rectangle is one inch longer than it is wide. Its diameter is five inches. What are its dimensions?

L = W + 1

Geometric Problems Examples

The diagonal formula for a rectangle is D 2 = L 2 + W 2 . In this example, D = 5 and L = W + 1. D 2 = L 2 + W 2 becomes 5 2 = ( W + 1) 2 + W 2 .

Geometric Problems Examples

The width is 3 inches and the length is 3 + 1 = 4 inches.

Find practice problems and solutions at Word Problems Involving Geometric Shapes Practice Problems - Set 1.

Triangles and Miscellaneous Shapes

Triangle Formulas

Geometric Problems

  • Area Geometric Problems BH
  • Perimeter P = a + b + c (for any triangle)
  • Pythagorean Theorem a 2 + b 2 = c 2 (for right triangles only)

Miscellaneous Shapes

  • Volume of a right circular cylinder V = π r 2 h , where r is the cylinder’s radius, h is the cylinder’s height
  • Surface area of a sphere (ball) is SA = 4π r 2 , where r is the sphere’s radius
  • Area of a circle A = π r 2 , where r is the circle’s radius
  • Volume of a rectangular box V = LWH , where L is the box’s length, W is the box’s width, and H is the box’s height

In many of the examples and practice problems, there will be two solutions to the equation but only one solution to the geometric problem. The extra solutions come from solving quadratic equations.

Examples

Example 1:

The area of a triangle is 40 in 2 . Its height is four-fifths the length of its base. What are its base and height?

The area is 40 and Geometric Problems Examples so the formula Geometric Problems Examples becomes Geometric Problems Examples .

Geometric Problems Examples

The triangle’s base is 10 inches long and its height is Geometric Problems Examples inches.

Example 2:

The hypotenuse of a right triangle is 34 feet. The sum of the lengths of the two legs is 46 feet. Find the lengths of the legs.

The sum of the lengths of the legs is 46 feet, so if a and b are the lengths of the legs, a + b = 46, so a = 46 – b . The hypotenuse is 34 feet so if c is the length of the hypotenuse, then the formula a 2 + b 2 = c 2 becomes (46 – b ) 2 + b 2 = 34 2.

Geometric Problems Examples

One leg is 30 feet long and the other is 46 – 30 = 16 feet long.

Example 3:

A can’s height is four inches and its volume is 28 cubic inches. What is the can’s radius?

The volume formula for a right circular cylinder is V = πr2h . The can’s volume is 28 cubic inches and its height is 4 inches, so V = πr2h becomes 28 = πr2 (4).

Geometric Problems Examples

The can’s radius is about 1.493 inches.

Example 4:

The volume of a box is 72 cm3. Its height is 3 cm. Its length is 1.5 times its width. What are the length and width of the box?

The formula for the volume of the box is V = LWH . The volume is 72, the height is 3 and the length is 1.5 times the width ( L = 1.5 W ) so the formula becomes 72 = (1.5 W ) W (3).

Geometric Problems Examples

The box’s width is 4 cm and its length is (1.5)(4) = 6 cm.

Example 5:

The surface area of a ball is 314 square inches. What is the ball’s diameter?

The formula for the surface area of a sphere is SA = 4πr2 . The area is 314, so the formula becomes 314 = 4πr2 .

Geometric Problems Examples

The radius of the ball is approximately 5 inches. The diameter is twice the radius, so the diameter is approximately 10 inches.

Example 6:

The manufacturer of a six-inch drinking cup is considering increasing its radius. The cup has straight sides (the top is the same size as the bottom). If the radius is increased by one inch, the new volume would be 169.6 cubic inches. What is the cup’s current radius?

The formula for the volume of a right circular cylinder is V = πr2h . The cup’s height is 6. If the cup’s radius is increased, the volume would be 169.6. Let x represent the cup’s current radius. Then the radius of the new cup would be x + 1. The volume formula becomes 169.6 = π( x + 1)2 6.

Geometric Problems Examples

The cup’s current radius is approximately 2 inches.

Find practice problems and solutions at Word Problems Involving Geometric Shapes Practice Problems - Set 2.

More practice problems for this concept can be found at: Algebra Quadratic Applications Practice Test.

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