By LearningExpress Editors
Updated on Sep 26, 2011
Answers
The following explanations show one way in which each problem can be solved. You may have another method for solving these problems.
 d. The area of a rectangle is length × width.
 b. The volume of a sphere is times π times the radius cubed.
 b. The area of a triangle is times the length of the base times the length of the height.
 b. The surface area of a sphere is four times π times the radius squared.
 d. The area of a circle is π times the radius squared.
 a. The volume of a cylinder is π times the radius squared, times the height of the cylinder.
 c. The perimeter of a square is four times the length of one side.
 c. The area of the base is π times radius squared. The area of the curved region is two times π times radius times height. Notice there is only one circular region since the storage tank would be on the ground. This area would not be painted.
 a. The area of a square is side squared or side times side.
 c. The circumference or distance around a circle is π times the diameter.
 a. The perimeter of a rectangle is two times the length plus two times the width.
 d. The volume of a cube is the length of the side cubed or the length of the side times the length of the side times the length of the side.
 a. The perimeter of a triangle is length of side a plus length of side b plus length of side c.
 c. The area of a rectangle is length times width. Therefore, the area of the racquetball court is equal to 40 ft times 20 ft or 800 ft^{2}. If you chose answer d, you found the perimeter or distance around the court.
 a. The width of the field, 180 ft, must be divided by the width of the mower, 2 ft. The result is that he must mow across the lawn 90 times. If you chose b, you calculated as if he were mowing the length of the field. If you chose c, you combined length and width, which would result in mowing the field twice.
 a. The area of the room is the sum of the area of four rectangular walls. Each wall has an area of length times width, or (9)(5.5), which equals 49.5 ft^{2}. Multiply this by 4 which equals 198 ft^{2}. If you chose b, you added 9 ft and 5.5 ft instead of multiplying.
 c. The ceiling fan follows a circular pattern, therefore area = π r^{2}. Area = (3.14)(25)^{2} = 1,962.5 in^{2}. If you chose a, the incorrect formula you used was π^{2}r. If you chose d, the incorrect formula you used was π d.
 c. To find the height of Heather's poodle, set up the following proportion: height of the building/shadow of the building = height of the poodle/shadow of the poodle or . Crossmultiply, 112.5 = 30x. Solve for x; 3.75 = x. If you chose d, the proportion was set up incorrectly as .
 b. The volume of a cylinder is π r^{2}h. Using a height of 4 ft and radius of 10 ft, the volume of the pool is (3.14)(10)^{2}(4) or 1,256 ft^{3}. If you chose a, you used π dh instead of π r^{2}h. If you chose c, you used the diameter squared instead of the radius squared.
 b. The connection of the pole with the ground forms the right angle of a triangle. The length of the pole is a leg within the right triangle. The distance between the stake and the pole is also a leg within the right triangle. The question is to find the length of the cable, which is the hypotenuse. Using the Pythagorean theorem: (36)^{2} + (15)^{2} = c^{2}; 1,296 + 225 = c^{2}; 1,521 = c^{2}; 39 = c.
More practice problems on geometry word problems can be found at:
 Geometry Word Problems Practice Questions Set 1 (You are here)
 Geometry Word Problems Practice Questions Set 2
 Geometry Word Problems Practice Questions Set 3
 Geometry Word Problems Practice Questions Set 4
 Geometry Word Problems Practice Questions Set 5
 Geometry Word Problems Practice Questions Set 6
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From 501 Math Word Problems. Copyright © 2003 by LearningExpress, LLC. All Rights Reserved.
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