Geometry Word Problems
The geometry problems in this set involve lines, angles, triangles, rectangles, squares, and circles. You will learn how to find length, perimeter, area, circumference, and volume, and how you can apply geometry to everyday problems.
- A sphere has a volume of 288π cm^{3}. Find its radius.
- 9.5 cm
- 7 cm
- 14 cm
- 6 cm
- Using the illustration provided below, if mABE = 4x + 5 and m CBD = 7x – 13, find the measure of ABE.
- 151°
- 73°
- 107°
- 29°
- Two angles are complementary. The measure of one angle is four times the measure of the other. Find the measure of the larger angle.
- 36°
- 72°
- 144°
- 18°
- If Gretta's bicycle has a 25-inch diameter wheel, how far will she travel in two turns of the wheel? (π = 3.14)
- 491 in
- 78.5 in
- 100 in
- 157 in
- Two angles are supplementary. The measure of one is 30 more than twice the measure of the other. Find the measure of the larger angle.
- 130°
- 20°
- 50°
- 70°
- Using the illustration provided, find the mAED. Given mBEC = 5x – 36 and mAED = 2x + 9.
- 141°
- 69°
- 111°
- 39°
- The measures of the angles of a triangle are in the ratio of 3:4:5. Find the measure of the largest angle.
- 75°
- 37.5°
- 45°
- 60°
- A mailbox opening is 4.5 inches high and 5 inches wide. What is the widest piece of mail able to fit in the mailbox without bending? Round answer to the nearest tenth.
- 9.5 inches
- 2.2 inches
- 6.7 inches
- 8.9 inches
- The figure below represents the cross section of a pipe inch thick that has an inside diameter of 3 inches. Find the area of the shaded region in terms of π.
- 8.75πin^{2}
- 3.25πin^{2}
- 7πin^{2}
- 1.75πin^{2}
- Using the same cross section of pipe from question 464, answer the following question. If the pipe is 18 inches long, what is the volume of the shaded region in terms of π?
- 31.5πin^{3}
- 126πin^{3}
- 157.5 in^{3}
- 58.5 in^{3}
- A person travels 10 miles due north, 4 miles due west, 5 miles due north, and 12 miles due east. How far is that person from the starting point?
- 23 miles northeast
- 13 miles northeast
- 17 miles northeast
- 17 miles northwest
- Using the illustration provided, find the area of the shaded region in terms of π.
- 264 – 18π
- 264 – 36π
- 264 – 12π
- 18π – 264
- Find how many square centimeters of paper are needed to create a label on a cylindrical can 45 cm tall with a circular base having diameter of 20 cm. Leave answer in terms of π.
- 450π cm^{2}
- 4,500π cm^{2}
- 900π cm^{2}
- 9,000π cm^{2}
- Using the illustration provided below, if the measure AEB = 5x + 40 and BEC = x + 20, find mDEC.
- 40°
- 25°
- 140°
- 65°
- The structural support system for a bridge is shown in the illustration provided. is parallel to is parallel to is parallel to . Find CGE.
- 46°
- 52°
- 82°
- 98°
- Find the area of the shaded portions, where . Leave answer in terms of π.
- 25π – 72
- 25π – 48
- 25π – 8
- 100π – 48
- Find the area of the shaded region in terms of π.
- 12 – 9π
- 36 – 9π
- 36 – 4.5π
- 2π – 16
- On a piece of machinery, the centers of two pulleys are 3 feet apart, and the radius of each pulley is 6 inches. How long a belt (in feet) is needed to wrap around both pulleys?
- (6 + .5π) ft
- (6 + .25π) ft
- (6 + 12π) ft
- (6 + π) ft
- Find the measure of each angle of a regular 15-sided polygon to the nearest tenth.
- 24°
- 12°
- 128.6°
- 156°
- A sand pile is shaped like a cone as illustrated below. How many cubic yards of sand are in the pile. Round to the nearest tenth. (π = 3.14)
- 5,358.9 yd^{3}
- 595.4 yd^{3}
- 198.5 yd^{3}
- 793.9 yd^{3}
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 volume of a sphere is found by using the formula π r^{3}. Since the volume is 288π cm^{3} and we are asked to find the radius, we will set up the following equation: π r^{3} = 288π. To solve for r, multiply both sides by 3; 4π r^{3} = 864π. Divide both sides by π; 4r^{3} = 864. Divide both sides by 4; r^{3} = 216. Take the cube root of both sides; r = 6. If you chose a, the formula for volume of a sphere was incorrect; π r^{3} was used instead of π r^{3}. If you chose c, near the end of calculations you mistakenly took the square root of 216 rather than the cube root.
- d. ABE and CBD are vertical angles that are equal in measurement. Solve the following equation for x: 4x + 5 = 7x – 13. Subtract 4x from both sides; 5 = 3x – 13. Add 13 to both sides; 18 = 3x. Divide both sides by 3; 6 = x or x = 6. To solve for ABE substitute x = 6 into the expression 4x + 5 and simplify; 4(6) + 5 equals 24 + 5 or 29. ABE equals 29°. If you chose a, you solved for ABC or EBD. If you chose b, you assumed the angles were supplementary and set the sum of the two angles equal to 180.
- b. If two angles are complementary, the sum of the measurement of the angles is 90°. 1 is represented by x. 2 is represented by 4x. Solve the following equation for x: x + 4x = 90. Simplify; 5x = 90. Divide both sides by 5; x = 18. The larger angle is 4x or 4(18), which equals 72°. If you chose a, the original equation was set equal to 180 rather than 90 and you solved for the smaller angle. If you chose c, the original equation was set equal to 180 rather than 90, and you solved for the larger angle. If you chose d, you solved the original equation correctly; however, you solved for the smaller of the two angles.
- d. To find how far the wheel will travel, find the circumference of the wheel multiplied by 2. The formula for the circumference of the wheel is π d. Since the diameter of the wheel is 25 inches, the circumference of the wheel is 25π. Multiply this by 2, (2)(25π) or 50π. Finally, substitute 3.14 for π 50(3.14) = 157 inches, the distance the wheel traveled in two turns. If you chose a, you used the formula for area of a circle rather than circumference. If you chose b, the distance traveled was one rotation, not two.
- a. If two angles are supplementary, the sum of the measurement of the angles is 180°. 1 is represented by x. 2 is represented by 2x + 30. Solve the following equation for x; x + 2x + 30 = 180. Simplify; 3x + 30 = 180. Subtract 30 from both sides; 3x = 150. Divide both sides by 3; x = 50. The larger angle is 2x + 30 or 2(50) + 30, which equals 130°. If you chose b, the equation was set equal to 90 rather than 180 and you solved for the smaller angle. If you chose c, x was solved for correctly; however, this was the smaller of the two angles. If you chose d, the original equation was set equal to 90 rather than 180, yet you continued to solve for the larger angle.
- d. AED and BEC are vertical angles that are equal in measurement. Solve the following equation for x: 5x – 36 = 2x + 9. Subtract 2x from both sides of the equation; 3x – 36 = 9. Add 36 to both sides of the equation; 3x = 45. Divide both sides by 3; x = 15. To solve for AED substitute x = 15 into the expression 2x + 9 and simplify. 2(15) + 9 equals 39. AED equals 39°. If you chose a, you solved for the wrong angle, either AEB or DEC. If you chose b, you assumed the angles were supplementary and set the sum of the angles equal to 180°. If you chose c, it was the same error as choice b.
- a. The sum of the measures of the angles of a triangle is 180°. Using this fact we can establish the following equation: 3x + 4x + 5x = 180. Simplifying; 12x = 180. Divide both sides by 12; x = 15. The largest angle is represented by 5x. Therefore, 5x, or 5(15), equals 75, the measure of the largest angle. If you chose b, the original equation was set equal to 90 rather than 180. If you chose c, this was the smallest angle within the triangle. If you chose d, this was the angle whose measurement lies between the smallest and largest angles.
- c. The widest piece of mail will be equal to the length of the diagonal of the mailbox. The width, 4.5 in, will be a leg of the right triangle. The height, 5 in, will be another leg of the right triangle. We will solve for the hypotenuse, which is the diagonal of the mailbox, using the Pythagorean theorem; a2 + b^{2}2 = c^{2} or 4.5^{2} + 5^{2} = c^{2}. Solve for c, 20.25 + 25 = c^{2}; 45.25 = c^{2}; c = 6.7. If you chose a, you assigned the legs the values of 4.5 and 10; 10 is incorrect. If you chose b, you assigned the legs the values of 5 and 10. Again, 10 is incorrect.
- d. To find the area of the cross section of pipe, we must find the area of the outer circle minus the area of the inner circle. To find the area of the outer circle, we will use the formula area = π r^{2}. The outer circle has a diameter of 4(3 + + ) and a radius of 2; therefore, the area = π2^{2} or 4π. The inner circle has a radius of 1.5; therefore, the area = π(1.5)^{2} or 2.25π. The difference, 4π – 2.25π or 1.75π is the area of the cross section of pipe. If you chose a, you used the outer circle's radius of 3 and the inner circle's radius of . If you chose b you used the outer circle's radius of and the inner circle's radius of 3. If you chose c, you used the outer radius of 4 and the inner radius of 3.
- a. To find the volume of the pipe with a known cross section and length of 18 inches, simply multiply the area of the cross section times the length of the pipe. The area of the cross section obtained from the previous question was 1.75π in^{2}. The length is 18 inches. Therefore, the volume is 1.75 in^{2} times 18 inches or 31.5π in^{3}. If you chose b, you multiplied choice c from the previous question by 18. If you chose c, you multiplied choice a from the previous question by 18. If you chose d, you multiplied choice b from the previous question by 18.
- c. Sketching an illustration would be helpful for this problem. Observe that point A is the starting point and point B is the ending point. After sketching the four directions, we connect point A to point B. We can add to the illustration the total distance traveled north as well as the total distance traveled east. This forms a right triangle, given the distance of both legs, with the hypotenuse to be solved. Using the Pythagorean theorem, a^{2} + b^{2} = c^{2}, or 8^{2} + 15^{2} = c^{2}; 64 + 225 = c^{2}; 289 = c^{2}; c = 17. If you chose a, you mistakenly traveled 4 miles due east instead of due west. If you chose b, you labeled the triangle incorrectly by assigning 15 to the hypotenuse rather than a leg. If you chose d, you solved the problem correctly but chose the wrong heading, northwest instead of northeast.
- b. The area of the shaded region is the area of a rectangle, 22 by 12, minus the area of a circle with a diameter of 12. The area of the rectangle is (22)(12) = 264. The area of a circle with diameter 12 and a radius of 6, is π(6)^{2} = 36π. The area of the shaded region is 264 – 36π. If you chose a, the formula for area of a circle was incorrect, π r^{2}. If you chose c, the formula for area of a circle was incorrect, π d. If you chose d, this was the reverse of choice a—area of the circle minus area of the rectangle.
- c. To find the area of the label, we will use the formula for the surface area of a cylinder, area = π dh, which excludes the top and bottom. Substituting d = 20 and h = 45, the area of the label is π(20)(45) or 900π cm^{2}. If you chose a, you used an incorrect formula for area, area = π rh. If you chose b, you used an incorrect formula for area, area = π r^{2}h.
- c. The sum of the measurement of AEB and BEC is 180°. Solve the following equation for x: 5x + 40 + x + 20 = 180. Simplify; 6x + 60 = 180. Subtract 60 from both sides; 6x = 120. Divide both sides by 6; x = 20. DEC and AEB are vertical angles that are equal in measurement. Therefore, if we find the measurement of AEB, we also know the measure of DEC. To solve for AEB, substitute x = 20 into the equation 5x + 40 or 5(20) + 40, which equals 140°. DEC is also 140°. If you chose a, you solved for BEC. If you chose b or d, the original equation was set equal to 90 rather than 180. In choice b, you then solved for BEC. In choice d, you solved for DEC.
- d. Two parallel lines cut by a transversal form corresponding angles that are congruent or equal in measurement. BAE is corresponding to CFE. Therefore CFE = 46°. CDF is corresponding to BEF. Therefore, BEF = 52°. The sum of the measures of the angles within a triangle is 180°. CFE + BEF + FGE = 180°. Using substitution, 46 + 52 + FGE = 180. Simplify; 98 + FGE = 180. Subtract 98 from both sides; FGE = 82°. FGE and CGE are supplementary angles. If two angles are supplementary, the sum of their measurements equals 180°. Therefore, FGE + CGE = 180. Using substitution, 82 + CGE = 180. Subtract 82 from both sides; CGE = 98°. If you chose a, you solved for CFE. If you chose b, you solved for BEF. If you chose c, you solved for FGE.
- b. To find the area of the shaded region, we must find the area of the circle minus the area of the rectangle. The formula for the area of a circle is π r^{2}. The radius is or (10), which is 5. The area of the circle is π(5^{2}) or 25π. The formula for the area of a rectangle is length × width. Using the fact that the rectangle is divided into two triangles with width of 6 and hypotenuse of 10, and using the Pythagorean theorem, we will find the length; a^{2} + b^{2} = c^{2}; a^{2} + 6^{2} = 10^{2}; a^{2} + 36 = 100; a^{2} = 64; a = 8. The area of the rectangle is length × width or 6 × 8 = 48. Finally, to answer the question, the area of the shaded region is the area of the circle – the area of the rectangle, or 25π – 48. If you chose a, the error was in the use of the Pythagorean theorem, 6^{2} + 10^{2} = c^{2}. If you chose c, the error was in finding the area of the rectangle. If you chose d, you used the wrong formula for area of a circle, π d^{2}.
- b. The area of the shaded region is equal to the area of the square minus the area of the two semicircles. The area of the square is s^{2} or 6^{2}, which equals 36. The area of the two semicircles is equal to the area of one circle. Area = π r^{2} or π(3)^{2} or 9π. Therefore, the area of the shaded region is 36 – 9π. If you chose a, you calculated the area of the square incorrectly as 12. If you chose c, you used an incorrect formula for the area of two semicircles, π r^{2}.
- d. To solve for the length of the belt, begin with the distance from the center of each pulley, 3 ft, and multiply by 2; (3)(2) or 6 ft. Secondly, you need to know that the distance of two semicircles with the same radius is equivalent to the circumference of one circle. Therefore C = π d or (12p) inches. Since the units are in feet, and not inches, convert (12p) inches to feet or (1p)ft. Now add these two values together, (6 + 1p)ft, to determine the length of the belt around the pulleys. If you chose a or b, you used an incorrect formula for circumference of a circle. Recall: Circumference = π d. If you chose c, you forgot to convert the unit from inches to feet.
- d. To find the measure of an angle of any regular polygon, we use the formula × 180, where n is the number of sides. Using 15 as the value for n, × 180 = × 180 or 156. Alternatively, we compute the exterior angle of a regular polygon using the formula , which in this case is = 24. Since the angle we want and the exterior angle are supplementary, we again have the answer 180° – 24° = 156°. If you chose a, you simply divided 360 (which is the sum of the exterior angles) by 15. If you chose b, you divided 180 by 15.
- c. To find how many cubic yards of sand are in the pile, we must find the volume of the pile in cubic feet and convert the answer to cubic yards. The formula for volume of a cone is V = (height)(Area of the base). The area of the base is found by using the formula Area = πr^{2}<. The area of the base of the sand pile is π(16)^{2} or 803.84 ft^{2}. The height of the pile is 20 feet. The volume of the pile in cubic feet is (803.84)(20) or 5,358.93 ft^{3}. To convert to cubic yards, divide 5,358.93 by 27 because 1 yard = 3 feet and 1 yd^{3} means 1 yd × 1 yd × 1 yd which equals 3 ft × 3 ft × 3 ft or 27 ft^{3}. The answer is 198.5 yd^{3}. If you chose a, you did not convert to cubic yards. If you chose b, you converted incorrectly by dividing 5,358.93 by 9 rather than 27. If you chose d, the area of the base formula was incorrect. Area of a circle does not equal πd^{2}.
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