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.
- Find the surface area of the monument to the nearest meter.
- 13,820 m^{2}
- 13,451 m^{2}
- 3,455 m^{2}
- 13,543 m^{2}
- An inground pool is filling with water. The shallow end is 3 ft deep and gradually slopes to the deepest end, which is 10 ft deep. The width of the pool is 15 ft and the length is 30 ft. What is the volume of the pool?
- 1,575 ft^{3}
- 4,500 ft^{3}
- 2,925 ft^{3}
- 1,350 ft^{3}
For questions 3 and 4, refer to the following illustration:
- In a periscope, a pair of mirrors is mounted parallel to each other as shown. The path of light becomes a transversal. If 2 measures 50°, what is the measurement of 3?
- 50°
- 40°
- 130°
- 310°
- Given that 2 measures 50°, what is the measurement of 4?
- 50°
- 40°
- 130°
- 85°
- The angle measure of the base angles of an isosceles triangle are represented by x and the vertex angle is 3x + 20. Find the measure of a base angle.
- 116°
- 40°
- 32°
- 14°
- Using the information from question 440, find the measure of the vertex angle of the isosceles triangle.
- 32°
- 62°
- 58°
- 116°
- In parallelogram ABCD, A = 5x + 2 and C = 6x –4. Find the measure of A.
- 32°
- 6°
- 84.7°
- 44°
- The longer base of a trapezoid is three times the shorter base. The nonparallel sides are congruent. The nonparallel side is 5 cm more that the shorter base. The perimeter of the trapezoid is 40 cm. What is the length of the longer base?
- 15 cm
- 5 cm
- 10 cm
- 21 cm
- The measure of the angles of a triangle are represented by 2x + 15, x + 20, and 3x + 25. Find the measure of the smallest angle within the triangle.
- 40°
- 85°
- 25°
- 55°
- Suppose ABCD is a rectangle. IF to the nearest tenth.
- 4.0
- 5.8
- 11.7
- 8.0
- The perimeter of the parallelogram is 32 cm. What is the length of the longer side?
- 9 cm
- 10 cm
- 6 cm
- 12 cm
- A door is 6 feet and 6 inches tall and 36 inches wide. What is the widest piece of sheetrock that can fit through the door? Round to the nearest inch.
- 114 in
- 86 in
- 85 in
- 69 in
- The width of a rectangle is 20 cm. The diagonal is 8 cm more than the length. Find the length of the rectangle.
- 20
- 23
- 22
- 21
- The measures of two complementary angles are in the ratio of 5:7. Find the measure of the smallest angle.
- 75°
- 37.5°
- 52.5°
- 105°
- In parallelogram ABCD, m A = 3x + 10 and mD = 2x + 30, find the mA.
- 70°
- 40°
- 86°
- 94°
- Using the diagram below and the fact that A + B + C + D = 320, find mE.
- 80°
- 40°
- 25°
- 75°
- The base of a triangle is 4 times as long as its height. If together they measure 95 cm, what is the area of the triangle?
- 1,444 cm^{2}
- 100 cm^{2}
- 722 cm^{2}
- 95 cm^{2}
- One method of finding the height of an object is to place a mirror on the ground and then position yourself so that the top of the object can be seen in the mirror. How high is a structure if a person who is 160 cm tall observes the top of a structure when the mirror is 100 m from the structure and the person is 8 m from the mirror?
- 50,000 cm
- 20,000 cm
- 2,000 cm
- 200 cm
- Suppose ABCD is a parallelogram; B = 120 and 2 = 40. Find m 4.
- 50°
- 40°
- 20°
- 30°
- The length and width of a rectangle together measure 130 yards. Their difference is 8 yards. What is the area of the rectangle?
- 4,209 yd^{2}
- 130 yd^{2}
- 3,233 yd^{2}
- 4,270 yd^{2}
Answers
The following explanations show one way in which each problem can be solved. You may have another method for solving these problems.
- a. The surface area of the monument is the sum of 4 sides of a trapezoidal shape plus 4 sides of a triangular shape. The trapezoid DFCA has a height of 152.5m (), b_{1} = 33.6 , and b_{2} = 10.5 . The area is h(b_{1} + b_{2}) or (152.5)(33.6 + 10.5) which equals 3,362.625 m^{2}. The triangle DGF has b = 10.5 and h = 17.6. The area is bh or (10.5)(17.6) which equals 92.4 m^{2}. The sum of 4 trapezoidal regions, (4)(3,362.625) = 13,450.5 m^{2}, plus 4 triangular regions, 4(92.4) = 369.6 m^{2}, is 13,820.1 m^{2}. Rounding this answer to the nearest meter is 13,820 m^{2}. If you chose b, you found the area of the trapezoidal regions only. If you chose c, you found the area of one trapezoidal region and one triangular region. If you chose d, you found the area of 4 trapezoidal regions and one triangular region.
- c. The volume of a rectangular solid is length times width times height. First, calculate what the volume would be if the entire pool had a depth of 10 ft. The volume would be (10)(30)(15) or 4,500 ft^{3}. Now subtract the area under the sloped plane, a triangular solid. The volume of the region is (base)(height)(depth) or (7)(30)(15) or 1,575 ft^{3}. Subtract: 4,500 ft^{3} minus 1,575 ft^{3} results in 2,925 ft^{3} as the volume of the pool. If you chose a, this is the volume of the triangular solid under the sloped plane in the pool. If you chose b, you did not calculate the slope of the pool, but rather a pool that is consistently 10 feet deep.
2 and 3 are alternate interior angles that are congruent. If 2 measures 50°, 3 is also 50°. If you chose b, your mistake was assuming 2 and 3 are complementary angles. If you chose c, your mistake was assuming 2 and 3 are supplementary angles.
- b. Knowing that 4 + 3 + the right angle placed between 4 and 3, equals 180 and the fact that 3 = 50, we simply subtract 180 – 90 – 50, which equals 40. If you chose a, you assumed that 3 and 4 are vertical angles. If you chose c, you assumed that 3 and 4 are supplementary.
- a. Two parallel lines by by a transversal from alternatate interior angles that are congruent. The two parallel lines are formed by the mirrors, and the path of light is the transversal. Therefore, 2 and 3 are alternate interior angles that are congruent. If 2 measures 50°, 3 is also 50°. If you chose b, your mistake was assumming 2 and 3 are supplementary angles.
- c. The sum of the measures of the angles of a triangle is 180. The question is asking us to solve for x. The equation is x + x + 3x + 20 = 180. Simplifying the equation, 5x + 20 = 180. Subtract 20 from each side; 5x = 160. Divide each side by 5; x = 32. If you chose a, you solved for the vertex angle. If you chose b, you wrote the original equation incorrectly as x + 3x + 20 = 180. If you chose d, you wrote the original equation incorrectly as x + x + 3x + 10 = 90.
- d. Since we solved for x in the previous question, simply substitute x = 32 into the equation for the vertex angle, 3x + 20. The result is 116°. If you chose a, you solved for the base angle. If you chose b, the original equation was written incorrectly as x + x + 3x + 20 = 90.
- a. Opposite angles of a parallelogram are equal in measure. Using this fact, A = C or 5x + 2 = 6x – 4. Subtract 5x from both sides; 2 = x – 4. Add 4 to both sides; 6 = x. Now substitute x = 6 into the expression for A; 6(6) – 4 = 36 – 4 or 32. If you chose b, you solved for x, not the angle. If you chose c, you assumed the angles were supplementary. If you chose d, you assumed the angles were complementary.
- a. The two bases of the trapezoid are represented by x and 3x. The nonparallel sides are each x + 5. Setting up the equation for the perimeter will allow us to solve for x; x + 3x + x + 5 + x + 5 = 40. Simplify to 6x + 10 = 40. Subtract 10 from both sides; 6x = 30. Divide both sides by 6; x = 5. The longer base is represented by 3x. Using substitution, 3x or (3)(5) equals 15, the longer base. If you chose b, you solved for the shorter base. If you chose c, you solved for the nonparallel side. If you chose d, the original equation was incorrect, x + x + 5 + 3x = 40.
- a. The sum of the measures of the angles of a triangle is 180. Using this information, we can write the equation 2x + 15 + x + 20 + 3x + 25 = 180. Simplify the equation; 6x + 60 = 180. Subtract 60 from both sides; 6x = 120. Divide both sides by 6; x = 20. Now substitute 20 for x in each expression to find the smallest angle. The smallest angle is found using the expression x + 20; 20 + 20 = 40. If you chose b, this was the largest angle within the triangle. If you chose c, the original equation was incorrectly written as 2x + 15 + x + 20 + 3x + 25 = 90. If you chose d, this was the angle that lies numerically between the smallest and largest angle measurements.
- b. and are the legs of a right triangle. is the hypotenuse and is equal to of . Solving for the hypotenuse, we use the Pythagorean theorem, a^{2} + b^{2} = c^{2}; 10^{2} + 6^{2} = ; 100 + 36 =; 136 = . = 11.66; of = 5.8. If you chose a, you assigned 10 as the length of the hypotenuse. If you chose d, the initial error was the same as choice a. In addition, you solved for and not .
- b. The perimeter of a parallelogram is the sum of the lengths of all four sides. Using this information and the fact that opposite sides of a parallelogram are equal, we can write the following equation: x + x + + 32. Simplify to 2x + 3x + 2 = 32. Simplify again; 5x + 2 = 32. Subtract 2 from both sides; 5x = 30. Divide both sides by 5; x = 6. The longer base is represented by . Using substitution, equals 10. If you chose c, you solved for the shorter side.
- b. To find the width of the piece of sheetrock that can fit through the door, we recognize it to be equal to the length of the diagonal of the door frame. If the height of the door is 6 ft 6 in, this is equivalent to 78 inches. Using the Pythagorean theorem, a = 78 and b = 36, we will solve for c. (78)^{2} + (36)^{2} = c^{2}. Simplify: 6,084 + 1,296 = c^{2}; 7,380 = c^{2}. Take the square root of both sides, c = 86. If you chose a, you added 78 + 36. If you chose c, you rounded incorrectly. If you chose d, you assigned 78 inches as the hypotenuse, c.
- d. To find the length of the rectangle, we will use the Pythagorean theorem. The width, a, is 20. The diagonal, c, is x + 8. The length, b, is x; a^{2} + b^{2} = c^{2}; 20^{2} + x^{2} = (x + 8)^{2}. After multiplying the two binomials (using FOIL), 400 + x^{2} = x^{2} + 16x + 64. Subtract x^{2} from both sides; 400 = 16x + 64. Subtract 64 from both sides; 336 = 16x. Divide both sides by 16; 21 = x. If you chose a, you incorrectly determined the diagonal to be 28.
- b. Two angles are complementary if their sum is 90°. Using this fact, we can establish the following equation: 5x + 7x = 90. Simplify; 12x = 90. Divide both sides of the equation by 12; x = 7.5. The smallest angle is represented by 5x. Therefore 5x = 5(7.5) or 37.5, the smallest angle measurement. If you chose a, the original equation was set equal to 180 rather than 90. If you chose c, you solved for the largest angle. If you chose d, the original equation was set equal to 180 and you solved for the largest angle as well.
- d. Adjacent angles in a parallelogram are supplementary. A and D are adjacent angles. Therefore, A + D = 180; 3x + 10 + 2x + 30 = 180. Simplifying, 5x + 40 = 180. Subtract 40 from both sides, 5x = 140. Divide both sides by 5; x = 28. A = 3x + 10 or 3(28) + 10 which equals 94. If you chose a, you assumed A = D. If you chose b, you assumed A + D = 90. If you chose c, you solved for D instead of A.
- b. The sum of the measures of the exterior angles of any polygon is 360°. Therefore, if the sum of four of the five angles equals 320, to find the fifth simply subtract 320 from 360, which equals 40. If you chose a, you divided 325 by 4, assuming all four angles are equal in measure and assigned this value to the fifth angle, E.
- c. This problem requires two steps. First, determine the base and height of the triangle. Second, determine the area of the triangle. To determine the base and height we will use the equation x + 4x = 95. Simplifying, 5x = 95. Divide both sides by 5, x = 19. By substitution, the height is 19 and the base is 4(19) or 76. The area of the triangle is found by using the formula area = base × height. Therefore, the area = (76)(19) or 722 cm^{2}. If you chose a, the area formula was incorrect. Area = base × height, not base × height. If you chose b, the original equation x + 4x = 95 was simplified incorrectly as 4x^{2} = 95.
- c. To solve for the height of the structure, solve the following proportion: . Cross-multiply, 8x = 16,000. Divide both sides by 8; x = 2,000. If you chose b or d, you made a decimal error.
- c. In parallelogram ABCD, 2 is equal in measurement to 5. 2 and 5 are alternate interior angles, which are congruent. If B is 120, then B + 5 + 4 = 180. Adjacent angles in a parallelogram are supplementary. Therefore, 40 + 120 + x = 180. Simplifying, 160 + x = 180. Subtract 160 from both sides; x = 20. If you chose a, you assumed 4 + 5 = 90. If you chose d, you assumed 4 is (4 + 5).
- a. There are two ways of solving this problem. The first method requires a linear equation with one variable. The second method requires a system of equations with two variables. Let the length of the rectangle equal x. Let the width of the rectangle equal x + 8. Together they measure 130 yards. Therefore, x + x + 8 = 130. Simplify, 2x + 8 = 130. Subtract 8 from both sides, 2x = 122. Divide both sides by 2; x = 61. The length of the rectangle is 61, and the width of the rectangle is 61 + 8 or 69; 61 × 69 = 4,209. The second method of choice is to develop a system of equations using x and y. Let x = the length of the rectangle and let y = the width of the rectangle. Since the sum of the length and width of the rectangle is 130, we have the equation x + y = 130. The difference is 8, so we have the equation x – y = 8. If we add the two equations vertically, we get 2x = 122. Divide both sides by 2: x = 61. The length of the rectangle is 61. Substitute 61 into either equation; 61 + y = 130. Subtract 61 from both sides, giving you y = 130 – 61 = 69. To find the area of the rectangle, we use the formula length × width or (61)(69) = 4,209. If you chose b, you added 61 to 69 rather than multiplied. If you chose c, the length is 61 but the width was decreased by 8 to 53.
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