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.
- Mark intends to tile a kitchen floor, which is 9 ft by 11 ft. How many 6-inch tiles are needed to tile the floor?
- 60
- 99
- 396
- 449
- A framed print measures 36 in by 22 in. If the print is enclosed by a 2-inch matting, what is the length of the diagonal of the print? Round to the nearest tenth. See illustration.
- 36.7 in
- 39.4 in
- 26.5 in
- 50 in
- A 20-foot light post casts a shadow 25 feet long. At the same time, a building nearby casts a shadow 60 feet long. How tall is the building?
- 48 ft
- 75 ft
- 8 ft
- 95 ft
- Barbara is wrapping a wedding gift that is contained within a rectangular box 20 in by 18 in by 4 in. How much wrapping paper will she need?
- 512 in^{2}
- 1,440 in^{2}
- 1,024 in^{2}
- 92 in^{2}
- Mark is constructing a walkway around his inground pool. The pool is 20 ft by 40 ft and the walkway is intended to be 4 ft wide. What is the area of the walkway?
- 224 ft^{2}
- 416 ft^{2}
- 256 ft^{2}
- 544 ft^{2}
- The picture frame shown below has outer dimensions of 8 in by 10 in and inner dimensions of 6 in by 8 in. Find the area of section A of the frame.
- 18 in^{2}
- 14 in^{2}
- 7 in^{2}
- 9 in^{2}
For questions 7 and 8, use the following illustration.
- John is planning to purchase an irregularly shaped plot of land. Referring to the diagram, find the total area of the land.
- 6,400 m^{2}
- 5,200 m^{2}
- 4,500 m^{2}
- 4,600 m^{2}
- Using the same illustration, determine the perimeter of the plot of land.
- 260 m
- 340 m
- 360 m
- 320 m
- A weather vane is mounted on top of an 18 ft pole. If a 22 ft guy wire is staked to the ground to keep the pole perpendicular, how far is the stake from the base of the pole?
- 160 ft
- 38 ft
- A surveyor is hired to measure the width of a river. Using the illustration provided, determine the width of the river.
- 48 ft
- 8 ft
- 35 ft
- 75 ft
- A publishing company is designing a book jacket for a newly published textbook. Find the area of the book jacket, given that the front cover is 8 in wide by 11 in high, the binding is 1.5 in by 11 in and the jacket will extend 2 inches inside the front and rear covers.
- 236.5 in^{2}
- 192.5 in^{2}
- 188 in^{2}
- 232 in^{2}
- A Norman window is to be installed in a new home. Using the dimensions marked on the illustration, find the area of the window to the nearest tenth of an inch. (π = 3.14)
- 2,453.3 in^{2}
- 2,806.5 in^{2}
- 147.1 in^{2}
- 2,123.6 in^{2}
- A surveyor is hired to measure the distance of the opening of a bay. Using the illustration and various measurements determined on land, find the distance of the opening of the bay.
- 272.7 yds
- 82.5 yds
- 27.5 yds
- 205 yds
- A car is initially 200 meters due west of a roundabout (traffic circle). If the car travels to the roundabout, continues halfway around the circle, exits due east, then travels an additional 160 meters, what is the total distance the car has traveled? Refer to diagram.
- 862.4 m
- 611.2 m
- 502.4 m
- 451.2 m
- Steve Fossett is approaching the shores of Australia on the first successful solo hot air balloon ride around the world. His balloon, the Bud Light™ Spirit of Freedom, is being escorted by a boat (directly below him) that is 108 meters away. The boat is 144 meters from the shore. How far is Fossett's balloon from the shore?
- 252 m
- 95.2 m
- 126 m
- 180 m
- Computer monitors are measured by their diagonals. If a monitor is advertised to be 17 in, what is the actual viewing area, assuming the screen is square? (Round to the nearest tenth.)
- 289 in^{2}
- 90.25 in^{2}
- 144.4 in^{2}
- 144.5 in^{2}
- An elevated cylindrical shaped water tower is in need of paint. If the radius of the tower is 10 ft and the tower is 40 ft tall, what is the total area to be painted? (π = 3.14)
- 1,570 ft^{2}
- 2,826 ft^{2}
- 2,575 ft^{2}
- 3,140 ft^{2}
- A sinking ship signals to the shore for help. Three individuals spot the signal from shore. The first individual is directly perpendicular to the sinking ship and 20 meters inland. The second individual is also 20 meters inland but 100 meters to the right of the first individual. The third is also 20 meters inland but 125 meters to the right of the first individual. How far off shore is the sinking ship? See illustration.
- 60 meters
- 136 meters
- 100 meters
- 80 meters
- You are painting the surface of a silo that has a radius of 8 ft and height of 50 ft. What is the total surface area to be painted? Assume the top of the silo is a sphere and sets on the ground. Refer to the illustration.
- 2,913.92 ft^{2}
- 1,607.68 ft^{2}
- 2,612.48 ft^{2}
- 3,315.84 ft^{2}
The Washington Monument is located in Washington D.C. Use the following illustration, which represents one of four identical sides, to answer questions 435 and 436
- Find the height of the Washington Monument to the nearest tenth of a meter.
- 157.8 m
- 169.3 m
- 170.1 m
- 192.2 m
Answers
The following explanations show one way in which each problem can be solved. You may have another method for solving these problems.
- c. Since the tiles are measured in inches, convert the area of the floor to inches as well. The length of the floor is 9 ft × 12 in per foot = 108 in. The width of the floor is 11 ft × 12 in per foot = 132 in. The formula for area of a rectangle is length × width. Therefore, the area of the kitchen floor is 108 in × 132 in or 14,256 in^{2}. The area of one tile is 6 in × 6 in or 36 in^{2}. Finally, divide the total number of square inches by 36 in^{2} or 14,256 in^{2} divided by 36 in^{2} = 396 tiles.
- a. If a framed print is enclosed by a 2–inch matting, the print is 4 in less in length and height. Therefore, the picture is 32 in by 18 in. These measurements along with the diagonal form a right triangle. Using the Pythagorean theorem, solve for the diagonal. 32^{2} + 18^{2} = c^{2}; 1,024 + 324 = c^{2}; 1,348 = c^{2}; 36.7 = c. If you chose b, you reduced the print 2 inches less than the frame in length and height rather than 4 inches.
- a. To find the height of the building set up the following proportion: or . Cross-multiply: 1,200 = 25x. Solve for x by dividing both sides by 25; x = 48. If you chose b, you set up the proportion incorrectly as . If you chose c, you set up the proportion incorrectly as .
- c. The surface area of the box is the sum of the areas of all six sides. Two sides are 20 in by 18 in or (20)(18) = 360 in^{2}. Two sides are 18 in by 4 in or (18)(4) = 72 in^{2}. The last two sides are 20 in by 4 in or (20)(4) = 80 in^{2}. Adding up all six sides: 360 in^{2} + 360 in^{2} + 77 in^{2} + 77 in^{2} + 80 in^{2} + 80 in^{2} = 1,024 in^{2}, is the total area. If you chose a, you did not double all sides. If you chose b, you calculated the volume of the box.
- d. The area of the walkway is equal to the entire area (area of the walkway and pool) minus the area of the pool.The area of the entire region is length times width. Since the pool is 20 feet wide and the walkway adds 4 feet onto each side, the width of the rectangle formed by the walkway and pool is 20 + 4 + 4 = 28 feet. Since the pool is 40 feet long and the walkway adds 4 feet onto each side, the length of the rectangle formed by the walkway and pool is 40 + 4 + 4 = 48 feet. Therefore, the area of the walkway and pool is (28)(48) = 1,344 ft^{2}. The area of the pool is (20)(40) = 800 ft^{2}. 1,344 ft^{2} – 800 ft^{2} = 544 ft^{2}. If you chose c, you extended the entire area's length and width by 4 feet instead of 8 feet.
- c. The area described as section A is a trapezoid. The formula for the area of a trapezoid is h(b_{1} + b_{2}). The height of the trapezoid is 1 inch, b_{1} is 6 inches, and b_{2} is 8 inches. Using these dimensions, area = (1)(6 + 8) or 7 in^{2}. If you chose b, you used a height of 2 inches rather than 1 inch. If you chose d, you found the area of section B or D.
- b. To find the total area, add the area of region A plus the area of region B plus the area of region C. The area of region A is length times width or (100)(40) = 4,000 m^{2}. Area of region B is bh or (40)(30) = 600 m^{2}. The area of region C is bh or (30)(40) = 600 m^{2}. The combined area is the sum of the previous areas or 4,000 + 600 + 600 = 5,200 m^{2}. If you chose a, you miscalculated the area of a triangle as bh instead of bh. If you chose c, you found only the area of the rectangle. If you chose d, you found the area of the rectangle and only one of the triangles.
- c. To find the perimeter, we must know the length of all sides. According to the diagram, we must find the length of the hypotenuse for the triangular regions B and C. Using the Pythagorean theorem for triangular region B, 30^{2} + 40^{2} = c^{2}; 900 + 1,600 = c^{2}; 2,500 = c^{2}; 50 m = c. The hypotenuse for triangular region C is also 50 m since the legs are 30 m and 40 m as well. Now adding the length of all sides, 40 m + 100 m + 30 m + 50 m + 30 m + 50 m + 60 m = 360 m, the perimeter of the plot of land. If you chose a, you did not calculate in the hypotenuse on either triangle. If you chose b, you miscalculated the hypotenuse as having a length of 40 m. If you chose d, you miscalculated the hypotenuse as having a length of 30 m.
- d. The 18 ft pole is perpendicular to the ground forming the right angle of a triangle. The 22 ft guy wire represents the hypotenuse. The task is to find the length of the remaining leg in the triangle. Using the Pythagorean theorem: 18^{2} + b^{2} = 222; 324 + b^{2} = 484; b^{2} = 160; b = or . If you chose a, you did not take the square root.
- c. ΔABD is similar to ΔECD. Using this fact, the following proportion is true: . Cross-multiply, 2,400 = 32(40 + x); 2,400 = 1,280 + 32x. Subtract 1,280; 1,120 = 32x; divide by 32; x = 35 feet.
- a. The area of the front cover is length times width or (8)(11) = 88 in^{2}. The rear cover is the same as the front, 88 in^{2}. The area of the binding is length times width or (1.5)(11) = 16.5 in^{2}. The extension inside the front cover is length times width or (2)(11) = 22 in^{2}. The extension inside the rear cover is also 22 in^{2}. The total area is the sum of all previous areas or 88 in^{2} + 88 in^{2} + 16.5 in^{2} + 22 in^{2} + 22 in^{2} or 236.5 in^{2}. If you chose b, you did not calculate the extensions inside the front and rear covers. If you chose c, you miscalculated the area of the binding as (1.5)(8) and omitted the extensions inside the front and rear covers. If you chose d, you miscalculated the area of the binding as (1.5)(8) only.
- a. To find the area of the rectangular region, multiply length times width or (30)(70), which equals 2,100 in^{2}. To find the area of the semi–circle, multiply times π r^{2} or π(15)^{2} which equals 353.25 in^{2}. Add the two areas together, 2,100 plus 353.25 or 2,453.3, rounded to the nearest tenth, for the area of the entire window. If you chose b, you included the area of a circle, not a semi–circle.
- b. ΔACE and ΔBCD are similar triangles. Using this fact, the following proportion is true: . Cross-multiply, 100x = 8,250. Divide by 100 to solve for x; x = 82.5 yards. If you chose a or c, you set up the proportion incorrectly.
- b. The question requires us to find the distance around the semi–circle. This distance will then be added to the distance traveled before entering the roundabout, 200 m, and the distance traveled after exiting the roundabout, 160 m. According to the diagram, the diameter of the roundabout is 160 m. The distance or circumference of half a circle is π d, (3.14)(160) or 251.2 m. The total distance or sum is 200 m + 160 m + 251.2 m = 611.2 m. If you chose a, you included the distance around the entire circle. If you chose c, you found the distance around the circle. If you chose d, you did not include the distance after exiting the circle, 160 m.
- d. The boat is the triangle's right angle. The distance between the balloon and the boat is 108 meters, one leg. The distance between the boat and the land, 144 meters, is the second leg. The distance between the balloon and the land, which is what we are finding, is the hypotenuse. Using the Pythagorean theorem: 108^{2} + 144^{2} = c^{2}; 11,664 + 20,736 = c^{2}; 32,400 = c^{2}; c = 180 m.
- d. Since the monitor is square, the diagonal and length of the sides of the monitor form an isosceles right triangle. The question requires one to find the length of one leg to find the area. Using the Pythagorean theorem: s^{2} + s^{2} = 17^{2}; 2s^{2} = 289. Divide by 2; s^{2} = 144.5. The area of the square monitor is s^{2}; thus the area is 144.5 in^{2}. If you chose a, you simply squared the diagonal or which does not give you the area of the monitor.
- b. To find the surface area of a cylinder, use the following formula: surface area = 2π r^{2} + π dh. Therefore, the surface area = 2(3.14)(10)^{2} + (3.14)(20)(40) or 3,140 ft^{2}. If you chose b, you found the surface area of the circular top and forgot about the bottom of the water tower. However, the bottom of the tower would need painting since the tank is elevated.
- d. Using the concept of similar triangles, ΔCDB is similar to ΔCEA, so set up the following proportion: . Cross-multiply, 25x + 500 = 2,500. Subtract 500; 25x = 2,000; Divide by 25; x = 80. If you chose b, the proportion was set up incorrectly as .
- a. To find the total surface area of the silo, add the surface area of the cylinder to the surface area of of the sphere. To find the area of the cylinder, use the formula π hd or (3.14)(50)(16) which equals 2,512 ft^{2}. The area of a sphere is ()(4)π r^{2}. Using a radius of 8 ft, the area is ()(4)π(8)^{2} = 401.92 ft^{2}. Adding the area of the cylinder plus the sphere is 2,512 + 401.92 = 2,913.92 ft2. If you chose b, your miscalculation was in finding the area of the sphere. You used the diameter rather than the radius. If you chose d, you found the surface area of the entire sphere, not just half.
- c. The height of the monument is the sum of plus . Therefore, the height is 152.5 + 17.6 or 170.1 meters. If you chose a, you added + . If you chose b, you added + .
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