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 area of the regular octagon with the following measurements.
- 224 square units
- 112 square units
- 84 square units
- 169 square units
- Two sides of a picture frame are glued together to form a corner. Each side is cut at a 45-degree angle. Using the illustration provided, find the measure of A.
- 45°
- 90°
- 115°
- 135°
- Find the total area of the shaded regions, if the radius of each circle is 5 cm. Leave answer in terms of π.
- 1,200 – 300π cm^{2}
- 300 – 300π cm^{2}
- 300 π – 1,200 cm^{2}
- 300 π – 300 cm^{2}
- The road from town A to town B travels at a direction of N23°E. The road from town C to town D travels at a direction of S48°E. The roads intersect at location E. Find the measure of BED, at the point of intersection.
- 71°
- 23°
- 109°
- 48°
- The figure provided below represents a hexagonal-shaped nut. What is the measure of ABC?
- 120°
- 135°
- 108°
- 144°
- If the lengths of all sides of a box are doubled, how much is the volume increased?
- 2 times
- 4 times
- 6 times
- 8 times
- If the radius of a circle is tripled, the circumference is
- multiplied by 3.
- multiplied by 6.
- multiplied by 9.
- multiplied by 12.
- If the diameter of a sphere is doubled, the surface area is
- multiplied by 4.
- multiplied by 2.
- multiplied by 3.
- multiplied by 8.
- If the diameter of a sphere is tripled, the volume is
- multiplied by 3.
- multiplied by 27.
- multiplied by 9.
- multiplied by 6.
- If the radius of a cone is doubled, the volume is
- multiplied by 2.
- multiplied by 4.
- multiplied by 6.
- multiplied by 8.
- If the radius of a cone is halved, the volume is
- multiplied by .
- multiplied by .
- multiplied by .
- multiplied by .
- If the radius of a right cylinder is doubled and the height is halved, its volume
- remains the same.
- is multiplied by 2.
- is multiplied by 4.
- is multiplied by .
- If the radius of a right cylinder is doubled and the height is tripled, its volume is
- multiplied by 12.
- multiplied by 2.
- multiplied by 6
- multiplied by 3.
- If each interior angle of a regular polygon has a measure of 150 degrees, how many sides does it have?
- 10
- 11
- 12
- 13
- A box is 30 cm long, 8 cm wide and 12 cm high. How long is the diagonal ? Round to the nearest tenth.
- 34.5 cm
- 32.1 cm
- 35.2 cm
- 33.3 cm
- Find the area of the shaded region. Leave answer in terms of π.
- 16.5π
- 30π
- 3π
- 7.5π
- A round tower with a 40 meter circumference is surrounded by a security fence that is 8 meters from the tower. How long is the security fence in terms of π?
- (40 + 16π) meters
- (40 + 8π) meters
- 48π meters
- 56π meters
- The figure below is two overlapping rectangles. Find the sum of 1 + 2 + 3 + 4.
- 360°
- 90°
- 180°
- 540°
- A solid is formed by cutting the top off of a cone with a slice parallel to the base, and then cutting a cylindrical hole into the resulting solid. Find the volume of the hollow solid in terms of π.
- 834π cm^{3}
- 2,880π cm^{3}
- 891π cm^{3}
- 1,326π cm^{3}
- A rectangular container is 5 cm wide and 15 cm long, and contains water to a depth of 8 cm. An object is placed in the water and the water rises 2.3 cm. What is the volume of the object?
- 92 cm^{3}
- 276 cm^{3}
- 172.5 cm^{3}
- 312.5 cm^{3}
- A concrete retaining wall is 120 feet long with ends shaped as shown. How many cubic yards of concrete are needed to construct the wall?
- 217.8 yd^{3}
- 5,880 yd^{3}
- 653.3 yd^{3}
- 49 yd^{3}
- A spherical holding tank whose diameter to the outer surface is 20 feet is constructed of steel 1 inch thick. How many cubic feet of steel is needed to construct the holding tank? Round to the nearest integer value. (π = 3.14)
- 78 ft^{3}
- 104 ft^{3}
- 26 ft^{3}
- 125 ft^{3}
- How many cubic inches of lead are there in the pencil? Round to the nearest thousandth. (π = 3.14)
- .061 in^{3}
- .060 in^{3}
- .062 in^{3}
- .063 in^{3}
- A cylindrical hole with a diameter of 4 inches is cut through a cube. The edge of the cube is 5 inches. Find the volume of the hollowed solid in terms of π.
- 125 – 80π
- 125 – 20π
- 80π – 125
- 20π – 125
- Find the area of the region.
- 478 units^{2}
- 578 units^{2}
- 528 units^{2}
- 428 units^{2}
- From a stationary point directly in front of the center of a bull's eye, Kim aims two arrows at the bull's eye. The first arrow nicks one point on the edge of the bull's eye; the second strikes the center of the bull's eye. Kim knows the second arrow traveled 20 meters since she knows how far she is from the target. If the bull's eye is 4 meters wide, how far did the first arrow travel? You may assume that the arrows traveled in straight-line paths and that the bull's eye is circular. Round answer to the nearest tenth.
- 19.9 meters
- 24 meters
- 22 meters
- 20.1 meters
Answers
The following explanations show one way in which each problem can be solved.
You may have another method for solving these problems.
- b. Observe that the octagon can be subdivided into 8 congruent triangles. Since each triangle has a base of 4 and a height of 7, the area of each traingle can be found using the formula, area = base × height. To find the area of the octagon, we will find the area of a triangle and multiply it by 8. The area of one triangle is (4)(7) or 14. Multiply this value times 8; (14)(8) = 112. This is the area of the octagon. If you chose a, you used an incorrect formula for area of a triangle. Area = base × height was used rather than area = base × height. If you chose d, you mistakenly divided the octagon into 6 triangles instead of 8 triangles.
Another way to solve this problem is to use the formula for area of a regular polygon. That formula is area = Pa, where P is the perimeter of the polygon and a is the apothem. If we know that the octagon is regular and each side is 4, that means the perimeter is 8 × 4 = 32. The apothem is the segment drawn from the center of the regular polygon and perpendicular to a side of the polygon; in this case it is 7. We substitute in our given values and get (32)(7) = 112.
- d. The sum of the measures of the angles of a quadrilateral is 360°. In the quadrilateral, three of the four angle measurements are known. They are 45° and two 90°angles. To find A, subtract these three angles from 360°, or 360 – 90 – 90 – 45 = 135°. This is the measure of angle A. If you chose a, you assumed A and the 45° angle are complementary angles.
- a. To find the total area of the shaded region, we must find the area of the rectangle minus the sum of the areas of all circles. The area of the rectangle is length × width. Since the rectangle is 4 circles long and 3 circles wide, and each circle has a diameter of 10 cm (radius of 5 cm × 2), the rectangle is 40 cm long and 30 cm wide; (40)(30) = 1,200 cm^{2}. The area of one circle is πr^{2} or π(5)2 = 25π. Multiply this value times 12, since we are finding the area of 12 circles, (12)(25)π = 300π. The difference is 1,200 – 300π cm^{2}, the area of the shaded region. If you chose b, the area of the rectangle was incorrectly calculated as (20)(15). If you chose c, you reversed the area of the circles minus the area of the rectangle. If you chose d, you reversed choice b as the area of the circles minus the area of the rectangle.
- c. Referring to the illustration, NEB = 23° and DES = 48°. Since NEB + BED + DES = 180; using substitution, 23 + BED + 48 = 180. Simplify; 72 + BED = 180. Subtract 72 from both sides; BED = 109°. If you chose a, you added 23 + 48 to total 71. If you chose b, you assumed BED = NEB. If you chose d, you assumed BED = DES.
- a. The measure of an angle of a regular polygon of n sides is × 180. Since a hexagon has 6 sides, to find the measure of ABC, substitute n = 6 and simplify. The measure of ABC is × 180 or 120°. If you chose b, you assumed a hexagon has 8 sides. If you chose c, you assumed a hexagon has 5 sides. If you chose d, you assumed a hexagon has 10 sides.
- d. The volume of a box is found by multiplying length × length × length or l × l × l = l^{3}. If the length is doubled, the new volume is (2l) × (2l) × (2l) or 8(l^{3}). When we compare the two expressions, we can see that the difference is a factor of 8. Therefore, the volume has been increased by a factor of 8.
- a. The formula for finding the circumference of a circle is πd. If the radius is tripled, the diameter is also tripled. The new circumference is π3d. Compare this expression to the original formula; with a factor of 3, the circumference is multiplied by 3.
- a. The formula for the surface area of a sphere is 4πr^{2}. If the diameter is doubled, this implies that the radius is also doubled. The formula then becomes 4π(2r)^{2}. Simplifying this expression, 4π(4r^{2}) equals 16πr^{2}. Compare 4πr^{2} to 16πr^{2}; 16πr^{2} is 4 times greater than 4πr^{2}. Therefore, the surface area is four times as great.
- b. If the diameter of a sphere is tripled, the radius is also tripled. The formula for the volume of a sphere is πr^{3}. If the radius is tripled, volume = π(3r)^{3} which equals π(27r^{3}) or (27)πr^{3}. Compare this equation for volume with the original formula; with a factor of 27, the volume is now 27 times as great.
- b. The formula for the volume of a cone is πr^{2}h. If the radius is doubled, then volume = π(2r)^{2}h or π4r^{2}h. Compare this expression to the original formula; with a factor of 4, the volume is multiplied by 4.
- a. The formula for the volume of a cone is πr^{2}h. If the radius is halved, the new formula is π(r)^{2}h or π()r^{2}h. Compare this expression to the original formula; with a factor of , the volume is multiplied by .
- b. The volume of a right cylinder is πr^{2}h. If the radius is doubled and the height halved, the new volume is π(2r)^{2}(h) or π4r^{2}(h) or 2πr^{2}h. Compare this expression to the original formula; with a factor of 2, the volume is multiplied by 2.
- a. The formula for finding the volume of a right cylinder is volume = πr^{2}h. If the radius is doubled and the height is tripled, the formula has changed to π(2r)2(3h). Simplified, π4r^{2}3h or π12r^{2}h. Compare this expression to the original formula; with a factor or 12, the volume is now multiplied by 12.
- c. The measure of an angle of a regular polygon of n sides is × 180. Since each angle measures 150°, we will solve for n, the number of sides. Using the formula 150 = × 180, solve for n. Multiply both sides by n, 150n = (n – 2)180. Distribute by 180, 150n = 180n – 360. Subtract 180n from both sides, –30n = –360. Divide both sides by –30, n = 12. The polygon has 12 sides. Alternately, if each angle is 150°, each exterior angle is 180° – 150° = 30°. The measure of an exterior angle of a rectangular polygon of n sides is . We now solve for n by solving = 30. Multiplying both sides by n, 360 = 30n. Dividing both sides by 30, n = 12.
- d. This problem requires two steps. First, find the diagonal of the base of the box. Second, using this value, find the length of the diagonal . To find the diagonal of the base, use 30 cm as a leg of a right triangle, 8 cm as the second leg, and solve for the hypotenuse. Using the Pythagorean theorem, 302 + 82 = c^{2}; 900 + 64 = c^{2}; 964 = c^{2}; c = 31.05. Now consider this newly obtained value as a leg of a right triangle, 12 cm as the second leg, and solve for the hypotenuse, ; 31.052 + 122 = ; 964 + 144 = ; 1,108 = . = 33.3. If you chose a, you used 30 and 12 as the measurements of the legs. If you chose b, you solved the first triangle correctly; however, you used 8 as the measure of one leg of the second triangle, which is incorrect.
- c. To find the area of the shaded region, we must find the area of the circle with diameter AC, minus the area of the circle with diameter BC, plus the area of the circle with diameter AB. To find the area of the circle with diameter AC, we use the formula area = πr^{2}. Since the diameter is 6, the radius is 3; therefore, the area is π3^{2} or 4.5π. To find the area of the circle with diameter BC, we again use the formula area = πr^{2}. Since the diameter is 4, the radius is 2; therefore the area is π2^{2} or 2π. To find the area of the circle with diameter AB we use the formula area = πr^{2}. Since the diameter is 2, the radius is 1; therefore the area is π. Finally, 4.5π – 2π + .5π = 3π, the area of the shaded region. If you chose a or b, in the calculations you mistakenly used πd^{2} as the area formula rather than πr^{2}.
- a. This problem has three parts. First, we must find the diameter of the existing tower. Secondly, we will increase the diameter by 16 meters for the purpose of the fence. Finally, we will find the circumference using this new diameter. This will be the length of the fence. The formula for circumference of a circle is πd. This formula, along with the fact that the tower has a circumference of 40 meters, gives us the following formula: 40 = πd. To solve for d, the diameter, divide both sides by π. D = the diameter of the existing tower. Now increase the diameter by 16 meters; + 16 to get the diameter of the fenced in section. Finally, use this value for d in the equation πd or π( + 16) meters. Simplify by distributing π through the expression; (40 + 16π) meters. This is the length of the security fence. If you chose b, you added 8 to the circumference of the tower rather than 16. If you chose c, you merely added 8 to the circumference of the tower.
- a. Using the illustration, 2 = a. 2 and a are vertical angles. 1 and a are supplementary, since c + d + 1 + a = 360° (the total number of degrees in a quadrilateral), 90 + 90 + 1 + a = 360. Simplifying, 180 + 1 + a = 360. Subtract 180 from both sides; 1 + a = 180. Since a = 2, using substitution, 1 + 2 = 180. Using similar logic, 4 = b. 4 and b are vertical angles. 3 and b are supplementary. e + f + b + 3 = 360 or 90 + 90 + b + 3 = 360. Simplifying, 180 + b + 3 = 360. Subtract 180 from both sides, b + 3 = 180. Since b = 4, using substitution, 3 + 4 = 180. Finally, adding 1 + 2 = 180 to 3 + 4 = 180, we can conclude 1 + 2 + 3 + 4 = 360.
- a. To find the volume of the hollowed solid, we must find the volume of the original cone minus the volume of the smaller cone sliced from the original cone minus the volume of the cylindrical hole. The volume of the original cone is found by using the formula V = πr^{2}h. Using the values r = 9 and h = 40, substitute and simplify to find the volume = π(9)^{2}(40) or 1,080π cm^{2}. The volume of the smaller cone is found by using the formula V = πr^{2}h. Using the values r = 3 and h = 19, substitute and simplify to find the volume = π(3)^{2}(19) or 57π cm^{3}. The volume of the cylinder is found by using the formula V = πr^{2}h. Using the values r = 3 and h = 21, substitute and simplify to find the volume = π(3)^{2}(21) or 189π cm^{3}. Finally, calculate the volume of the hollow solid; 1,080π – 57π – 189π or 834π cm^{3}. If you chose b, you used an incorrect formula for the volume of a cone, V = πr^{2}h. If you chose c, you subtracted the volume of the large cone minus the volume of the cylinder. If you chose d, you added the volumes of all three sections.
- c. To find the volume of the object, we must find the volume of the water that is displaced after the object is inserted. Since the container is 5 cm wide and 15 cm long, and the water rises 2.3 cm after the object is inserted, the volume of the displaced water can be found by multiplying length by width by depth: (5)(15)(2.3) = 172.5 cm^{3}.
- a. To find how many cubic yards of concrete are needed to construct the wall, we must determine the volume of the wall. The volume of the wall is calculated by finding the surface area of the end and multiplying it by the length of the wall, 120 ft. The surface area of the wall is found by dividing it into three regions, calculating each region's area, and adding them together. The regions are labeled A, B, and C. To find the area of region A, multiply the length (3) times the height (10) for an area of 30 ft^{2}. To find the area of region B, multiply the length (5) times the height (3) for an area of 15 ft^{2}. To find the area of region C, multiply times the base (2) times the height (4) for an area of 4 ft^{2}. The surface area of the end is 30 ft^{2} + 15 ft^{2} + 4 ft^{2} or 49 ft^{2}. Multiply 49 ft^{2} by the length of the wall 120 ft; 5,880 ft^{3} is the volume of the wall. The question, however, asks for the answer in cubic yards. To convert cubic feet to cubic yards, divide 5,880 ft^{3} by 27 ft^{3}, the number of cubic feet in one cubic yard, which equals 217.8 yd^{3}. If you chose b, you did not convert to yd^{3}. If you chose c, the conversion to cubic yards was incorrect. You divided 5,880 by 9 rather than 27. If you chose d, you found the area of the end of the wall and not the volume of the wall.
- b. To find the volume of the sphere we must find the volume of the outer sphere minus the volume of the inner sphere. The formula for volume of a sphere is πr^{3}. The volume of the outer sphere is π(120)^{3}. Here the radius is 10 feet (half the diameter) multiplied by 12 (converted to inches), or 120 inches. The volume equals 7,234,560 in^{3}. The volume of the inner sphere is π(119)^{3} or 7,055,199. (This is rounded to the nearest integer value.) The difference of the volumes is 7,234,560 – 7,055,199 or 179,361 in^{3}. This answer is in cubic inches, and the question is asking for cubic feet. Since one cubic foot equals 1,728 cubic inches, we simply divide 179,361 by 1,728, which equals 104, rounded to the nearest integer value. As an alternative to changing units to inches only to have to change them back into feet again, keep units in feet. The radius of the outer sphere is 10 feet and the radius of the inner sphere is one inch less than 10 feet, which is 9 and feet, or 9.917 feet. Use the formula for volume of a sphere: and find the difference in the volumes. If you chose a, you used an incorrect formula for the volume of a sphere, V = πr^{3}. If you chose c, you also used an incorrect formula for the volume of a sphere, V = πr^{3}. If you chose d, you found the correct answer in cubic inches; however, your conversion to cubic feet was incorrect.
- c. To solve this problem, we must find the volume of the sharpened tip and add this to the volume of the remaining lead that has a cylindrical shape. To find the volume of the sharpened point, we will use the formula for finding the volume of a cone, πr^{2}h. Using the values r = .0625 (half the diameter) and h = .25, the volume = π(.0625)^{2}(.25) or .002 in^{3}. To find the volume of the remaining lead, we will use the formula for finding the volume of a cylinder, πr^{2}h. Using the values r = .0625 and h = 5, the volume = π(.0625)^{2}(5) or .0613. The sum is .001 + .0613 or .0623 in^{3}, the volume of the lead. If you chose a, this is the volume of the lead without the sharpened tip. If you chose b, you subtracted the volumes calculated.
- b. To find the volume of the hollowed solid, we must find the volume of the cube minus the volume of the cylinder. The volume of the cube is found by multiplying length × width × height or (5)(5)(5) equals 125 in^{3}. The value of the cylinder is found by using the formula πr^{2}h. In this question, the radius of the cylinder is 2 and the height is 5. Therefore, the volume is π(2)^{2}(5) or 20π. The volume of the hollowed solid is 125 – 20π. If you chose a, you made an error in the formula of a cylinder, using πd^{2}h rather than πr^{2}h. If you chose c, this was choice a reversed. This is the volume of the cylinder minus the volume of the cube. If you chose d, you found the reverse of choice b.
- b. Refer to the diagram to find the area of the shaded region. One method is to enclose the figure into a rectangle, and subtract the area of the unwanted regions from the area of the rectangle. The unwanted regions have been labeled A through F. The area of region A is (15)(4) = 60. The area of region B is (5)(10) = 50. The area of region C is (20)(5) = 100. The area of region D is (17)(3) = 51. The area of region E is (20)(5) = 100. The area of region F is (10)(5) = 50. The area of the rectangle is (23)(43) = 989. The area of the shaded region is 989 – 60 – 50 – 100 – 51 – 100 – 50 = 578. If you chose a, c or d, you omitted one or more of the regions A through F.
- d. The shape formed by the paths of the two arrows and the radius of the bull's eye is a right triangle. The radius of the bull's eye is one leg and the distance the second arrow traveled is the second leg. The distance the first arrow traveled is the hypotenuse. To find the distance the first arrow traveled, use the Pythagorean theorem where 2 meters (half the diameter of the target) and 20 meters are the lengths of the legs and the length of the hypotenuse is missing. Therefore, a^{2} + b^{2} = c^{2} and a = 2 and b = 20, so 2^{2} + 20^{2} = c^{2}. Simplify: 4 + 400 = c^{2}. Simplify: 404 = c^{2}. Find the square root of both sides: 20.1 = c. So the first arrow traveled about 20.1 meters. If you chose c, you added the two lengths together without squaring. If you chose b, you added Kim's distance from the target to the diameter of the target. If you chose a, you let 20 meters be the hypotenuse of the right triangle instead of a leg and you used the radius of the target.
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