Math 202: Homework 6 Solutions

Transcription

1 Math 202: Homework 6 Solutions 1. ( 11.3 #6) (a) 10 in 4 in 4 in 4 in Notice that each of the bases is a square, so each of the four lateral sides are congruent. Hence, S = 2(4 2 ) + 4(4 10) = = 192 in 2 (b) 16 mm 16 mm 12 mm 16 mm 20 mm 16 mm 20 mm All of the dimensions are different, so the only congruent faces we have are opposite faces: S = 2(12 16) + 2(16 20) + 2(12 20) = = 1504 mm 2 1

2 2. ( 11.3 #10) (a) 30 mm 20 mm 20 mm S = = = 1600 mm 2 (b) 9 in 5.2 in 6 in 2

3 S = ( ) = = 96.6 in 2 3. ( 11.3 #12) 1 cm 3 cm 3 cm 6 cm 1 cm 5 cm 5 cm 1 cm Notice that the bases are congruent trapezoids, and all other faces are rectangles. Hence, ( ) 1 S = 2 (1 + 5) = ( 11.3 #14) = 102 cm 2 (a) The prism in the description should have a width of 4 ft, a height of 4 ft, and a length of 8 ft. According to our formula, then S = 2lw + 2lh + 2wh = 2(8)(4) + 2(8)(4) + 2(4)(4) = = 160 ft 2 (b) The pyramid described has a base area of 93.5 cm 2, and it has 6 lateral faces, 3

4 5. ( 11.3 #16) each with a base of 6 cm and a height of 6 cm. According to our formula, then S = area of base + 6 area of lateral side = = = cm 2 (a) False. The surface area of a pyramid also includes the area of its base. (b) True. This is for the same reason that doubling the dimensions of a prism quadruples its surface area. 6. ( 11.3 #18) (a) Notice that the pyramid and the prism are glued together along a similar base, so we want to omit this base in our calculation of the surface area. Then we only have the four lateral faces of the pyramid, the four lateral faces of the prism (which happen to be congruent), and a single base of the prism to consider: S = (3 7) = = 123 units 2 (b) Notice that the pyramid is glued along its base to a portion of one of the lateral sides of the prism, so we will need to omit the base of the pyramid and this portion of the lateral side in our calculation of the surface area. Hence, we need to consider the four lateral faces of the pyramid, the three full lateral faces of the prism, the portion of the final lateral face of the prism, and the two bases of the prism: S = (12 1.5) (1 1.5) = = 68 units 2 7. ( 11.3 #20) Recall that, for a square pyramid with base b and slant height h, the formula for the surface area is S = area of base + 4 (area of lateral face) = b bh 4

5 Inserting the information we ve been given into this equation, we get 176 = (8)h. Now we can solve this equation for h: 176 = (8)h 176 = h = 16h 112 = 16h = h 7 = h Hence, the slant height of the square pyramid is 7 inches. 8. ( 11.3 #34) (a) The original solid is a full rectangular prism with width 2 cm, length 7 cm, and height 3 cm. Hence, S = 2lw + 2lh + 2wh = 2(7)(2) + 2(7)(3) + 2(2)(3) = = 82 cm 2 (b) Consider first what has been taken away from our old faces. We have lost a 1 3 rectangle from the front face, and we have lost a 1 3 rectangle from the top face. However, looking in the cavity that our incision has left, we have gained two 1 3 rectangles and two 1 1 rectangles (on the sides). Hence, to find our new surface area, we can take away from and add to our old surface area: S = 82 2 (1 3) + 2 (1 3) + 2 (1 1) = = 84 cm 2 (c) The surface area of our solid has actually increased by taking out a chunk. 5

6 9. ( 11.3 #36) First, let s figure out what the side length, s, of the original cube was. We know that that the surface area of a cube is 6s 2, so S = 6s = 6s = s2 49 = s 2 7 = s So the side length of the original cube is 7 in. Just as in #34, we will consider what faces we have lost and what faces we have gained from the operation of cutting the hole out of the cube. First, we have lost a rectangle in the top face and a rectangle in the bottom face. However, we have gained four faces through the middle of the cube. Two of these are rectangles, and two of them are 2 7 rectangles. Hence, our new surface area is S = (2 1.5) + 2(1.5 7) + 2(2 7) = = 337 in ( 11.3 #38) The total area of the lateral faces of a pyramid is always greater than the area of the base. Imagine if the vertex at the top of some pyramid was very close to the base (so the height of the pyramid is very small). Then the total area of the lateral faces would be very close to the area of the base, but it should be a little bigger. (Imagine trying to smash this pyramid flat onto the base. The base would be covered up, but there would be some folds.) Now imagine a pyramid whose top vertex is very far away from the base. The total area of the lateral faces has only increased, and the base has stayed the same, so it should always be true that the total area of the lateral faces is greater than the area of the base. 11. ( 11.4 #6) Counting the number of cubes on the top of the prism, we get 20 cubes. There are three layers of 20 cubes, so there should be a total of 60 cubes. This would mean the prism has a volume of 60 units ( 11.4 #8) (a) In the case of a rectangular prism, we can choose any face as the base of the prism. Let s choose the bottom face. Then the area of the base is = 6

7 21.6 in 2, and the height is 6 in. Hence, = (21.6)(6) = in 3 (b) In this case, we must choose the front or back face as our base. Then the area of the base is 1 2 (10)(4) = 20 ft2, and the height is 5.5 ft. Hence, = (20)(5.5) = 110 ft ( 11.4 #10) In this case, the base has been identified as the bottom face, so the area of the base is 18 8 = 144 cm 2. Since we know the area of the base and the volume of the prism, we can use our formula to figure out the height: So the height of the prism is 9 cm. 14. ( 11.4 #14) 1296 = 144h = h 9 = h (a) Since the base is a rectangle, the area of the base is = 195 in 2. The height of the pyramid has been given as 10 in, so V = 1 3 Bh = 1 3 (195)(10) = 650 in 3 (b) Since the base is a triangle, the area of the base is 1 2 (6)(5) = 15 in2. The height of the pyramid has been given as 7 ft, so V = 1 3 Bh = 1 3 (15)(7) = 35 ft 3 7

8 (c) The area of the base has been given as 75 m 2, and the height is 7.4 m, so 15. ( 11.4 #16) V = 1 3 Bh = 1 3 (75)(7.4) = 185 m 3 (a) The solid consists of two hexagonal prisms stacked on top of each other. To find the volume of the entire solid, we can find each separately and add them together. The hexagon on the bottom has a given base of 210 cm 2, and its height is 8 cm. Hence, V 1 = Bh = (210)(8) = 1680 cm 3 The hexagon on the top has a given base of 42 cm 2, and its height is 3.6 cm. Hence, V 2 = Bh = (42)(3.6) = cm 3 Now, as we said, the volume of the entire solid is the sum of these two volumes: V = V 1 + V 2 = = cm 3 (b) The solid consists of a rectangular prism with a rectangular pyramid removed from the top of it. We can find the volume of the entire solid by finding the volume of each and subtracting. The rectangular prism has a rectangular base of area 9 7 = 63 in 2, and its height is 4 in. Hence, V 1 = Bh = (63)(4) = 252 in 3 8

9 The rectangular pyramid has the same base and height, so we can immediately say that V 2 = 1 3 Bh = 1 3 (63)(4) = 84 in 3 Now, as we said, the volume of the entire solid is the volume of the prism minus the volume of the pyramid: V = V 1 V 2 = = 168 in 3 (c) Notice that the solid is simply a prism, where the front and back faces are our bases. The trick is then to find the area of this base. One way to do this is to notice that it is a trapezoid with b 1 = 8, b 2 = 13, and h = 5, but there is a rectangle missing. Hence, the area of the base is B = 1 2 (b 1 + b 2 )h (1.5 3) = 1 (8 + 13)(5) = = 48 m 2 Now, since the height of the prism is 10 m, we need only plug into the formula for volume of a prism: = (48)(10) = 480 m ( 11.4 #18) The base of this prism is the triangle on the top or bottom of the block of cheese. Hence, the area of the base is 1 2 (12)(15) = 90 cm2, and since the height is 6 cm, we know = (90)(6) = 540 cm 3 9

10 17. ( 11.4 #20) The polyhedron is a triangular prism that looks something like 4 cm 6 cm 8 cm The base is a triangle, so its area is = 24 cm2. The height of the prism is 4 cm, so = (24)(4) = 96 cm ( 11.4 #24) The student used the wrong formula for the volume of a pyramid, involving a factor of 1 2 instead of a factor of 3 1. I would encourage the student to review the formula for and the reasoning behind the volume of a pyramid. 19. ( 11.4 #28) (a) The base of the prism is a rectangle, so the area of the base is = 441 in 2. The height is also given as 14 in, so = (441)(14) = 6174 in 3 (b) Since a single gallon has a volume of 231 cubic inches, then 21 gallons have a volume of = 4851 in 3. If this gasoline is in the tank, it will take the form of a rectangular prism with the same base as the tank. Hence, we can use the same formula to find how much of the height of the tank it would take up: 4851 = 441h = h 11 = h So the height of 21 gallons of gasoline in the tank would be 11 inches. 20. ( 11.4 #29) First, we must find the height of each box, and then we can find its surface area. We use the formula for volume to find the height: 10

11 Box = (9.5)(15)h 285 = 142.5h = h 2 = h Box = (10)(14.25)h 285 = 142.5h = h 2 = h Box = (8)(14.25)h 285 = 114h = h 2.5 = h And now we can find their respective surface areas. Recall that they have no tops: Box 1 Box 3 S = lw + 2lh + 2wh S = lw + 2lh + 2wh = (15)(9.5) + 2(15)(2) + 2(9.5)(2) = (14.25)(8) + 2(14.25)(2.5) + 2(8)(2.5) = = = = Box 2 S = lw + 2lh + 2wh = (14.25)(10) + 2(14.25)(2) + 2(10)(2) = = Hence, Box 3 uses the least amount of cardboard, and Box 1 uses the greatest amount. 11