The surface area of the textbook can be found by adding the areas of all its faces.
The textbook has 6 faces: 2 faces with dimensions 2 in. × 14 in., 2 faces with dimensions 14 in. × 9 in., and 2 faces with dimensions 9 in. × 2 in.
The area of each face is determined by multiplying its length by its width.
Area of first face: 2 in. × 14 in. = 28 in.²
Area of second face: 2 in. × 14 in. = 28 in.²
Area of third face: 14 in. × 9 in. = 126 in.²
Area of fourth face: 14 in. × 9 in. = 126 in.²
Area of fifth face: 9 in. × 2 in. = 18 in.²
Area of sixth face: 9 in. × 2 in. = 18 in.²
The total surface area is the sum of the areas of all the faces:
28 in.² + 28 in.² + 126 in.² + 126 in.² + 18 in.² + 18 in.² = 344 in.²
Therefore, the surface area of the textbook is 344 in.².
Find the surface area of a textbook that is 2 in. × 14 in. × 9 in.(1 point)
Responses
344 in.2
344 in. squared
50 in.2
50 in. squared
172 in.2
172 in. squared
252 in.2
11 answers
Sue is wrapping books to give as gifts. Book A has a length of 6.5 inches, a width of 1 inch, and a height of 8.5 inches. Book B has a length of 5.5 inches, a width of 2 inches, and a height of 8 inches. Based on surface area, which book will require less wrapping paper?(1 point)
Responses
Book B will require less wrapping paper because its surface area is 1.5 in.2
less than the surface area of Book A.
Book B will require less wrapping paper because its surface area is 1.5 in.2
less than the surface area of Book A.
Book A will require less wrapping paper because its surface area is 32.75 in.2
less than the surface area of Book B.
Book A will require less wrapping paper because its surface area is 32.75 in.2
less than the surface area of Book B.
Book B will require less wrapping paper because its surface area is 0.5 in.2
less than the surface area of Book A.
Book B will require less wrapping paper because its surface area is 0.5 in.2
less than the surface area of Book A.
Book A will require less wrapping paper because its surface area is 1.5 in.2
less than the surface area of Book B.
Responses
Book B will require less wrapping paper because its surface area is 1.5 in.2
less than the surface area of Book A.
Book B will require less wrapping paper because its surface area is 1.5 in.2
less than the surface area of Book A.
Book A will require less wrapping paper because its surface area is 32.75 in.2
less than the surface area of Book B.
Book A will require less wrapping paper because its surface area is 32.75 in.2
less than the surface area of Book B.
Book B will require less wrapping paper because its surface area is 0.5 in.2
less than the surface area of Book A.
Book B will require less wrapping paper because its surface area is 0.5 in.2
less than the surface area of Book A.
Book A will require less wrapping paper because its surface area is 1.5 in.2
less than the surface area of Book B.
To determine which book requires less wrapping paper, we need to calculate the surface area of each book and compare them.
For Book A:
Length = 6.5 inches
Width = 1 inch
Height = 8.5 inches
To find the surface area, we need to calculate the areas of all the faces and then sum them up.
Area of the first face: 6.5 in. × 1 in. = 6.5 in.²
Area of the second face: 6.5 in. × 1 in. = 6.5 in.²
Area of the third face: 1 in. × 8.5 in. = 8.5 in.²
Area of the fourth face: 1 in. × 8.5 in. = 8.5 in.²
Area of the fifth face: 6.5 in. × 8.5 in. = 55.25 in.²
Area of the sixth face: 6.5 in. × 8.5 in. = 55.25 in.²
Total surface area of Book A: 6.5 in.² + 6.5 in.² + 8.5 in.² + 8.5 in.² + 55.25 in.² + 55.25 in.² = 140 in.²
For Book B:
Length = 5.5 inches
Width = 2 inches
Height = 8 inches
Calculating the surface area using the same method,
Area of the first face: 5.5 in. × 2 in. = 11 in.²
Area of the second face: 5.5 in. × 2 in. = 11 in.²
Area of the third face: 2 in. × 8 in. = 16 in.²
Area of the fourth face: 2 in. × 8 in. = 16 in.²
Area of the fifth face: 5.5 in. × 8 in. = 44 in.²
Area of the sixth face: 5.5 in. × 8 in. = 44 in.²
Total surface area of Book B: 11 in.² + 11 in.² + 16 in.² + 16 in.² + 44 in.² + 44 in.² = 142 in.²
Therefore, Book A has a surface area of 140 in.², while Book B has a surface area of 142 in.².
The correct response is:
"Book A will require less wrapping paper because its surface area is 2 in.² less than the surface area of Book B."
For Book A:
Length = 6.5 inches
Width = 1 inch
Height = 8.5 inches
To find the surface area, we need to calculate the areas of all the faces and then sum them up.
Area of the first face: 6.5 in. × 1 in. = 6.5 in.²
Area of the second face: 6.5 in. × 1 in. = 6.5 in.²
Area of the third face: 1 in. × 8.5 in. = 8.5 in.²
Area of the fourth face: 1 in. × 8.5 in. = 8.5 in.²
Area of the fifth face: 6.5 in. × 8.5 in. = 55.25 in.²
Area of the sixth face: 6.5 in. × 8.5 in. = 55.25 in.²
Total surface area of Book A: 6.5 in.² + 6.5 in.² + 8.5 in.² + 8.5 in.² + 55.25 in.² + 55.25 in.² = 140 in.²
For Book B:
Length = 5.5 inches
Width = 2 inches
Height = 8 inches
Calculating the surface area using the same method,
Area of the first face: 5.5 in. × 2 in. = 11 in.²
Area of the second face: 5.5 in. × 2 in. = 11 in.²
Area of the third face: 2 in. × 8 in. = 16 in.²
Area of the fourth face: 2 in. × 8 in. = 16 in.²
Area of the fifth face: 5.5 in. × 8 in. = 44 in.²
Area of the sixth face: 5.5 in. × 8 in. = 44 in.²
Total surface area of Book B: 11 in.² + 11 in.² + 16 in.² + 16 in.² + 44 in.² + 44 in.² = 142 in.²
Therefore, Book A has a surface area of 140 in.², while Book B has a surface area of 142 in.².
The correct response is:
"Book A will require less wrapping paper because its surface area is 2 in.² less than the surface area of Book B."
Use the image to answer the question.
An illustration shows a 3 D box with length of 27 centimeters labeled as front, a width of 20 centimeters, and height of 20 centimeters.
A box needs to be decorated to use as a prop in a play. The bottom and the back of the box do not need to be decorated. What is the surface area of the box that needs to be decorated?
(1 point)
Responses
1,080 cm
1,080 cm
67 cm2
67 cm squared
2,960 cm2
2,960 cm squared
1,880 cm2
1,880 cm squared
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An illustration shows a 3 D box with length of 27 centimeters labeled as front, a width of 20 centimeters, and height of 20 centimeters.
A box needs to be decorated to use as a prop in a play. The bottom and the back of the box do not need to be decorated. What is the surface area of the box that needs to be decorated?
(1 point)
Responses
1,080 cm
1,080 cm
67 cm2
67 cm squared
2,960 cm2
2,960 cm squared
1,880 cm2
1,880 cm squared
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To find the surface area of the box that needs to be decorated, we need to calculate the areas of the front, top, and sides of the box.
Front:
Length = 27 cm
Height = 20 cm
Area of the front: 27 cm x 20 cm = 540 cm²
Top:
Length = 27 cm
Width = 20 cm
Area of the top: 27 cm x 20 cm = 540 cm²
Sides (two sides):
Width = 20 cm
Height = 20 cm
Area of each side: 20 cm x 20 cm = 400 cm²
Area of both sides: 400 cm² x 2 = 800 cm²
The surface area of the box that needs to be decorated is the sum of the areas of the front, top, and sides:
540 cm² + 540 cm² + 800 cm² = 1,880 cm²
Therefore, the correct response is:
"1,880 cm²"
Front:
Length = 27 cm
Height = 20 cm
Area of the front: 27 cm x 20 cm = 540 cm²
Top:
Length = 27 cm
Width = 20 cm
Area of the top: 27 cm x 20 cm = 540 cm²
Sides (two sides):
Width = 20 cm
Height = 20 cm
Area of each side: 20 cm x 20 cm = 400 cm²
Area of both sides: 400 cm² x 2 = 800 cm²
The surface area of the box that needs to be decorated is the sum of the areas of the front, top, and sides:
540 cm² + 540 cm² + 800 cm² = 1,880 cm²
Therefore, the correct response is:
"1,880 cm²"
Find the surface area of a rectangular pyramid with these measurements: l = 8 yd., w = 4 yd., and h = 2 yd.(1 point)
Responses
72.52 yd.2
72.52 yd. 2
43.31 yd.2
43.31 yd. 2
66 yd.2
66 yd. 2
55.55 yd.2
Responses
72.52 yd.2
72.52 yd. 2
43.31 yd.2
43.31 yd. 2
66 yd.2
66 yd. 2
55.55 yd.2
To find the surface area of a rectangular pyramid, we need to calculate the areas of its base and all its four triangular faces and sum them up.
Given measurements:
Length (l) = 8 yd.
Width (w) = 4 yd.
Height (h) = 2 yd.
Area of the base:
Base length (l) = 8 yd.
Base width (w) = 4 yd.
Area of the base = l * w = 8 yd. * 4 yd. = 32 yd²
Area of the four triangular faces:
Each triangular face has a base length equal to the width of the pyramid (w = 4 yd.) and a height equal to the height of the pyramid (h = 2 yd.).
Area of each triangular face = (1/2) * base * height = (1/2) * 4 yd. * 2 yd. = 4 yd²
Total area of the four triangular faces = 4 * (area of each triangular face) = 4 * 4 yd² = 16 yd²
The surface area of the rectangular pyramid is the sum of the base area and the total area of the four triangular faces:
Surface area = Area of base + Total area of four triangular faces
Surface area = 32 yd² + 16 yd² = 48 yd²
Therefore, the correct response is:
"48 yd²."
Given measurements:
Length (l) = 8 yd.
Width (w) = 4 yd.
Height (h) = 2 yd.
Area of the base:
Base length (l) = 8 yd.
Base width (w) = 4 yd.
Area of the base = l * w = 8 yd. * 4 yd. = 32 yd²
Area of the four triangular faces:
Each triangular face has a base length equal to the width of the pyramid (w = 4 yd.) and a height equal to the height of the pyramid (h = 2 yd.).
Area of each triangular face = (1/2) * base * height = (1/2) * 4 yd. * 2 yd. = 4 yd²
Total area of the four triangular faces = 4 * (area of each triangular face) = 4 * 4 yd² = 16 yd²
The surface area of the rectangular pyramid is the sum of the base area and the total area of the four triangular faces:
Surface area = Area of base + Total area of four triangular faces
Surface area = 32 yd² + 16 yd² = 48 yd²
Therefore, the correct response is:
"48 yd²."
Use the formula for the surface area of a square pyramid to find the surface area if the base edge length is 5 yards and the height is 25 yards. Round to the nearest whole number.(1 point)
Responses
1,298 square yards
1,298 square yards
251 square yards
251 square yards
276 square yards
276 square yards
151 square yards
Responses
1,298 square yards
1,298 square yards
251 square yards
251 square yards
276 square yards
276 square yards
151 square yards
The formula for the surface area of a square pyramid is:
Surface Area = Base Area + (1/2) * Perimeter of Base * Slant Height
Given measurements:
Base Edge Length = 5 yards
Height = 25 yards
Calculating the Base Area:
Base Area = (Base Edge Length)^2 = (5 yards)^2 = 25 square yards
Calculating the Perimeter of Base:
Perimeter of Base = 4 * Base Edge Length = 4 * 5 yards = 20 yards
Calculating the Slant Height:
We can use the Pythagorean Theorem to calculate the slant height using the base edge length and the height of the pyramid:
Slant Height = sqrt((Height)^2 + (1/2 * Base Edge Length)^2) = sqrt((25 yards)^2 + (1/2 * 5 yards)^2) = sqrt(625 + 6.25) = sqrt(631.25) ≈ 25.14 yards
Plugging in the values into the formula:
Surface Area = Base Area + (1/2) * Perimeter of Base * Slant Height
Surface Area = 25 square yards + (1/2) * 20 yards * 25.14 yards
Surface Area = 25 square yards + 10 * 25.14 square yards
Surface Area ≈ 25 square yards + 251.4 square yards ≈ 276.4 square yards
Rounding to the nearest whole number, the surface area of the square pyramid is approximately 276 square yards.
Therefore, the correct response is:
"276 square yards."
Surface Area = Base Area + (1/2) * Perimeter of Base * Slant Height
Given measurements:
Base Edge Length = 5 yards
Height = 25 yards
Calculating the Base Area:
Base Area = (Base Edge Length)^2 = (5 yards)^2 = 25 square yards
Calculating the Perimeter of Base:
Perimeter of Base = 4 * Base Edge Length = 4 * 5 yards = 20 yards
Calculating the Slant Height:
We can use the Pythagorean Theorem to calculate the slant height using the base edge length and the height of the pyramid:
Slant Height = sqrt((Height)^2 + (1/2 * Base Edge Length)^2) = sqrt((25 yards)^2 + (1/2 * 5 yards)^2) = sqrt(625 + 6.25) = sqrt(631.25) ≈ 25.14 yards
Plugging in the values into the formula:
Surface Area = Base Area + (1/2) * Perimeter of Base * Slant Height
Surface Area = 25 square yards + (1/2) * 20 yards * 25.14 yards
Surface Area = 25 square yards + 10 * 25.14 square yards
Surface Area ≈ 25 square yards + 251.4 square yards ≈ 276.4 square yards
Rounding to the nearest whole number, the surface area of the square pyramid is approximately 276 square yards.
Therefore, the correct response is:
"276 square yards."
What is the surface area of a square pyramid where the base edge is 8 ft. and the slant height is 16 ft.?(1 point)
Responses
512 ft.2512 ft. squared
256 ft.2256 ft. squared
320 ft.2
320 ft. squared
128 ft.2
Responses
512 ft.2512 ft. squared
256 ft.2256 ft. squared
320 ft.2
320 ft. squared
128 ft.2
The surface area of a square pyramid can be calculated using the formula:
Surface Area = Base Area + (1/2) * Perimeter of Base * Slant Height
Given measurements:
Base Edge Length = 8 ft.
Slant Height = 16 ft.
Calculating the Base Area:
The area of a square base is given by the formula:
Base Area = (Base Edge Length)^2 = (8 ft.)^2 = 64 ft²
Calculating the Perimeter of Base:
Since the base is a square, all sides have the same length.
Perimeter of Base = 4 * Base Edge Length = 4 * 8 ft. = 32 ft.
Plugging in the values into the formula:
Surface Area = Base Area + (1/2) * Perimeter of Base * Slant Height
Surface Area = 64 ft² + (1/2) * 32 ft. * 16 ft.
Surface Area = 64 ft² + 16 ft. * 16 ft.
Surface Area = 64 ft² + 256 ft²
Surface Area = 320 ft²
Therefore, the correct response is:
"320 ft²."
Surface Area = Base Area + (1/2) * Perimeter of Base * Slant Height
Given measurements:
Base Edge Length = 8 ft.
Slant Height = 16 ft.
Calculating the Base Area:
The area of a square base is given by the formula:
Base Area = (Base Edge Length)^2 = (8 ft.)^2 = 64 ft²
Calculating the Perimeter of Base:
Since the base is a square, all sides have the same length.
Perimeter of Base = 4 * Base Edge Length = 4 * 8 ft. = 32 ft.
Plugging in the values into the formula:
Surface Area = Base Area + (1/2) * Perimeter of Base * Slant Height
Surface Area = 64 ft² + (1/2) * 32 ft. * 16 ft.
Surface Area = 64 ft² + 16 ft. * 16 ft.
Surface Area = 64 ft² + 256 ft²
Surface Area = 320 ft²
Therefore, the correct response is:
"320 ft²."
1 answer