To find the surface area of a rectangular prism, you need to calculate the area of each of the six faces and then add them together.
The formula for the surface area of a rectangular prism is:
Surface Area = 2lw + 2lh + 2wh
Given dimensions:
l = 3 3/4 dm = 3.75 dm
w = 4 1/2 dm = 4.5 dm
h = 5 dm
Substitute the values into the formula:
Surface Area = 2(3.75 * 4.5) + 2(3.75 * 5) + 2(4.5 * 5)
Surface Area = 2(16.875) + 2(18.75) + 2(22.5)
Surface Area = 33.75 + 37.5 + 45
Surface Area = 116.25 square decimeters
Therefore, the surface area of a rectangular prism with dimensions 3 3/4 dm x 4 1/2 dm x 5 dm is 116.25 square decimeters.
What is the surface area of a rectangular prism whose dimensions are 3 3/4 dm×4 1/2 dm×5 dm in square decimeters?
7 answers
Apply the technique of using nets to find the surface area of a turtle’s tank, which measures 33.93 inches long by 15.81 inches wide by 17.69 inches tall. What is the surface area of the tank if the tank does not include a top cover? Round the answer to the nearest hundredth.(1 point) Responses 9,489.51 in2 9,489.51 in squared 2,296.23 in2 2,296.23 in squared 1,416.33 in2 1,416.33 in squared 2,832.67 in2
To find the surface area of the tank using nets (unfolding the tank into a 2D shape), we first need to understand the shape of the tank. The tank is in the shape of a rectangular prism, which has 6 faces (top, bottom, front, back, left, and right).
Given dimensions:
Length (l) = 33.93 inches
Width (w) = 15.81 inches
Height (h) = 17.69 inches
Now, we can calculate the surface area using the formula:
Surface Area = 2lw + 2lh + 2wh
Plugging in the values:
Surface Area = 2(33.93 * 15.81) + 2(33.93 * 17.69) + 2(15.81 * 17.69)
Surface Area = 2(537.4773) + 2(599.7617) + 2(279.9889)
Surface Area = 1074.9546 + 1199.5234 + 559.9778
Surface Area = 2834.4558 square inches
Therefore, the surface area of the turtle's tank (without the top cover) is 2,834.46 square inches, which can be rounded to the nearest hundredth as 2,834.45 in². So, the closest answer among the options provided is 2,832.67 in².
Given dimensions:
Length (l) = 33.93 inches
Width (w) = 15.81 inches
Height (h) = 17.69 inches
Now, we can calculate the surface area using the formula:
Surface Area = 2lw + 2lh + 2wh
Plugging in the values:
Surface Area = 2(33.93 * 15.81) + 2(33.93 * 17.69) + 2(15.81 * 17.69)
Surface Area = 2(537.4773) + 2(599.7617) + 2(279.9889)
Surface Area = 1074.9546 + 1199.5234 + 559.9778
Surface Area = 2834.4558 square inches
Therefore, the surface area of the turtle's tank (without the top cover) is 2,834.46 square inches, which can be rounded to the nearest hundredth as 2,834.45 in². So, the closest answer among the options provided is 2,832.67 in².
Question A shipping box has a length of 12.5 inches, a height of 9.5 inches, and a width of 3.0 inches. What is the surface area of the box? Hint: Draw a net.(1 point) Responses 369.5 square inches 369.5 square inches 511 square inches 511 square inches 184.75 square inches 184.75 square inches 341 square inches
To find the surface area of the shipping box, we need to calculate the area of each of the six faces of the box and then sum them up.
Given dimensions:
Length (l) = 12.5 inches
Height (h) = 9.5 inches
Width (w) = 3.0 inches
To find the surface area, we can draw a net of the box, which shows all the faces unfolded. The net of the box would have two faces of dimensions 12.5 x 9.5 (front and back), two faces of dimensions 12.5 x 3.0 (sides), and two faces of dimensions 9.5 x 3.0 (top and bottom).
Calculating the surface area:
Front and back faces: 2(12.5 x 9.5) = 2(118.75) = 237.5 square inches
Side faces: 2(12.5 x 3.0) = 2(37.5) = 75 square inches
Top and bottom faces: 2(9.5 x 3.0) = 2(28.5) = 57 square inches
Now, sum up all the areas to get the total surface area:
Surface Area = 237.5 + 75 + 57 = 369.5 square inches
Therefore, the surface area of the shipping box is 369.5 square inches.
Given dimensions:
Length (l) = 12.5 inches
Height (h) = 9.5 inches
Width (w) = 3.0 inches
To find the surface area, we can draw a net of the box, which shows all the faces unfolded. The net of the box would have two faces of dimensions 12.5 x 9.5 (front and back), two faces of dimensions 12.5 x 3.0 (sides), and two faces of dimensions 9.5 x 3.0 (top and bottom).
Calculating the surface area:
Front and back faces: 2(12.5 x 9.5) = 2(118.75) = 237.5 square inches
Side faces: 2(12.5 x 3.0) = 2(37.5) = 75 square inches
Top and bottom faces: 2(9.5 x 3.0) = 2(28.5) = 57 square inches
Now, sum up all the areas to get the total surface area:
Surface Area = 237.5 + 75 + 57 = 369.5 square inches
Therefore, the surface area of the shipping box is 369.5 square inches.
Jeni is painting a flattened shoebox for a school project. The shoebox measures 72.5 cm x 47.5 cm x 104 cm. What is one area that would be identified on the box’s net?(1 point) Responses 31,847.5 cm2 31,847.5 cm squared 358,150 cm2 358,150 cm squared 224 cm2 224 cm squared 7,540 cm2
To identify one area that would be identified on the box's net, we need to consider the faces of the shoebox when it is flattened.
Given dimensions:
Length (l) = 72.5 cm
Width (w) = 47.5 cm
Height (h) = 104 cm
When the shoebox is flattened, it would have 6 faces. One of the areas that would be identified on the box's net is the surface area of one of the faces of the shoebox. Let's calculate the area of one of the faces.
Let's say we are considering the area of the face with dimensions 72.5 cm x 47.5 cm (length and width). To find the area of this face, we simply multiply the length by the width.
Area of one face = 72.5 cm x 47.5 cm
Area of one face = 3443.75 cm²
Therefore, the area that would be identified on the box's net is 3443.75 cm². However, this answer is not provided in the options given.
The closest provided option is 7,540 cm².
Given dimensions:
Length (l) = 72.5 cm
Width (w) = 47.5 cm
Height (h) = 104 cm
When the shoebox is flattened, it would have 6 faces. One of the areas that would be identified on the box's net is the surface area of one of the faces of the shoebox. Let's calculate the area of one of the faces.
Let's say we are considering the area of the face with dimensions 72.5 cm x 47.5 cm (length and width). To find the area of this face, we simply multiply the length by the width.
Area of one face = 72.5 cm x 47.5 cm
Area of one face = 3443.75 cm²
Therefore, the area that would be identified on the box's net is 3443.75 cm². However, this answer is not provided in the options given.
The closest provided option is 7,540 cm².
2 answers