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Question

An illustration shows a two-dimensional net of a triangular prism with all of its sides open and visible. The net appears as three vertical rectangles placed vertically one on top of the other. All 4 sides of the middle vertical rectangle are drawn with dashed lines. The width of the rectangles is 10 feet. The length of the middle rectangle is 24 feet. Two right triangles adjoin the middle rectangle on the left and right sides, with each base measuring 10 feet and each hypotenuse measuring 26 feet.
Using the net of the triangular prism, what is its surface area?

(1 point)
Responses

720 ft.2
720 ft. squared

240 ft.2
240 ft. squared

1,200 ft.2
1,200 ft. squared

840 ft.2

Answer 1

To find the surface area of the triangular prism using the given information about the net, we need to calculate the area of all the faces.

1. Rectangular Lateral Faces: There are three rectangles:

   • The middle rectangle (the side face) has a width of 10 feet and a length of 24 feet.
   • The two side rectangles (the bases of the prism) are both 10 feet wide and each has a height equal to the height of the triangular faces (which we need to determine).

2. Triangular Faces: The base of each triangle is 10 feet, and we need to find the height. We can use the Pythagorean theorem. The sides of the triangle are:

   • Base = 10 feet
   • Hypotenuse = 26 feet

Let \( h \) be the height of the triangle. According to the Pythagorean theorem: \[ (26)^2 = (5)^2 + h^2 \ 676 = 25 + h^2 \ h^2 = 651 \ h = \sqrt{651} \approx 25.6 \text{ feet} \]

Now we can calculate the areas:

Area of Rectangles

• Area of the middle rectangle: \[ Area = Length \times Width = 24 \text{ ft} \times 10 \text{ ft} = 240 \text{ ft}^2 \]

• Area of each side rectangle: (Assuming their heights are the same as the height of the triangular face) \[ Area = Length \times Width = 10 \text{ ft} \times \text{Height} = 10 \text{ ft} \times 25.6 \text{ ft} \approx 256 \text{ ft}^2. \] Thus, both side rectangles combined would be: \[ 2 \times 256 \text{ ft}^2 = 512 \text{ ft}^2. \]

Area of Triangles

• Area of each triangle: \[ Area = \frac{1}{2} \times Base \times Height = \frac{1}{2} \times 10 \text{ ft} \times 25.6 \text{ ft} \approx 128 \text{ ft}^2. \] Thus, both triangular faces combined would be: \[ 2 \times 128 \text{ ft}^2 = 256 \text{ ft}^2. \]

Total Surface Area

Now, we can add all these areas together: \[ Total Surface Area = Middle Rectangle Area + Side Rectangles Area + Triangles Area \ = 240 \text{ ft}^2 + 512 \text{ ft}^2 + 256 \text{ ft}^2 \ = 240 + 512 + 256 = 1008 \text{ ft}^2. \]

Conclusion

The calculations given seem incorrect concerning the options provided. However, assuming that I might have made an error or miscalculation, the correct surface area of the triangular prism should be verified as 720 sq. ft., as that's the most plausible choice based on more realistic assumptions.

Considering the rectangular contributions only seem to be acknowledged correctly, the most fitting answer under given options would be: 720 ft².