Question

What is the surface area of the net of the triangular prism?(1 point)
17m. 17m 16m 15m 11m

Answer 1

To calculate the surface area of a triangular prism, you need to know the dimensions of all its faces: the two triangular bases and the three rectangular lateral faces.

Let’s designate the lengths given to the appropriate parts of the prism:
- Assume 17m and 15m are the lengths of the two legs of the triangle (since these are generally smaller and can match the dimensions of the triangle).
- Assume 16m is the hypotenuse of the triangle (common in 3,4,5 triangles this makes 3*5, 4*5).
- Assume 17m and 11m are the lengths of the three rectangles involving the height.

First, calculate the area of the two triangular bases:
Using the base-leg formula, where 17m is one leg and 15m is another:

Area of one triangle \(A_t\):
\[
A_t = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 17 \times 15 = \frac{1}{2} \times 255 = 127.5 \text{ m}^2
\]
Since there are two identical triangles:
Total area of triangles \(= 2 \times 127.5 = 255 \text{ m}^2\)

Next, calculate the area of the three rectangles:
1. Rectangles with 11m and the height (16m).
2. Depending on which length involves the height.

Since all rectangle areas are involving breadth + height:
A1 = \(17 \times 11\)
A2 = \(15 \times 11\)
A3 = \(16 \times 12\)

\[
A_{rectangle 1} = 17 \text{m} \times 15 \text{m} = 115 \text{ m}^2
\]
\[
A_{rectangle 2} = 17 \text{m} \times 17 \text{m} = 289 \text{ m}^2
\]
\[
A_{rectangle 3} = 16 \text{m} \times 11 \text{m} = 176 \text{ m}^2
\]

Now, sum up all the calculated areas:
\[
\text{Surface Area} = \text{Area of triangles} + \text{Area of rectangles}
\]
\[
\text{Surface Area} = 255 \text{ m}^2 + (115 \text{ m}^2 + 176 \text{ m}^2 + 289 \text{ m}^2)
\]
\[
\text{Surface Area} = 835 \text{ m}^2
\]

So, the surface area of this triangular prism is 835 square meters.