Ask Question
1 June, 15:33

A triangular prism is 16 millimeters long and has a triangular face with a base of 18 millimeters and a height of 12 millimeters. The other two sides of the triangle are each 15 millimeters. What is the surface area of the triangular prism?

+4
Answers (1)
  1. 1 June, 15:56
    0
    Answer: The surface area of the triangular prism is 984mm^2

    Step-by-step explanation:

    A triangular prism normally has five components which are two triangles and three rectangles all fixed together to form a single shape.

    To calculate the surface area of this triangular prism, we have to calculate the separate areas of the two triangles and the respective areas of the three rectangles and add them all up. The result thus, becomes the surface area of the shape (the triangular prism).

    The area of a triangle = 1/2 * base * height.

    Where the base is 18mm and the height is 12mm

    So, the area of one triangle = 1/2 * 12 * 18

    = 108mm^2

    Since the prism has two triangles, the area of the two triangles should be:-

    108 * 2 = 216mm^2

    Now, among the rectangles that made up the prism, two are identical. The formula for calculating the area of a rectangle = length * width;

    Where: the length of one of the identical rectangles is 16mm and the width is 15mm.

    The area of one of the two identical rectangles is = 16 * 15 = 240mm^2

    The area of the two identical rectangles is then = 240 * 2 = 480mm^2

    In the same manner, the length of the other rectangle is = 18mm and the width is 16mm

    The area of this other rectangle is then = 18 * 16 = 288mm^2

    The surface area of the prism is then:

    288mm^2 + 480mm^2 + 216mm^2

    = 984mm^2
Know the Answer?
Not Sure About the Answer?
Find an answer to your question 👍 “A triangular prism is 16 millimeters long and has a triangular face with a base of 18 millimeters and a height of 12 millimeters. The other ...” in 📗 Mathematics if the answers seem to be not correct or there’s no answer. Try a smart search to find answers to similar questions.
Search for Other Answers