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A design engineer built a rotating isosceles triangular prism in the middle of a park for people to enjoy. If the rotating prism has the following dimensions as shown in the image below, what is the total surface area of the rotating prism?

A design engineer built a rotating isosceles triangular prism in the middle of a park for people to enjoy. If the rotating prism has the following dimensions as class=

Answer :

bexwhite2610

Answer:

0.5 X 3 X 2 = 3

3 X 2 = 6 ft^2

4 X 3 = 12 ft^2

4 X 2.5 = 10 X 2 = 20 ft^2

6 + 12 + 20 = 38ft^2

lhmarianateixeira

In this exercise we have to use the knowledge of rotation and prism to calculate the total area of ​​the prism that has undergone a rotation, so we find that this will correspond to:

[tex]A= 38\ ft^2[/tex]

Knowing that the dimensions of the rotating prism are:

  • H: [tex]2 \ ft[/tex]
  • B: [tex]3 \ ft[/tex]
  • W: [tex]0.5 \ ft[/tex]

From the data we will use the formula that of the area, that is:

[tex]A= H*W*B[/tex]

Replacing the values ​​in the informed formula, we find that:

[tex]0.5 *3 * 2 = 3\\3 * 2 = 6 ft^2\\4 * 3 = 12 ft^2\\4 * 2.5 = 10 * 2 = 20 ft^2\\6 + 12 + 20 = 38ft^2[/tex]

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