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Ray purchases a paperweight in the shape of a rectangular pyramid and wants to cover it in decorative paper. He creates the net of the figure.

Ray only wants to cover the lateral area of his paper weight. How many square inches of paper will he need? Round your answer to the nearest square inch.

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Answer: Ray will need approximately 40 square inches of decorative paper to cover the lateral area of his paperweight. Rounded to the nearest square inch, he will need 40 square inches.

Explanation:

To find the lateral area of the rectangular pyramid, we need to find the area of each triangular face and add them up.

Let's label the dimensions of the rectangular pyramid:

The base of the rectangular pyramid has dimensions 4 inches by 6 inches.

The height of the rectangular pyramid is 3 inches.

First, we need to find the slant height of the pyramid, which is the hypotenuse of each triangular face. We can use the Pythagorean theorem to find it:

slant height = √( (1.5)^2 + 3^2 ) = √( 2.25 + 9 ) = √11.25 ≈ 3.354

Now we can find the area of each triangular face using the formula:

area = 0.5 x base x height

The base of each triangular face is the width of the rectangular base, which is 6 inches. The height of each triangular face is the slant height we just found, which is approximately 3.354 inches.

So, the area of each triangular face is:

area = 0.5 x 6 x 3.354 = 10.062 square inches (rounded to three decimal places)

There are four triangular faces, so the total lateral area of the pyramid is:

total lateral area = 4 x 10.062 = 40.248 square inches (rounded to three decimal places)

Therefore, Ray will need approximately 40 square inches of decorative paper to cover the lateral area of his paperweight. Rounded to the nearest square inch, he will need 40 square inches.

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