Question

A parallelogram is cut out of a 12-inch by 8-inch sheet of paper. There are four right triangle remnants. Two have the dimensions 2 inches by 9 inches, and the other two have the dimensions 3 inches by 6 inches. The resulting parallelogram has a base of approximately 9.22 inches. Complete the following steps to calculate the altitude of the parallelogram using area methods. The area of the sheet of paper is square inches. The combined area of the triangle cutouts is square inches. The area of the parallelogram is square inches. The altitude of the parallelogram rounded to two decimals is square inches.

Answer 1

Answer:

96

36

60

6.51

Step-by-step explanation:

Answer 2

The area of a shape is the amount of space it occupies.

• The area of the paper is 96 square inches
• The combined area of the triangle cutouts is 36 square inches
• The area of the parallelogram is 60 square inches
• The altitude of the parallelogram is 6.51 inches

The dimension of the paper is given as: 12-inch by 8-inch

So, its area is:

$\mathbf{A_1 =12 \times 8}$

$\mathbf{A_1 =96}$

The dimensions of the 4 right triangles are: Two 2 inches by 9 inches, and two 3 inches by 6 inches

So, the combined area is:

$\mathbf{A_2 = 2 \times \frac 12 \times 2 \times 9 + 2 \times \frac 12 \times 3 \times 6}$

$\mathbf{A_2 = 36}$

The area of the parallelogram is the difference between the areas of the paper and the four right triangles.

So, we have:

$\mathbf{A_3 = A_1 - A_2}$

$\mathbf{A_3 = 96 - 36}$

$\mathbf{A_3 = 60}$

The area of a parallelogram is:

$\mathbf{Area = Base \times Altitude}$

The base is given as 9.22

So, we have:

$\mathbf{60 = 9.22\times Altitude}$

Divide both sides by 9.22

$\mathbf{6.51 = Altitude}$

Rewrite as:

$\mathbf{Altitude = 6.51}$

Hence, the altitude of the parallelogram is 6.51 inches

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