How to find the area of a parallelogram - SSAT Middle Level Math

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Question

Parallelogram

Note: Figure NOT drawn to scale

In the above diagram,

Give the area of the parallelogram.

Answer

The area of a parallelogram is its base multiplied by its height - represented by and here:

Note that the value of is irrelevant.

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Question

Find the area of the following parallelogram:

Isee_mid_question_42

Note: The formula for the area of a parallelogram is .

Answer

The base of the parallelogram is 10, while the height is 5.

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Question

Find the area:

Question_5

Answer

The area of a parallelogram can be determined using the following equation:

Therefore,

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Question

Parallelogram

Find the area of the given parallelogram if .

Answer

In order to find the area of a parallelogram, we need to find the product of the base length and height.

Notice that only two of the given values were needed to slove this problem.

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Question

A given parallelogram has a base in length, a height in length, and a side of length opposite the height. What is the area of the parallelogram?

Answer

The formula for the area of a parallelogram is , with base and height represented by and , respectively. Substituting values from the question:

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Question

A parallelogram has a height of in length, a side of length opposite the height, and a base of . What is the area of the parallelogram?

Answer

Given base and height , .

Substituting the values from our question:

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Question

A parallelogram has a base of length , a height of length , and a side of length . What is the area of the parallelogram?

Answer

Given base and height , .

Substituting the values from our question:

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Question

Find the area of a parallelogram with a height of , a base of , and a side length of .

Answer

The area of a parallelogram with height and base can be found with the equation . Consequently:

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Question

Find the area of a parallelogram with a height of , base of , and a side length of .

Answer

The area of a parallelogram with height and base can be found with the equation . Consequently:

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