How to find the length of the side of a parallelogram - ACT Math

Card 0 of 54

Question

In parallelogram , $\overline{AB} = 10$ and $\measuredangle A = 60^{\circ}$ . Find $\overline{CD}$ .

Answer

In a parallelogram, opposite sides are congruent. Thus,

$\overline{AB} = \overline{CD}$

$10 = \overline{CD}$

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Question

In parallelogram , $\overline{BC} = 30$ and $\measuredangle B = 120^{\circ}$ . Find $\overline{AD}$ .

Answer

In a parallelogram, opposite sides are congruent.

$\overline{BC} = \overline{AD}$

$30 = \overline{AD}$

Compare your answer with the correct one above

Question

Parallelogram has an area of . If , find $\overline{BC}$ .

Answer

The area of a parallelogram is given by:

In this problem, the height is given as and the area is . Both $\overline{AD}$ and $\overline{BC}$ are bases.

$375 = \left ( 15\right )\left ( \overline{BC} \right )$

$\overline{BC} = 25$

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Question

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

Answer

The formula for the area of a parallelogram is:

By plugging in the given values, we get:

$204 = b\cdot12$

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Question

is a parallelogram. Find $\overline{AB}$ .

Answer

$\overline{AB}$ is the hypotenuse of the right triangle formed when we draw the height of the parallelogram. Because it is a right triangle, we can use SOH CAH TOA to solve for $\overline{AB}$ . With respect to $\measuredangle A$ , we know the opposite side of the triangle and we are looking for the hypotenuse. Thus, we can use the sine function to solve for $\overline{AB}$ .

$\sin \left ( \theta\right ) = \frac{\textup{opposite}}{\textup{hypotenuse}}$

$\sin\left ( 42^{\circ}\right ) = \frac{16}{\overline{AB}}$

$\overline{AB} = \frac{16}{\sin\left ( 42^{\circ}\right )}$

$\overline{AB} = 23.9$

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Question

is a parallelogram with an area of . Find $\overline{AB}$ .

Answer

In order to find $\overline{AB}$ , we must first find . The formula for the area of a parallelogram is:

We are given as the area and as the base.

$510 = 34\cdot h$

Now, we can use trigonometry to solve for $\overline{AB}$ . With respect to $\measuredangle A$ , we know the opposite side of the right triangle and we are looking for the hypotenuse. Thus, we can use the sine function.

$\sin\left ( \theta\right ) = \frac{\textup{opposite}}{\textup{hypotenuse}}$

$\sin\left ( 62^{\circ}\right ) = \frac{15}{\overline{AB}}$

$\overline{AB} = \frac{15}{\sin \left ( 62^{\circ}\right )} = 17$

Compare your answer with the correct one above

Question

In parallelogram , $\overline{AB} = 10$ and $\measuredangle A = 60^{\circ}$ . Find $\overline{CD}$ .

Answer

In a parallelogram, opposite sides are congruent. Thus,

$\overline{AB} = \overline{CD}$

$10 = \overline{CD}$

Compare your answer with the correct one above

Question

In parallelogram , $\overline{BC} = 30$ and $\measuredangle B = 120^{\circ}$ . Find $\overline{AD}$ .

Answer

In a parallelogram, opposite sides are congruent.

$\overline{BC} = \overline{AD}$

$30 = \overline{AD}$

Compare your answer with the correct one above

Question

Parallelogram has an area of . If , find $\overline{BC}$ .

Answer

The area of a parallelogram is given by:

In this problem, the height is given as and the area is . Both $\overline{AD}$ and $\overline{BC}$ are bases.

$375 = \left ( 15\right )\left ( \overline{BC} \right )$

$\overline{BC} = 25$

Compare your answer with the correct one above

Question

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

Answer

The formula for the area of a parallelogram is:

By plugging in the given values, we get:

$204 = b\cdot12$

Compare your answer with the correct one above

Question

is a parallelogram. Find $\overline{AB}$ .

Answer

$\overline{AB}$ is the hypotenuse of the right triangle formed when we draw the height of the parallelogram. Because it is a right triangle, we can use SOH CAH TOA to solve for $\overline{AB}$ . With respect to $\measuredangle A$ , we know the opposite side of the triangle and we are looking for the hypotenuse. Thus, we can use the sine function to solve for $\overline{AB}$ .

$\sin \left ( \theta\right ) = \frac{\textup{opposite}}{\textup{hypotenuse}}$

$\sin\left ( 42^{\circ}\right ) = \frac{16}{\overline{AB}}$

$\overline{AB} = \frac{16}{\sin\left ( 42^{\circ}\right )}$

$\overline{AB} = 23.9$

Compare your answer with the correct one above

Question

is a parallelogram with an area of . Find $\overline{AB}$ .

Answer

In order to find $\overline{AB}$ , we must first find . The formula for the area of a parallelogram is:

We are given as the area and as the base.

$510 = 34\cdot h$

Now, we can use trigonometry to solve for $\overline{AB}$ . With respect to $\measuredangle A$ , we know the opposite side of the right triangle and we are looking for the hypotenuse. Thus, we can use the sine function.

$\sin\left ( \theta\right ) = \frac{\textup{opposite}}{\textup{hypotenuse}}$

$\sin\left ( 62^{\circ}\right ) = \frac{15}{\overline{AB}}$

$\overline{AB} = \frac{15}{\sin \left ( 62^{\circ}\right )} = 17$

Compare your answer with the correct one above

Question

In parallelogram , $\overline{AB} = 10$ and $\measuredangle A = 60^{\circ}$ . Find $\overline{CD}$ .

Answer

In a parallelogram, opposite sides are congruent. Thus,

$\overline{AB} = \overline{CD}$

$10 = \overline{CD}$

Compare your answer with the correct one above

Question

In parallelogram , $\overline{BC} = 30$ and $\measuredangle B = 120^{\circ}$ . Find $\overline{AD}$ .

Answer

In a parallelogram, opposite sides are congruent.

$\overline{BC} = \overline{AD}$

$30 = \overline{AD}$

Compare your answer with the correct one above

Question

Parallelogram has an area of . If , find $\overline{BC}$ .

Answer

The area of a parallelogram is given by:

In this problem, the height is given as and the area is . Both $\overline{AD}$ and $\overline{BC}$ are bases.

$375 = \left ( 15\right )\left ( \overline{BC} \right )$

$\overline{BC} = 25$

Compare your answer with the correct one above

Question

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

Answer

The formula for the area of a parallelogram is:

By plugging in the given values, we get:

$204 = b\cdot12$

Compare your answer with the correct one above

Question

is a parallelogram. Find $\overline{AB}$ .

Answer

$\overline{AB}$ is the hypotenuse of the right triangle formed when we draw the height of the parallelogram. Because it is a right triangle, we can use SOH CAH TOA to solve for $\overline{AB}$ . With respect to $\measuredangle A$ , we know the opposite side of the triangle and we are looking for the hypotenuse. Thus, we can use the sine function to solve for $\overline{AB}$ .

$\sin \left ( \theta\right ) = \frac{\textup{opposite}}{\textup{hypotenuse}}$

$\sin\left ( 42^{\circ}\right ) = \frac{16}{\overline{AB}}$

$\overline{AB} = \frac{16}{\sin\left ( 42^{\circ}\right )}$

$\overline{AB} = 23.9$

Compare your answer with the correct one above

Question

is a parallelogram with an area of . Find $\overline{AB}$ .

Answer

In order to find $\overline{AB}$ , we must first find . The formula for the area of a parallelogram is:

We are given as the area and as the base.

$510 = 34\cdot h$

Now, we can use trigonometry to solve for $\overline{AB}$ . With respect to $\measuredangle A$ , we know the opposite side of the right triangle and we are looking for the hypotenuse. Thus, we can use the sine function.

$\sin\left ( \theta\right ) = \frac{\textup{opposite}}{\textup{hypotenuse}}$

$\sin\left ( 62^{\circ}\right ) = \frac{15}{\overline{AB}}$

$\overline{AB} = \frac{15}{\sin \left ( 62^{\circ}\right )} = 17$

Compare your answer with the correct one above

Question

In parallelogram , $\overline{AB} = 10$ and $\measuredangle A = 60^{\circ}$ . Find $\overline{CD}$ .

Answer

In a parallelogram, opposite sides are congruent. Thus,

$\overline{AB} = \overline{CD}$

$10 = \overline{CD}$

Compare your answer with the correct one above

Question

In parallelogram , $\overline{BC} = 30$ and $\measuredangle B = 120^{\circ}$ . Find $\overline{AD}$ .

Answer

In a parallelogram, opposite sides are congruent.

$\overline{BC} = \overline{AD}$

$30 = \overline{AD}$

Compare your answer with the correct one above

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