Parallelograms

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Parallelogram1

Note: Figure NOT drawn to scale.

Give the perimeter of Parallelogram in the above diagram.

$34 + 34 \sqrt{2}$

$34 + 34 \sqrt{3}$

$68 \sqrt{2}$

$68 \sqrt{3}$

Explanation

By the 45-45-90 Theorem, the lengths of the legs of $\bigtriangleup BDC$ are equal, so

Its hypotenuse has measure $\sqrt{2}$ that of the common measure of its legs, so

$BC = CD \cdot \sqrt{2} = 17 \sqrt{2}$

The perimeter of the parallelogram is

$=17+ 17 \sqrt{2}+ 17 + 17\sqrt{2}= 34 + 34 \sqrt{2}$

Parallelogram

Note: Figure NOT drawn to scale.

To the nearest tenth, give the perimeter of Parallelogram in the above diagram.

Explanation

$\frac{BD}{CD} = \tan 65^{\circ}$

$\frac{CD}{BD} = \frac{1}{\tan 65^{\circ}}$

$CD= \frac{BD}{\tan 65^{\circ}} \approx \frac{10}{2.1445} \approx 4.66$

$\frac{BD}{BC} = \sin 65^{\circ}$

$\frac{BC}{BD} = \frac{1}{\sin 65^{\circ}}$

$BC= \frac{BD}{\sin 65^{\circ}} \approx \frac{10}{0.9063} \approx 11.03$

The perimeter of the parallelogram is

$\approx 4.66 + 11.03 + 4.66 + 11.03 \approx 31.4$

Parallelogram1

Note: Figure NOT drawn to scale.

Give the perimeter of Parallelogram in the above diagram.

$34 + 34 \sqrt{2}$

$34 + 34 \sqrt{3}$

$68 \sqrt{2}$

$68 \sqrt{3}$

Explanation

By the 45-45-90 Theorem, the lengths of the legs of $\bigtriangleup BDC$ are equal, so

Its hypotenuse has measure $\sqrt{2}$ that of the common measure of its legs, so

$BC = CD \cdot \sqrt{2} = 17 \sqrt{2}$

The perimeter of the parallelogram is

$=17+ 17 \sqrt{2}+ 17 + 17\sqrt{2}= 34 + 34 \sqrt{2}$

Parallelogram_10

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

There is insufficient information to solve the problem.

Explanation

$\overline{AB}$ 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 $\overline{AB}$ .

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

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

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

$\overline{AB} = 23.9$

Parallelogram

Note: Figure NOT drawn to scale.

To the nearest tenth, give the perimeter of Parallelogram in the above diagram.

Explanation

$\frac{BD}{CD} = \tan 65^{\circ}$

$\frac{CD}{BD} = \frac{1}{\tan 65^{\circ}}$

$CD= \frac{BD}{\tan 65^{\circ}} \approx \frac{10}{2.1445} \approx 4.66$

$\frac{BD}{BC} = \sin 65^{\circ}$

$\frac{BC}{BD} = \frac{1}{\sin 65^{\circ}}$

$BC= \frac{BD}{\sin 65^{\circ}} \approx \frac{10}{0.9063} \approx 11.03$

The perimeter of the parallelogram is

$\approx 4.66 + 11.03 + 4.66 + 11.03 \approx 31.4$

Parallelogram_10

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

There is insufficient information to solve the problem.

Explanation

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

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

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

$\overline{AB} = 23.9$

Parallelogram_9

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

There is insufficient information to solve the problem.

Explanation

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.

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

$\overline{BC} = 25$

Parallelogram_9

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

There is insufficient information to solve the problem.

Explanation

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.

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

$\overline{BC} = 25$

Parallelogram_12

is a parallelogram. Find the length of diagonal $\overline{BD}$ .

Explanation

To find the length of the diagonal, we can consider only the triangle and use the law of cosines to find the length of the unknown side.

The Law of Cosines:

$c^2 = a^2 + b^2 - 2ab\cos \left ( C\right )$

Where is the length of the unknown side, and are the lengths of the known sides, and is the angle between and .

From the problem:

$\overline{BD}^2 = \overline{AB}^2 + \overline{DA}^2 - 2\left ( \overline{AB} \right )\left (\overline{DA} \right )\cos \left ( \measuredangle A\right )$

$\overline{BD}^2 = 35^2 + 82 ^2 - 2\cdot35\cdot82\cdot \cos \left ( 37^{\circ}\right )$

$\overline{BD}^2 = 3365$

$\overline{BD} = 58$

Parallelogram_12

is a parallelogram. Find the length of diagonal $\overline{BD}$ .

Explanation

To find the length of the diagonal, we can consider only the triangle and use the law of cosines to find the length of the unknown side.

The Law of Cosines:

$c^2 = a^2 + b^2 - 2ab\cos \left ( C\right )$

Where is the length of the unknown side, and are the lengths of the known sides, and is the angle between and .

From the problem:

$\overline{BD}^2 = \overline{AB}^2 + \overline{DA}^2 - 2\left ( \overline{AB} \right )\left (\overline{DA} \right )\cos \left ( \measuredangle A\right )$

$\overline{BD}^2 = 35^2 + 82 ^2 - 2\cdot35\cdot82\cdot \cos \left ( 37^{\circ}\right )$

$\overline{BD}^2 = 3365$

$\overline{BD} = 58$

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