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Intermediate Geometry : Quadrilaterals

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Example Questions

Example Question #91 : Parallelograms

Find the area of a parallelogram.

Possible Answers:

109.92 $\displaystyle 109.92$

112.23 $\displaystyle 112.23$

120.54 $\displaystyle 120.54$

107.33 $\displaystyle 107.33$

Correct answer:

107.33 $\displaystyle 107.33$

Explanation:

Recall how to find the area of a parallelogram:

$\text{Area}=\text{base}\times\text{height}$ $\displaystyle \text{Area}=\text{base}\times\text{height}$

From the given parallelogram, we will need to use the Pythagorean Theorem to find the length of the height of the parallelogram.

$\text{Hypotenuse}^2=\text{Triangle base}^2+\text{height}^2$ $\displaystyle \text{Hypotenuse}^2=\text{Triangle base}^2+\text{height}^2$

$\text{height}^2=\text{Hypotenuse}^2-\text{Triangle base}^2$ $\displaystyle \text{height}^2=\text{Hypotenuse}^2-\text{Triangle base}^2$

$\text{height}=\sqrt{\text{Hypotenuse}^2-\text{Triangle base}^2}$ $\displaystyle \text{height}=\sqrt{\text{Hypotenuse}^2-\text{Triangle base}^2}$

Plug in the given values to find the length of the height.

$\text{height}=\sqrt{9^2-6^2}=\sqrt{81-36}=\sqrt{45}$ $\displaystyle \text{height}=\sqrt{9^2-6^2}=\sqrt{81-36}=\sqrt{45}$

Now, use the height to find the area of the parallelogram.

$\text{Area}=16\times\sqrt{45}=107.33$ $\displaystyle \text{Area}=16\times\sqrt{45}=107.33$

Remember to round to $\displaystyle 2$ places after the decimal.

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Example Question #92 : Parallelograms

Find the area of the parallelogram.

Possible Answers:

$\displaystyle 42$

$\displaystyle 40$

$\displaystyle 39$

$\displaystyle 48$

Correct answer:

$\displaystyle 40$

Explanation:

Recall how to find the area of a parallelogram:

$\text{Area}=\text{base}\times\text{height}$ $\displaystyle \text{Area}=\text{base}\times\text{height}$

From the given parallelogram, we will need to use the Pythagorean Theorem to find the length of the height of the parallelogram.

$\text{Hypotenuse}^2=\text{Triangle base}^2+\text{height}^2$ $\displaystyle \text{Hypotenuse}^2=\text{Triangle base}^2+\text{height}^2$

$\text{height}^2=\text{Hypotenuse}^2-\text{Triangle base}^2$ $\displaystyle \text{height}^2=\text{Hypotenuse}^2-\text{Triangle base}^2$

$\text{height}=\sqrt{\text{Hypotenuse}^2-\text{Triangle base}^2}$ $\displaystyle \text{height}=\sqrt{\text{Hypotenuse}^2-\text{Triangle base}^2}$

Plug in the given values to find the length of the height.

$\text{height}=\sqrt{5^2-3^2}=\sqrt{25-9}=4$ $\displaystyle \text{height}=\sqrt{5^2-3^2}=\sqrt{25-9}=4$

Now, use the height to find the area of the parallelogram.

$\text{Area}=10\times4=40$ $\displaystyle \text{Area}=10\times4=40$

Remember to round to $\displaystyle 2$ places after the decimal.

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Example Question #261 : Quadrilaterals

Find the area of the parallelogram.

Possible Answers:

13.32 $\displaystyle 13.32$

14.14 $\displaystyle 14.14$

9.85 $\displaystyle 9.85$

15.08 $\displaystyle 15.08$

Correct answer:

14.14 $\displaystyle 14.14$

Explanation:

Recall how to find the area of a parallelogram:

$\text{Area}=\text{base}\times\text{height}$ $\displaystyle \text{Area}=\text{base}\times\text{height}$

From the given parallelogram, we will need to use the Pythagorean Theorem to find the length of the height of the parallelogram.

$\text{Hypotenuse}^2=\text{Triangle base}^2+\text{height}^2$ $\displaystyle \text{Hypotenuse}^2=\text{Triangle base}^2+\text{height}^2$

$\text{height}^2=\text{Hypotenuse}^2-\text{Triangle base}^2$ $\displaystyle \text{height}^2=\text{Hypotenuse}^2-\text{Triangle base}^2$

$\text{height}=\sqrt{\text{Hypotenuse}^2-\text{Triangle base}^2}$ $\displaystyle \text{height}=\sqrt{\text{Hypotenuse}^2-\text{Triangle base}^2}$

Plug in the given values to find the length of the height.

$\text{height}=\sqrt{3^2-1^2}=\sqrt{9-1}=\sqrt{8}$ $\displaystyle \text{height}=\sqrt{3^2-1^2}=\sqrt{9-1}=\sqrt{8}$

Now, use the height to find the area of the parallelogram.

$\text{Area}=5\times\sqrt{8}=14.14$ $\displaystyle \text{Area}=5\times\sqrt{8}=14.14$

Remember to round to $\displaystyle 2$ places after the decimal.

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Example Question #91 : Parallelograms

Find the area of the parallelogram.

Possible Answers:

298.50 $\displaystyle 298.50$

287.56 $\displaystyle 287.56$

291.11 $\displaystyle 291.11$

295.16 $\displaystyle 295.16$

Correct answer:

295.16 $\displaystyle 295.16$

Explanation:

Recall how to find the area of a parallelogram:

$\text{Area}=\text{base}\times\text{height}$ $\displaystyle \text{Area}=\text{base}\times\text{height}$

From the given parallelogram, we will need to use the Pythagorean Theorem to find the length of the height of the parallelogram.

$\text{Hypotenuse}^2=\text{Triangle base}^2+\text{height}^2$ $\displaystyle \text{Hypotenuse}^2=\text{Triangle base}^2+\text{height}^2$

$\text{height}^2=\text{Hypotenuse}^2-\text{Triangle base}^2$ $\displaystyle \text{height}^2=\text{Hypotenuse}^2-\text{Triangle base}^2$

$\text{height}=\sqrt{\text{Hypotenuse}^2-\text{Triangle base}^2}$ $\displaystyle \text{height}=\sqrt{\text{Hypotenuse}^2-\text{Triangle base}^2}$

Plug in the given values to find the length of the height.

$\text{height}=\sqrt{18^2-12^2}=\sqrt{324-144}=\sqrt{180}$ $\displaystyle \text{height}=\sqrt{18^2-12^2}=\sqrt{324-144}=\sqrt{180}$

Now, use the height to find the area of the parallelogram.

$\text{Area}=22\times\sqrt{180}=295.16$ $\displaystyle \text{Area}=22\times\sqrt{180}=295.16$

Remember to round to $\displaystyle 2$ places after the decimal.

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Example Question #92 : Parallelograms

Find the area of the parallelogram.

Possible Answers:

490.57 $\displaystyle 490.57$

455.96 $\displaystyle 455.96$

482.12 $\displaystyle 482.12$

402.19 $\displaystyle 402.19$

Correct answer:

455.96 $\displaystyle 455.96$

Explanation:

Recall how to find the area of a parallelogram:

$\text{Area}=\text{base}\times\text{height}$ $\displaystyle \text{Area}=\text{base}\times\text{height}$

From the given parallelogram, we will need to use the Pythagorean Theorem to find the length of the height of the parallelogram.

$\text{Hypotenuse}^2=\text{Triangle base}^2+\text{height}^2$ $\displaystyle \text{Hypotenuse}^2=\text{Triangle base}^2+\text{height}^2$

$\text{height}^2=\text{Hypotenuse}^2-\text{Triangle base}^2$ $\displaystyle \text{height}^2=\text{Hypotenuse}^2-\text{Triangle base}^2$

$\text{height}=\sqrt{\text{Hypotenuse}^2-\text{Triangle base}^2}$ $\displaystyle \text{height}=\sqrt{\text{Hypotenuse}^2-\text{Triangle base}^2}$

Plug in the given values to find the length of the height.

$\text{height}=\sqrt{20^2-13^2}=\sqrt{400-169}=\sqrt{231}$ $\displaystyle \text{height}=\sqrt{20^2-13^2}=\sqrt{400-169}=\sqrt{231}$

Now, use the height to find the area of the parallelogram.

$\text{Area}=30\times\sqrt{231}=455.96$ $\displaystyle \text{Area}=30\times\sqrt{231}=455.96$

Remember to round to $\displaystyle 2$ places after the decimal.

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Example Question #51 : How To Find The Area Of A Parallelogram

Find the area of the parallelogram.

Possible Answers:

90.21 $\displaystyle 90.21$

88.53 $\displaystyle 88.53$

109.44 $\displaystyle 109.44$

96.75 $\displaystyle 96.75$

Correct answer:

96.75 $\displaystyle 96.75$

Explanation:

Recall how to find the area of a parallelogram:

$\text{Area}=\text{base}\times\text{height}$ $\displaystyle \text{Area}=\text{base}\times\text{height}$

From the given parallelogram, we will need to use the Pythagorean Theorem to find the length of the height of the parallelogram.

$\text{Hypotenuse}^2=\text{Triangle base}^2+\text{height}^2$ $\displaystyle \text{Hypotenuse}^2=\text{Triangle base}^2+\text{height}^2$

$\text{height}^2=\text{Hypotenuse}^2-\text{Triangle base}^2$ $\displaystyle \text{height}^2=\text{Hypotenuse}^2-\text{Triangle base}^2$

$\text{height}=\sqrt{\text{Hypotenuse}^2-\text{Triangle base}^2}$ $\displaystyle \text{height}=\sqrt{\text{Hypotenuse}^2-\text{Triangle base}^2}$

Plug in the given values to find the length of the height.

$\text{height}=\sqrt{9^2-4^2}=\sqrt{81-16}=\sqrt{65}$ $\displaystyle \text{height}=\sqrt{9^2-4^2}=\sqrt{81-16}=\sqrt{65}$

Now, use the height to find the area of the parallelogram.

$\text{Area}=12\times\sqrt{65}=96.75$ $\displaystyle \text{Area}=12\times\sqrt{65}=96.75$

Remember to round to $\displaystyle 2$ places after the decimal.

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Example Question #91 : Parallelograms

Find the area of the parallelogram.

Possible Answers:

305.68 $\displaystyle 305.68$

298.72 $\displaystyle 298.72$

234.07 $\displaystyle 234.07$

299.14 $\displaystyle 299.14$

Correct answer:

298.72 $\displaystyle 298.72$

Explanation:

Recall how to find the area of a parallelogram:

$\text{Area}=\text{base}\times\text{height}$ $\displaystyle \text{Area}=\text{base}\times\text{height}$

From the given parallelogram, we will need to use the Pythagorean Theorem to find the length of the height of the parallelogram.

$\text{Hypotenuse}^2=\text{Triangle base}^2+\text{height}^2$ $\displaystyle \text{Hypotenuse}^2=\text{Triangle base}^2+\text{height}^2$

$\text{height}^2=\text{Hypotenuse}^2-\text{Triangle base}^2$ $\displaystyle \text{height}^2=\text{Hypotenuse}^2-\text{Triangle base}^2$

$\text{height}=\sqrt{\text{Hypotenuse}^2-\text{Triangle base}^2}$ $\displaystyle \text{height}=\sqrt{\text{Hypotenuse}^2-\text{Triangle base}^2}$

Plug in the given values to find the length of the height.

$\text{height}=\sqrt{14^2-8^2}=\sqrt{196-64}=\sqrt{132}$ $\displaystyle \text{height}=\sqrt{14^2-8^2}=\sqrt{196-64}=\sqrt{132}$

Now, use the height to find the area of the parallelogram.

$\text{Area}=26\times\sqrt{132}=298.72$ $\displaystyle \text{Area}=26\times\sqrt{132}=298.72$

Remember to round to $\displaystyle 2$ places after the decimal.

Report an Error

Example Question #91 : Parallelograms

Find the area of the parallelogram.

Note that the figure is not drawn to scale.

Possible Answers:

112

100

Correct answer:

Explanation:

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

A=bh $\displaystyle A=bh$

In this problem, the height is 6 (not 8). Substitute the known variables and solve.

A=bh $\displaystyle A=bh$

$A=14\times 6$ $\displaystyle A=14\times 6$

A=84 $\displaystyle A=84$

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Example Question #433 : Intermediate Geometry

If the height of the parallelogram is half of the length of the rectangle, then find the area of the shaded region in the figure.

Possible Answers:

102 $\displaystyle 102$

$\displaystyle 56$

$\displaystyle 65$

$\displaystyle 72$

Correct answer:

$\displaystyle 72$

Explanation:

In order to find the area of the shaded region, we must first find the areas of the rectangle and parallelogram.

Recall how to find the area of a rectangle:

$\text{Area of Rectangle}=\text{length}\times\text{width}$ $\displaystyle \text{Area of Rectangle}=\text{length}\times\text{width}$

Substitute in the given length and height to find the area of the rectangle.

$\text{Area of Rectangle}=8 \times 14$ $\displaystyle \text{Area of Rectangle}=8 \times 14$

$\text{Area of Rectangle}=112$ $\displaystyle \text{Area of Rectangle}=112$

Next, find the area of the parallelogram.

Recall how to find the area of a parallelogram:

$\text{Area of Parallelogram}=\text{base}\times\text{height}$ $\displaystyle \text{Area of Parallelogram}=\text{base}\times\text{height}$

We need to find the height of the parallelogram. From the question, we know the following relationship:

$\text{height}=\frac{\text{length}}{2}$ $\displaystyle \text{height}=\frac{\text{length}}{2}$

Substitute in the length of the rectangle to find the height of the parallelogram.

$\text{height}=\frac{8}{2}$ $\displaystyle \text{height}=\frac{8}{2}$

$\text{height}=4$ $\displaystyle \text{height}=4$

Now, substitute in the height and the given length of the base to find the area of the parallelogram.

$\text{Area of Parallelogram}=10 \times 4$ $\displaystyle \text{Area of Parallelogram}=10 \times 4$

$\text{Area of Parallelogram}=40$ $\displaystyle \text{Area of Parallelogram}=40$

Now, we are ready to find the area of the shaded region by subtracting the area of the parallelogram from the area of the rectangle.

$\text{Area of Shaded Region}=\text{Area of Rectangle}-\text{Area of Parallelogram}$ $\displaystyle \text{Area of Shaded Region}=\text{Area of Rectangle}-\text{Area of Parallelogram}$

$\text{Area of Shaded Region}=112-40$ $\displaystyle \text{Area of Shaded Region}=112-40$ =72 $\displaystyle =72$

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Example Question #1 : How To Find If Parallelograms Are Similar

A parallelogram has adjacent sides with the lengths of $\displaystyle 6$ and $\displaystyle 9$. Find a pair of possible adjacent side lengths for a similar parallelogram.

Possible Answers:

$\displaystyle 12$ and $\displaystyle 20$

$\displaystyle 12$ and $\displaystyle 18$

$\displaystyle 1$ and $\displaystyle 5$

$\displaystyle 3$ and $\displaystyle 6$

Correct answer:

$\displaystyle 12$ and $\displaystyle 18$

Explanation:

Since the two parallelogram are similar, each of the corresponding sides must have the same ratio.

The solution is:

$6:9=(6\div3):(9\div 3)=2:3$ $\displaystyle 6:9=(6\div3):(9\div 3)=2:3$

$\displaystyle 12:18=2:3$, (divide both numbers by the common divisor of $\displaystyle 6$).

$\displaystyle 6:9=12:18$

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Intermediate Geometry : Quadrilaterals

Study concepts, example questions & explanations for Intermediate Geometry

All Intermediate Geometry Resources

Example Questions

Example Question #91 : Parallelograms

Example Question #92 : Parallelograms

Example Question #261 : Quadrilaterals

Example Question #91 : Parallelograms

Example Question #92 : Parallelograms

Example Question #51 : How To Find The Area Of A Parallelogram

Example Question #91 : Parallelograms

Example Question #91 : Parallelograms

Example Question #433 : Intermediate Geometry

Example Question #1 : How To Find If Parallelograms Are Similar

All Intermediate Geometry Resources

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