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Intermediate Geometry : Plane Geometry

Study concepts, example questions & explanations for Intermediate Geometry

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8 Diagnostic Tests 250 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #92 : Parallelograms

Find the area of the parallelogram.

Possible Answers:

Correct answer:

Explanation:

Recall how to find the area of a parallelogram:

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

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

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

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

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

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

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

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

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

Remember to round to places after the decimal.

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Example Question #431 : Intermediate Geometry

Find the area of the parallelogram.

Possible Answers:

14.14

9.85

15.08

13.32

Correct answer:

14.14

Explanation:

Recall how to find the area of a parallelogram:

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

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

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

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

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

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

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

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

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

Remember to round to places after the decimal.

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Example Question #91 : Parallelograms

Find the area of the parallelogram.

Possible Answers:

298.50

287.56

291.11

295.16

Correct answer:

295.16

Explanation:

Recall how to find the area of a parallelogram:

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

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

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

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

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

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

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

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

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

Remember to round to places after the decimal.

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Example Question #92 : Parallelograms

Find the area of the parallelogram.

Possible Answers:

490.57

455.96

482.12

402.19

Correct answer:

455.96

Explanation:

Recall how to find the area of a parallelogram:

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

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

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

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

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

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

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

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

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

Remember to round to 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

88.53

109.44

96.75

Correct answer:

96.75

Explanation:

Recall how to find the area of a parallelogram:

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

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

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

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

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

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

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

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

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

Remember to round to places after the decimal.

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Example Question #261 : Quadrilaterals

Find the area of the parallelogram.

Possible Answers:

299.14

234.07

305.68

298.72

Correct answer:

298.72

Explanation:

Recall how to find the area of a parallelogram:

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

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

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

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

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

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

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

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

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

Remember to round to places after the decimal.

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Example Question #431 : Intermediate Geometry

Find the area of the parallelogram.

Note that the figure is not drawn to scale.

Possible Answers:

100

112

Correct answer:

Explanation:

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

A=bh

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

A=bh

$A=14\times 6$

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

Correct answer:

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:

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

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

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

$\text{Area of Rectangle}=112$

Next, find the area of the parallelogram.

Recall how to find the area of a parallelogram:

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

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

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

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

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

$\text{height}=4$

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

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

$\text{Area of Parallelogram}=40$

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.

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

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

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

A parallelogram has adjacent sides with the lengths of and . Find a pair of possible adjacent side lengths for a similar parallelogram.

Possible Answers:

and

Correct answer:

and

Explanation:

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

The solution is:

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

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

6:9=12:18

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

A parallelogram has adjacent sides with the lengths of and . Find a pair of possible adjacent side lengths for a similar parallelogram.

Possible Answers:

and

Correct answer:

and

Explanation:

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

The ratio of the first parallelogram is:

4:13

Applying this ratio we are able to find the lengths of a similar parallelogram.

$4:13=(4\times 3):(13\times3 )=12:39$

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Intermediate Geometry : Plane Geometry

Study concepts, example questions & explanations for Intermediate Geometry

All Intermediate Geometry Resources

Example Questions

Example Question #92 : Parallelograms

Example Question #431 : Intermediate Geometry

Example Question #91 : Parallelograms

Example Question #92 : Parallelograms

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

Example Question #261 : Quadrilaterals

Example Question #431 : Intermediate Geometry

Example Question #433 : Intermediate Geometry

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

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

All Intermediate Geometry Resources

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