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Intermediate Geometry : Quadrilaterals

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Example Questions

Example Question #81 : Parallelograms

In the figure, the area of the parallelogram is 150 . Find the length of the base.

Possible Answers:

Correct answer:

Explanation:

Recall how to find the area of a parallelogram:

$\text{Area of Parallelogram}=(\text{base})(\text{height})$

Now, substitute in the area, base, and height values that are given by the question.

x(x+5)=150

Expand this equation.

x^2+5x=150

x^2+5x-150=0

Now factor this equation.

(x-10)(x+15)=0

Solve for .

x=10, x=-5

Since lengths of bases and heights can only be positive, x=10 .

Notice that the length of the base is given by the expression x+5 . Substitute in the value of to find the length of the base.

$\text{Length of Base}=x+5$

$\text{Length of Base}=10+5$

$\text{Length of Base}=15$

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Example Question #251 : Quadrilaterals

In the figure, the area of the parallelogram is 500 . Find the length of the base.

Possible Answers:

Correct answer:

Explanation:

Recall how to find the area of a parallelogram:

$\text{Area of Parallelogram}=(\text{base})(\text{height})$

Now, substitute in the area, base, and height values that are given by the question.

x(x+5)=500

Expand this equation.

x^2+5x=500

x^2+5x-500=0

Now factor this equation.

(x-20)(x+25)=0

Solve for .

x=20, x=-25

Since lengths of bases and heights can only be positive, x=20 .

Notice that the length of the base is given by the expression x+5 . Substitute in the value of to find the length of the base.

$\text{Length of Base}=x+5$

$\text{Length of Base}=20+5$

$\text{Length of Base}=25$

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Example Question #40 : How To Find The Area Of A Parallelogram

In the figure, the area of the parallelogram is 164 . Find the length of the base.

Possible Answers:

Correct answer:

Explanation:

Recall how to find the area of a parallelogram:

$\text{Area of Parallelogram}=(\text{base})(\text{height})$

Now, substitute in the area, base, and height values that are given by the question.

x(x+37)=164

Expand this equation.

x^2+37x=164

x^2+37x-164=0

Now factor this equation.

(x-4)(x+41)=0

Solve for .

x=4, x=-41

Since lengths of bases and heights can only be positive, x=4 .

Notice that the length of the base is given by the expression x+37 . Substitute in the value of to find the length of the base.

$\text{Length of Base}=x+37$

$\text{Length of Base}=4+37$

$\text{Length of Base}=41$

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Example Question #81 : Parallelograms

In the figure, the area of the parallelogram is 256 . Find the length of the base.

Possible Answers:

Correct answer:

Explanation:

Recall how to find the area of a parallelogram:

$\text{Area of Parallelogram}=(\text{base})(\text{height})$

Now, substitute in the area, base, and height values that are given by the question.

x(x+24)=256

Expand this equation.

x^2+24x=256

x^2+24x-256=0

Now factor this equation.

(x-8)(x+32)=0

Solve for .

x=8, x=-32

Since lengths of bases and heights can only be positive, x=8 .

Notice that the length of the base is given by the expression x+24 . Substitute in the value of to find the length of the base.

$\text{Length of Base}=x+24$

$\text{Length of Base}=8+24$

$\text{Length of Base}=32$

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Example Question #82 : Parallelograms

In the figure, the area of the parallelogram is 128 . Find the length of the base.

Possible Answers:

Correct answer:

Explanation:

Recall how to find the area of a parallelogram:

$\text{Area of Parallelogram}=(\text{base})(\text{height})$

Now, substitute in the area, base, and height values that are given by the question.

x(x-8)=128

Expand this equation.

x^2-8x=128

x^2-8x-128=0

Now factor this equation.

(x-16)(x+8)=0

Solve for .

x=16, x=-8

Since lengths of bases and heights can only be positive, x=16 .

Notice that the length of the base is given by the expression x-8 . Substitute in the value of to find the length of the base.

$\text{Length of Base}=x-8$

$\text{Length of Base}=16-8$

$\text{Length of Base}=8$

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Example Question #421 : Plane Geometry

Find the area of the parallelogram.

Possible Answers:

169.48

184.09

199.23

186.23

Correct answer:

186.23

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.

$\text{height}=\sqrt{13^2-7^2}=\sqrt{169-49}=\sqrt{120}$

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

$\text{Area}=17\times\sqrt{120}=186.23$

Remember to round to places after the decimal.

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Example Question #252 : Quadrilaterals

Find the area of the parallelogram.

Possible Answers:

54.99

60.31

42.19

58.29

Correct answer:

54.99

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.

$\text{height}=\sqrt{5^2-2^2}=\sqrt{25-4}=\sqrt{21}$

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

$\text{Area}=12\times\sqrt{21}=54.99$

Remember to round to places after the decimal.

Report an Error

Example Question #253 : Quadrilaterals

Find the area of the parallelogram.

Possible Answers:

256

240

189

213

Correct answer:

240

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.

$\text{height}=\sqrt{15^2-9^2}=\sqrt{225-81}=12$

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

$\text{Area}=12\times20=240$

Remember to round to places after the decimal.

Report an Error

Example Question #254 : Quadrilaterals

Find the area of the parallelogram.

Possible Answers:

62.61

69.22

60.06

78.45

Correct answer:

62.61

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.

$\text{height}=\sqrt{6^2-4^2}=\sqrt{36-16}=\sqrt{20}$

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

$\text{Area}=16\times\sqrt{20}=62.61$

Remember to round to places after the decimal.

Report an Error

Example Question #255 : Quadrilaterals

Find the area of the parallelogram.

Possible Answers:

93.67

90.88

82.50

95.61

Correct answer:

93.67

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.

$\text{height}=\sqrt{8^2-5^2}=\sqrt{64-25}=\sqrt{39}$

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

$\text{Area}=15\times\sqrt{39}=93.67$

Remember to round to places after the decimal.

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Intermediate Geometry : Quadrilaterals

Study concepts, example questions & explanations for Intermediate Geometry

All Intermediate Geometry Resources

Example Questions

Example Question #81 : Parallelograms

Example Question #251 : Quadrilaterals

Example Question #40 : How To Find The Area Of A Parallelogram

Example Question #81 : Parallelograms

Example Question #82 : Parallelograms

Example Question #421 : Plane Geometry

Example Question #252 : Quadrilaterals

Example Question #253 : Quadrilaterals

Example Question #254 : Quadrilaterals

Example Question #255 : Quadrilaterals

All Intermediate Geometry Resources

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