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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 $\displaystyle 150$. Find the length of the base.

Possible Answers:

$\displaystyle 25$

$\displaystyle 10$

$\displaystyle 20$

$\displaystyle 15$

Correct answer:

$\displaystyle 15$

Explanation:

Recall how to find the area of a parallelogram:

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

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

$\displaystyle x(x+5)=150$

Expand this equation.

$\displaystyle x^2+5x=150$

$\displaystyle x^2+5x-150=0$

Now factor this equation.

$\displaystyle (x-10)(x+15)=0$

Solve for $\displaystyle x$.

$\displaystyle x=10, x=-5$

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

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

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

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

$\text{Length of Base}=15$ $\displaystyle \text{Length of Base}=15$

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Example Question #421 : Intermediate Geometry

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

Possible Answers:

$\displaystyle 20$

$\displaystyle 25$

$\displaystyle 15$

$\displaystyle 30$

Correct answer:

$\displaystyle 25$

Explanation:

Recall how to find the area of a parallelogram:

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

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

$\displaystyle x(x+5)=500$

Expand this equation.

$\displaystyle x^2+5x=500$

$\displaystyle x^2+5x-500=0$

Now factor this equation.

$\displaystyle (x-20)(x+25)=0$

Solve for $\displaystyle x$.

$\displaystyle x=20, x=-25$

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

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

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

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

$\text{Length of Base}=25$ $\displaystyle \text{Length of Base}=25$

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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 $\displaystyle 164$. Find the length of the base.

Possible Answers:

$\displaystyle 43$

$\displaystyle 41$

$\displaystyle 39$

$\displaystyle 37$

Correct answer:

$\displaystyle 41$

Explanation:

Recall how to find the area of a parallelogram:

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

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

$\displaystyle x(x+37)=164$

Expand this equation.

$\displaystyle x^2+37x=164$

$\displaystyle x^2+37x-164=0$

Now factor this equation.

$\displaystyle (x-4)(x+41)=0$

Solve for $\displaystyle x$.

$\displaystyle x=4, x=-41$

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

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

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

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

$\text{Length of Base}=41$ $\displaystyle \text{Length of Base}=41$

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Example Question #81 : Parallelograms

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

Possible Answers:

$\displaystyle 32$

$\displaystyle 30$

$\displaystyle 28$

$\displaystyle 38$

Correct answer:

$\displaystyle 32$

Explanation:

Recall how to find the area of a parallelogram:

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

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

$\displaystyle x(x+24)=256$

Expand this equation.

$\displaystyle x^2+24x=256$

$\displaystyle x^2+24x-256=0$

Now factor this equation.

$\displaystyle (x-8)(x+32)=0$

Solve for $\displaystyle x$.

$\displaystyle x=8, x=-32$

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

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

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

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

$\text{Length of Base}=32$ $\displaystyle \text{Length of Base}=32$

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Example Question #82 : Parallelograms

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

Possible Answers:

$\displaystyle 16$

$\displaystyle 12$

$\displaystyle 8$

$\displaystyle 4$

Correct answer:

$\displaystyle 8$

Explanation:

Recall how to find the area of a parallelogram:

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

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

$\displaystyle x(x-8)=128$

Expand this equation.

$\displaystyle x^2-8x=128$

$\displaystyle x^2-8x-128=0$

Now factor this equation.

$\displaystyle (x-16)(x+8)=0$

Solve for $\displaystyle x$.

$\displaystyle x=16, x=-8$

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

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

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

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

$\text{Length of Base}=8$ $\displaystyle \text{Length of Base}=8$

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Example Question #421 : Plane Geometry

Find the area of the parallelogram.

Possible Answers:

169.48 $\displaystyle 169.48$

184.09 $\displaystyle 184.09$

199.23 $\displaystyle 199.23$

186.23 $\displaystyle 186.23$

Correct answer:

186.23 $\displaystyle 186.23$

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.

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

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

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

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

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Example Question #421 : Plane Geometry

Find the area of the parallelogram.

Possible Answers:

54.99 $\displaystyle 54.99$

60.31 $\displaystyle 60.31$

42.19 $\displaystyle 42.19$

58.29 $\displaystyle 58.29$

Correct answer:

54.99 $\displaystyle 54.99$

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.

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

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

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

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

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Example Question #251 : Quadrilaterals

Find the area of the parallelogram.

Possible Answers:

189 $\displaystyle 189$

213 $\displaystyle 213$

240 $\displaystyle 240$

256 $\displaystyle 256$

Correct answer:

240 $\displaystyle 240$

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.

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

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

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

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

Report an Error

Example Question #41 : How To Find The Area Of A Parallelogram

Find the area of the parallelogram.

Possible Answers:

60.06 $\displaystyle 60.06$

78.45 $\displaystyle 78.45$

69.22 $\displaystyle 69.22$

62.61 $\displaystyle 62.61$

Correct answer:

62.61 $\displaystyle 62.61$

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.

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

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

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

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

Report an Error

Example Question #422 : Plane Geometry

Find the area of the parallelogram.

Possible Answers:

95.61 $\displaystyle 95.61$

82.50 $\displaystyle 82.50$

93.67 $\displaystyle 93.67$

90.88 $\displaystyle 90.88$

Correct answer:

93.67 $\displaystyle 93.67$

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.

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

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

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

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

Report an Error

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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 #421 : Intermediate Geometry

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 #421 : Plane Geometry

Example Question #251 : Quadrilaterals

Example Question #41 : How To Find The Area Of A Parallelogram

Example Question #422 : Plane Geometry

All Intermediate Geometry Resources

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