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ACT Math : Parallelograms

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Example Question #1 : How To Find The Perimeter Of A Parallelogram

Parallelogram2

Note: Figure NOT drawn to scale.

Give the perimeter of Parallelogram ABCD in the above diagram.

Possible Answers:

$30 \sqrt{2}$

$15 + 15\sqrt{3}$

$15 + 15\sqrt{2}$

$30 \sqrt{3}$

Correct answer:

$30 \sqrt{3}$

Explanation:

By the 30-60-90 Theorem, the length of the short leg of $\bigtriangleup BDC$ is the length of the long leg divided by $\sqrt{3}$ , so

$CD = \frac{BD}{\sqrt{3}} = \frac{15}{\sqrt{3}} = \frac{15\cdot \sqrt{3}}{\sqrt{3}\cdot \sqrt{3}}= \frac{15 \sqrt{3}}{3} = 5\sqrt{3}$

Its hypotenuse has twice the length of the short leg, so

$BC = 2 \cdot CD = 2 \cdot 5\sqrt{3}= 10 \sqrt{3}$

The perimeter of the parallelogram is

AB + BC + CD + DA

=CD + BC + CD + BC

$=5 \sqrt{3} + 10\sqrt{3} + 5 \sqrt{3} + 10\sqrt{3} = 30 \sqrt{3}$

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Example Question #2 : How To Find The Perimeter Of A Parallelogram

Parallelogram1

Note: Figure NOT drawn to scale.

Give the perimeter of Parallelogram ABCD in the above diagram.

Possible Answers:

$68 \sqrt{2}$

$34 + 34 \sqrt{2}$

$68 \sqrt{3}$

$34 + 34 \sqrt{3}$

Correct answer:

$34 + 34 \sqrt{2}$

Explanation:

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

CD = BD = 17

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

AB + BC + CD + DA

=CD + BC + CD + BC

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

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Example Question #3 : How To Find The Perimeter Of A Parallelogram

Parallelogram

Note: Figure NOT drawn to scale.

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

Possible Answers:

56.7

69.4

28.3

34.7

31.4

Correct answer:

31.4

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

AB + BC + CD + DA

=CD + BC + CD + BC

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

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Example Question #4 : How To Find The Perimeter Of A Parallelogram

Parallelogram1

In the above figure, Parallelogram ABCD has area 100. To the nearest tenth, what is its perimeter?

Possible Answers:

48.3

50.5

45.6

54.6

47.6

Correct answer:

48.3

Explanation:

By the 45-45-90 Theorem, BD = CD . Since and are its base and height:

$A = BD \cdot CD$

$100 = CD \cdot CD = \left (CD \right ) ^{2}$

CD =10

Also by the 45-45-90 Theorem,

$BC = CD \cdot \sqrt{2} = 10 \cdot 1.4142 \approx 14.14$

The perimeter of the parallelogram is

AB + BC + CD + DA

=CD + BC + CD + BC

$\approx 10+ 14.14 + 10 + 14.14$

$\approx 48.3$

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Example Question #1 : How To Find The Perimeter Of A Parallelogram

Parallelogram2

In the above figure, Parallelogram ABCD has area 100. To the nearest tenth, what is its perimeter?

Possible Answers:

45.6

54.6

50.5

47.6

48.3

Correct answer:

45.6

Explanation:

By the 30-60-90 Theorem,

$BD = CD\cdot \sqrt{3}$

The area of the parallelogram is the product of height and base , so

$A = BD \cdot CD$

$100 = CD\cdot \sqrt{3}\cdot CD$

$100 = \left (CD\right )^{2} \sqrt{3}$

$\left (CD\right )^{2} = \frac{100 }{\sqrt{3}}$

$\left (CD\right )^{2} \approx \frac{100 }{1.7321}$

$\left (CD\right )^{2} \approx 57.73$

$CD \approx \sqrt{57.73} \approx 7.60$

Also by the 30-60-90 Theorem,

$BD =2\cdot CD \approx 2 \cdot 7.60 = 15.20$

The perimeter of the parallelogram is

AB + BC + CD + DA

=CD + BC + CD + BC

$\approx 7.60 + 15.20 + 7.60 + 15.20 \approx 45.6$

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Example Question #6 : How To Find The Perimeter Of A Parallelogram

Parallelogram

Note: Figure NOT drawn to scale.

In the above figure, Parallelogram ABCD has area 100. To the nearest tenth, what is its perimeter?

Possible Answers:

28.7

70.7

44.2

46.0

46.6

Correct answer:

46.0

Explanation:

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

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

The area of the parallelogram is the product of height and base , so

$A = BD \cdot CD$

$100 = CD \cdot \tan 65^{\circ}\cdot CD$

$(CD)^{2} \cdot \tan 65^{\circ} = 100$

$(CD)^{2} = \frac{100}{\tan 65^{\circ} } = \frac{100}{2.1445} \approx 46.63$

$CD \approx \sqrt{46.63} \approx 6.83$

$\frac{CD}{BC} = \cos 65^{\circ}$

$\frac{BC} {CD}= \frac {1}{ \cos 65^{\circ}}$

$BC = \frac{CD}{ \cos 65^{\circ}}$

$BC \approx \frac{6.83}{0.4226} \approx 16.16$

The perimeter of the parallelogram is

AB + BC + CD + DA

=CD + BC + CD + BC

$\approx 6.83 + 16.16 + 6.83 +16.16$

$\approx 46.0$

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Example Question #1 : How To Find The Length Of The Diagonal Of A Parallelogram

If a rectangular plot measures $8 \:ft$ by $6\:ft$ , what is the length of the diagonal of the plot, in feet?

Possible Answers:

$7\:ft$

$9\:ft$

$13\:ft$

$10\:ft$

$11\:ft$

Correct answer:

$10\:ft$

Explanation:

To answer this question, we must find the diagonal of a rectangle that is $8\: ft$ by $6\:ft$ . Because a rectangle is made up of right angles, the diagonal of a rectangle creates a right triangle with two of the sides.

Because a right triangle is formed by the diagonal, we can use the Pythagorean Theorem, which is:

$a^{2}+b^{2}=c^{2}$

and each represent a different leg of the triangle and represents the length of the hypotenuse, which in this case is the same as the diagonal length.

We can then plug in our known values and solve for $c^{2}$

$a^{2}+b^{2}=c^{2}\rightarrow 8^{2}+6^{2}=c^{2}\rightarrow 64+36=c^{2}\rightarrow 100=c^{2}$

We now must take the square root of each side so that we can solve for

$\sqrt{100}=\sqrt{c^{2}}\rightarrow 10=c$

Therefore, the diagonal of the rectangle is $10\:ft$ .

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Example Question #2 : How To Find The Length Of The Diagonal Of A Parallelogram

Parallelogram_12

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

Possible Answers:

17.5

Correct answer:

Explanation:

To find the length of the diagonal, we can consider only the triangle ABD 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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Example Question #3 : How To Find The Length Of The Diagonal Of A Parallelogram

Parallelogram_13

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

Possible Answers:

108

136

112

102

Correct answer:

112

Explanation:

To find the length of the diagonal, we can consider only the triangle ACD 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:

$\overline{AC}^2 = \overline{CD}^2 + \overline{DA}^2 - 2\left ( \overline{CD} \right )\left (\overline{DA} \right )\cos \left ( \measuredangle D\right )$

$\overline{AC}^2 = 35^2 + 82 ^2 - 2\cdot35\cdot82\cdot \cos \left ( 143^{\circ}\right )$

$\overline{AC}^2 = 12553$

$\overline{AC} = 112$

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Example Question #4 : How To Find The Length Of The Diagonal Of A Parallelogram

Parallelogram_15

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

Possible Answers:

49.2

41.8

56.6

59.1

28.3

Correct answer:

49.2

Explanation:

To find the length of the diagonal, we can consider only the triangle ACD 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:

$\overline{AC}^2 = \overline{CD}^2 + \overline{DA}^2 - 2\left ( \overline{CD} \right )\left (\overline{DA} \right )\cos \left ( \measuredangle D\right )$

$\overline{AC}^2 = 12^2 + 40 ^2 - 2\cdot12\cdot40\cdot \cos \left ( 135^{\circ}\right )$

$\overline{AC}^2 = 2423$

$\overline{AC} = 49.2$

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ACT Math : Parallelograms

Study concepts, example questions & explanations for ACT Math

All ACT Math Resources

Example Questions

Example Question #1 : How To Find The Perimeter Of A Parallelogram

Example Question #2 : How To Find The Perimeter Of A Parallelogram

Example Question #3 : How To Find The Perimeter Of A Parallelogram

Example Question #4 : How To Find The Perimeter Of A Parallelogram

Example Question #1 : How To Find The Perimeter Of A Parallelogram

Example Question #6 : How To Find The Perimeter Of A Parallelogram

Example Question #1 : How To Find The Length Of The Diagonal Of A Parallelogram

Example Question #2 : How To Find The Length Of The Diagonal Of A Parallelogram

Example Question #3 : How To Find The Length Of The Diagonal Of A Parallelogram

Example Question #4 : How To Find The Length Of The Diagonal Of A Parallelogram

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