How to find the perimeter of a parallelogram - ACT Math

Card 0 of 63

Question

A parallelogram, with dimensions in cm, is shown below.

What is the perimeter of the parallelogram, in cm?

Answer

The triangle on the left side of the figure has a $30^{\circ}$ and a $90^{\circ}$ angle. Since all of the angles of a triangle must add up to $180^{\circ}$ , we can find the angle measure of the third angle:

$180-120=60^{\circ}$

Our third angle is $60^{\circ}$ and we have a triangle.

A triangle has sides that are in the corresponding ratio of $x-x\sqrt{3}-2x$ . In this case, the side opposite our $30^{\circ}$ angle is , so

We also now know that

$x\sqrt{3}=6\sqrt{3}$

Now we know all of our missing side lengths. The right and left side of the parallelogram will each be . The bottom and top will each be $20+6\sqrt{3}$ . Let's combine them to find the perimeter:

$2\cdot (12)+2\cdot (20+6\sqrt{3})$

$24+40+12\sqrt{3}$

$64+12\sqrt{3}$

Compare your answer with the correct one above

Question

Note: Figure NOT drawn to scale.

Give the perimeter of Parallelogram in the above diagram.

Answer

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

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

Compare your answer with the correct one above

Question

Note: Figure NOT drawn to scale.

Give the perimeter of Parallelogram in the above diagram.

Answer

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

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

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

Compare your answer with the correct one above

Question

Note: Figure NOT drawn to scale.

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

Answer

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

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

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

$\approx 6.83 + 16.16 + 6.83 +16.16$

$\approx 46.0$

Compare your answer with the correct one above

Question

Note: Figure NOT drawn to scale.

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

Answer

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

$\frac{CD}{BD} = \frac{1}{\tan 65^{\circ}}$

$CD= \frac{BD}{\tan 65^{\circ}} \approx \frac{10}{2.1445} \approx 4.66$

$\frac{BD}{BC} = \sin 65^{\circ}$

$\frac{BC}{BD} = \frac{1}{\sin 65^{\circ}}$

$BC= \frac{BD}{\sin 65^{\circ}} \approx \frac{10}{0.9063} \approx 11.03$

The perimeter of the parallelogram is

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

Compare your answer with the correct one above

Question

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

Answer

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

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

Compare your answer with the correct one above

Question

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

Answer

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

$A = BD \cdot CD$

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

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

$\approx 10+ 14.14 + 10 + 14.14$

$\approx 48.3$

Compare your answer with the correct one above

Question

A parallelogram, with dimensions in cm, is shown below.

What is the perimeter of the parallelogram, in cm?

Answer

The triangle on the left side of the figure has a $30^{\circ}$ and a $90^{\circ}$ angle. Since all of the angles of a triangle must add up to $180^{\circ}$ , we can find the angle measure of the third angle:

$180-120=60^{\circ}$

Our third angle is $60^{\circ}$ and we have a triangle.

A triangle has sides that are in the corresponding ratio of $x-x\sqrt{3}-2x$ . In this case, the side opposite our $30^{\circ}$ angle is , so

We also now know that

$x\sqrt{3}=6\sqrt{3}$

Now we know all of our missing side lengths. The right and left side of the parallelogram will each be . The bottom and top will each be $20+6\sqrt{3}$ . Let's combine them to find the perimeter:

$2\cdot (12)+2\cdot (20+6\sqrt{3})$

$24+40+12\sqrt{3}$

$64+12\sqrt{3}$

Compare your answer with the correct one above

Question

Note: Figure NOT drawn to scale.

Give the perimeter of Parallelogram in the above diagram.

Answer

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

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

Compare your answer with the correct one above

Question

Note: Figure NOT drawn to scale.

Give the perimeter of Parallelogram in the above diagram.

Answer

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

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

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

Compare your answer with the correct one above

Question

Note: Figure NOT drawn to scale.

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

Answer

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

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

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

$\approx 6.83 + 16.16 + 6.83 +16.16$

$\approx 46.0$

Compare your answer with the correct one above

Question

Note: Figure NOT drawn to scale.

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

Answer

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

$\frac{CD}{BD} = \frac{1}{\tan 65^{\circ}}$

$CD= \frac{BD}{\tan 65^{\circ}} \approx \frac{10}{2.1445} \approx 4.66$

$\frac{BD}{BC} = \sin 65^{\circ}$

$\frac{BC}{BD} = \frac{1}{\sin 65^{\circ}}$

$BC= \frac{BD}{\sin 65^{\circ}} \approx \frac{10}{0.9063} \approx 11.03$

The perimeter of the parallelogram is

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

Compare your answer with the correct one above

Question

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

Answer

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

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

Compare your answer with the correct one above

Question

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

Answer

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

$A = BD \cdot CD$

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

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

$\approx 10+ 14.14 + 10 + 14.14$

$\approx 48.3$

Compare your answer with the correct one above

Question

A parallelogram, with dimensions in cm, is shown below.

What is the perimeter of the parallelogram, in cm?

Answer

The triangle on the left side of the figure has a $30^{\circ}$ and a $90^{\circ}$ angle. Since all of the angles of a triangle must add up to $180^{\circ}$ , we can find the angle measure of the third angle:

$180-120=60^{\circ}$

Our third angle is $60^{\circ}$ and we have a triangle.

A triangle has sides that are in the corresponding ratio of $x-x\sqrt{3}-2x$ . In this case, the side opposite our $30^{\circ}$ angle is , so

We also now know that

$x\sqrt{3}=6\sqrt{3}$

Now we know all of our missing side lengths. The right and left side of the parallelogram will each be . The bottom and top will each be $20+6\sqrt{3}$ . Let's combine them to find the perimeter:

$2\cdot (12)+2\cdot (20+6\sqrt{3})$

$24+40+12\sqrt{3}$

$64+12\sqrt{3}$

Compare your answer with the correct one above

Question

Note: Figure NOT drawn to scale.

Give the perimeter of Parallelogram in the above diagram.

Answer

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

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

Compare your answer with the correct one above

Question

Note: Figure NOT drawn to scale.

Give the perimeter of Parallelogram in the above diagram.

Answer

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

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

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

Compare your answer with the correct one above

Question

Note: Figure NOT drawn to scale.

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

Answer

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

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

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

$\approx 6.83 + 16.16 + 6.83 +16.16$

$\approx 46.0$

Compare your answer with the correct one above

Question

Note: Figure NOT drawn to scale.

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

Answer

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

$\frac{CD}{BD} = \frac{1}{\tan 65^{\circ}}$

$CD= \frac{BD}{\tan 65^{\circ}} \approx \frac{10}{2.1445} \approx 4.66$

$\frac{BD}{BC} = \sin 65^{\circ}$

$\frac{BC}{BD} = \frac{1}{\sin 65^{\circ}}$

$BC= \frac{BD}{\sin 65^{\circ}} \approx \frac{10}{0.9063} \approx 11.03$

The perimeter of the parallelogram is

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

Compare your answer with the correct one above

Question

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

Answer

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

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

Compare your answer with the correct one above

Tap the card to reveal the answer