How to find the perimeter of a parallelogram - ISEE Upper Level Quantitative Reasoning

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Question

The area of the Parallelogram is . Give its perimeter in terms of .

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Answer

The height of the parallelogram is , and the base is . By the 45-45-90 Theorem, . Since the product of the height and the base of a parallelogram is its area,

$BD \cdot CD = A$

$\left (BD \right $)^{2}$ = X$

$BD = $\sqrt{X}$$

Also by the 45-45-90 Theorem,

$CD = AB = BD = $\sqrt{X}$$ , and

$AD = BC = $\sqrt{X}$ \cdot $\sqrt{2}$ = $\sqrt{2X}$$

The perimeter of the parallelogram is

$AB + CD + BC + AD = $\sqrt{X}$ + $\sqrt{X}$ + $\sqrt{2X}$+ $\sqrt{2X}$ = 2 $\sqrt{X}$ + 2 $\sqrt{2X}$$

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Question

Calculate the perimeter of the above parallelogram if $BD = 8$\sqrt{2}$$ .

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Answer

By the 45-45-90 Theorem,

$AB= CD = BD = 8$\sqrt{2}$$

$AD= BC = BD \cdot $\sqrt{2}$ = 8$\sqrt{2}$ \cdot $\sqrt{2}$ = 8 \cdot 2 = 16$

The perimeter of the parallelogram is

$AB + CD + BC + AD = 8 $\sqrt{2}$ + 8 $\sqrt{2}$ + 16+16 = 32+ 16 $\sqrt{2}$$

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Question

Calculate the perimeter of the above parallelogram if $BD = 30$\sqrt{6}$$ .

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Answer

By the 30-60-90 Theorem:

$AB = CD= $\frac{BD}{\sqrt{3}$} = $\frac{30 \sqrt{6}$}{$\sqrt{3}$}= 30$\sqrt{2}$$ , and

$BC = AD = 2\cdot AB = 2 \cdot 30$\sqrt{2}$ =60$\sqrt{2}$$

The perimeter of the parallelogram is

$AB + CD + BC + AD = 30$\sqrt{2}$+ 30$\sqrt{2}$+60$\sqrt{2}$+60$\sqrt{2}$= 180$\sqrt{2}$$

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Question

Find the perimeter of a parallelogram with a base of 6in and a side of length 8in.

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Answer

A parallelogram has 4 sides. A base (where the opposite side is equal) and a side (where the opposite side is equal). So, we will use the following formula:

where b is the base and s is the side of the parallelogram.

We know the base has a length of 6in. We also know the side has a length of 8in.

Knowing this, we can substitute into the formula. We get

$P = 6$\text{in}$ + 6$\text{in}$ + 8$\text{in}$ + 8$\text{in}$$

$P = 28$\text{in}$$

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Question

The area of the Parallelogram is . Give its perimeter in terms of .

Tap to reveal answer

Answer

The height of the parallelogram is , and the base is . By the 45-45-90 Theorem, . Since the product of the height and the base of a parallelogram is its area,

$BD \cdot CD = A$

$\left (BD \right $)^{2}$ = X$

$BD = $\sqrt{X}$$

Also by the 45-45-90 Theorem,

$CD = AB = BD = $\sqrt{X}$$ , and

$AD = BC = $\sqrt{X}$ \cdot $\sqrt{2}$ = $\sqrt{2X}$$

The perimeter of the parallelogram is

$AB + CD + BC + AD = $\sqrt{X}$ + $\sqrt{X}$ + $\sqrt{2X}$+ $\sqrt{2X}$ = 2 $\sqrt{X}$ + 2 $\sqrt{2X}$$

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Question

Calculate the perimeter of the above parallelogram if $BD = 8$\sqrt{2}$$ .

Tap to reveal answer

Answer

By the 45-45-90 Theorem,

$AB= CD = BD = 8$\sqrt{2}$$

$AD= BC = BD \cdot $\sqrt{2}$ = 8$\sqrt{2}$ \cdot $\sqrt{2}$ = 8 \cdot 2 = 16$

The perimeter of the parallelogram is

$AB + CD + BC + AD = 8 $\sqrt{2}$ + 8 $\sqrt{2}$ + 16+16 = 32+ 16 $\sqrt{2}$$

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Question

Calculate the perimeter of the above parallelogram if $BD = 30$\sqrt{6}$$ .

Tap to reveal answer

Answer

By the 30-60-90 Theorem:

$AB = CD= $\frac{BD}{\sqrt{3}$} = $\frac{30 \sqrt{6}$}{$\sqrt{3}$}= 30$\sqrt{2}$$ , and

$BC = AD = 2\cdot AB = 2 \cdot 30$\sqrt{2}$ =60$\sqrt{2}$$

The perimeter of the parallelogram is

$AB + CD + BC + AD = 30$\sqrt{2}$+ 30$\sqrt{2}$+60$\sqrt{2}$+60$\sqrt{2}$= 180$\sqrt{2}$$

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Question

Find the perimeter of a parallelogram with a base of 6in and a side of length 8in.

Tap to reveal answer

Answer

A parallelogram has 4 sides. A base (where the opposite side is equal) and a side (where the opposite side is equal). So, we will use the following formula:

where b is the base and s is the side of the parallelogram.

We know the base has a length of 6in. We also know the side has a length of 8in.

Knowing this, we can substitute into the formula. We get

$P = 6$\text{in}$ + 6$\text{in}$ + 8$\text{in}$ + 8$\text{in}$$

$P = 28$\text{in}$$

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Question

The area of the Parallelogram is . Give its perimeter in terms of .

Tap to reveal answer

Answer

The height of the parallelogram is , and the base is . By the 45-45-90 Theorem, . Since the product of the height and the base of a parallelogram is its area,

$BD \cdot CD = A$

$\left (BD \right $)^{2}$ = X$

$BD = $\sqrt{X}$$

Also by the 45-45-90 Theorem,

$CD = AB = BD = $\sqrt{X}$$ , and

$AD = BC = $\sqrt{X}$ \cdot $\sqrt{2}$ = $\sqrt{2X}$$

The perimeter of the parallelogram is

$AB + CD + BC + AD = $\sqrt{X}$ + $\sqrt{X}$ + $\sqrt{2X}$+ $\sqrt{2X}$ = 2 $\sqrt{X}$ + 2 $\sqrt{2X}$$

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Question

Calculate the perimeter of the above parallelogram if $BD = 8$\sqrt{2}$$ .

Tap to reveal answer

Answer

By the 45-45-90 Theorem,

$AB= CD = BD = 8$\sqrt{2}$$

$AD= BC = BD \cdot $\sqrt{2}$ = 8$\sqrt{2}$ \cdot $\sqrt{2}$ = 8 \cdot 2 = 16$

The perimeter of the parallelogram is

$AB + CD + BC + AD = 8 $\sqrt{2}$ + 8 $\sqrt{2}$ + 16+16 = 32+ 16 $\sqrt{2}$$

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Question

Calculate the perimeter of the above parallelogram if $BD = 30$\sqrt{6}$$ .

Tap to reveal answer

Answer

By the 30-60-90 Theorem:

$AB = CD= $\frac{BD}{\sqrt{3}$} = $\frac{30 \sqrt{6}$}{$\sqrt{3}$}= 30$\sqrt{2}$$ , and

$BC = AD = 2\cdot AB = 2 \cdot 30$\sqrt{2}$ =60$\sqrt{2}$$

The perimeter of the parallelogram is

$AB + CD + BC + AD = 30$\sqrt{2}$+ 30$\sqrt{2}$+60$\sqrt{2}$+60$\sqrt{2}$= 180$\sqrt{2}$$

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Question

Find the perimeter of a parallelogram with a base of 6in and a side of length 8in.

Tap to reveal answer

Answer

A parallelogram has 4 sides. A base (where the opposite side is equal) and a side (where the opposite side is equal). So, we will use the following formula:

where b is the base and s is the side of the parallelogram.

We know the base has a length of 6in. We also know the side has a length of 8in.

Knowing this, we can substitute into the formula. We get

$P = 6$\text{in}$ + 6$\text{in}$ + 8$\text{in}$ + 8$\text{in}$$

$P = 28$\text{in}$$

← Didn't Know|Knew It →

Question

The area of the Parallelogram is . Give its perimeter in terms of .

Tap to reveal answer

Answer

The height of the parallelogram is , and the base is . By the 45-45-90 Theorem, . Since the product of the height and the base of a parallelogram is its area,

$BD \cdot CD = A$

$\left (BD \right $)^{2}$ = X$

$BD = $\sqrt{X}$$

Also by the 45-45-90 Theorem,

$CD = AB = BD = $\sqrt{X}$$ , and

$AD = BC = $\sqrt{X}$ \cdot $\sqrt{2}$ = $\sqrt{2X}$$

The perimeter of the parallelogram is

$AB + CD + BC + AD = $\sqrt{X}$ + $\sqrt{X}$ + $\sqrt{2X}$+ $\sqrt{2X}$ = 2 $\sqrt{X}$ + 2 $\sqrt{2X}$$

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Question

Calculate the perimeter of the above parallelogram if $BD = 8$\sqrt{2}$$ .

Tap to reveal answer

Answer

By the 45-45-90 Theorem,

$AB= CD = BD = 8$\sqrt{2}$$

$AD= BC = BD \cdot $\sqrt{2}$ = 8$\sqrt{2}$ \cdot $\sqrt{2}$ = 8 \cdot 2 = 16$

The perimeter of the parallelogram is

$AB + CD + BC + AD = 8 $\sqrt{2}$ + 8 $\sqrt{2}$ + 16+16 = 32+ 16 $\sqrt{2}$$

← Didn't Know|Knew It →

Question

Calculate the perimeter of the above parallelogram if $BD = 30$\sqrt{6}$$ .

Tap to reveal answer

Answer

By the 30-60-90 Theorem:

$AB = CD= $\frac{BD}{\sqrt{3}$} = $\frac{30 \sqrt{6}$}{$\sqrt{3}$}= 30$\sqrt{2}$$ , and

$BC = AD = 2\cdot AB = 2 \cdot 30$\sqrt{2}$ =60$\sqrt{2}$$

The perimeter of the parallelogram is

$AB + CD + BC + AD = 30$\sqrt{2}$+ 30$\sqrt{2}$+60$\sqrt{2}$+60$\sqrt{2}$= 180$\sqrt{2}$$

← Didn't Know|Knew It →

Question

Find the perimeter of a parallelogram with a base of 6in and a side of length 8in.

Tap to reveal answer

Answer

A parallelogram has 4 sides. A base (where the opposite side is equal) and a side (where the opposite side is equal). So, we will use the following formula:

where b is the base and s is the side of the parallelogram.

We know the base has a length of 6in. We also know the side has a length of 8in.

Knowing this, we can substitute into the formula. We get

$P = 6$\text{in}$ + 6$\text{in}$ + 8$\text{in}$ + 8$\text{in}$$

$P = 28$\text{in}$$

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How to find the perimeter of a parallelogram - ISEE Upper Level Quantitative Reasoning

The area of the Parallelogram is . Give its perimeter in terms of .

The height of the parallelogram is , and the base is . By the 45-45-90 Theorem, . Since the product of the height and the base of a parallelogram is its area, Also by the 45-45-90 Theorem, , and The perimeter of the parallelogram is

Calculate the perimeter of the above parallelogram if .

By the 45-45-90 Theorem, The perimeter of the parallelogram is

Calculate the perimeter of the above parallelogram if .

By the 30-60-90 Theorem: , and The perimeter of the parallelogram is

Find the perimeter of a parallelogram with a base of 6in and a side of length 8in.

The area of the Parallelogram is . Give its perimeter in terms of .

The height of the parallelogram is , and the base is . By the 45-45-90 Theorem, . Since the product of the height and the base of a parallelogram is its area, Also by the 45-45-90 Theorem, , and The perimeter of the parallelogram is

Calculate the perimeter of the above parallelogram if .

By the 45-45-90 Theorem, The perimeter of the parallelogram is

Calculate the perimeter of the above parallelogram if .

By the 30-60-90 Theorem: , and The perimeter of the parallelogram is

Find the perimeter of a parallelogram with a base of 6in and a side of length 8in.

The area of the Parallelogram is . Give its perimeter in terms of .

The height of the parallelogram is , and the base is . By the 45-45-90 Theorem, . Since the product of the height and the base of a parallelogram is its area, Also by the 45-45-90 Theorem, , and The perimeter of the parallelogram is

Calculate the perimeter of the above parallelogram if .

By the 45-45-90 Theorem, The perimeter of the parallelogram is

Calculate the perimeter of the above parallelogram if .

By the 30-60-90 Theorem: , and The perimeter of the parallelogram is

Find the perimeter of a parallelogram with a base of 6in and a side of length 8in.

The area of the Parallelogram is . Give its perimeter in terms of .

The height of the parallelogram is , and the base is . By the 45-45-90 Theorem, . Since the product of the height and the base of a parallelogram is its area, Also by the 45-45-90 Theorem, , and The perimeter of the parallelogram is

Calculate the perimeter of the above parallelogram if .

By the 45-45-90 Theorem, The perimeter of the parallelogram is

Calculate the perimeter of the above parallelogram if .

By the 30-60-90 Theorem: , and The perimeter of the parallelogram is

Find the perimeter of a parallelogram with a base of 6in and a side of length 8in.

Calculate the perimeter of the above parallelogram if $BD = 8$\sqrt{2}$$ .

By the 45-45-90 Theorem,

$AB= CD = BD = 8$\sqrt{2}$$

$AD= BC = BD \cdot $\sqrt{2}$ = 8$\sqrt{2}$ \cdot $\sqrt{2}$ = 8 \cdot 2 = 16$

The perimeter of the parallelogram is

$AB + CD + BC + AD = 8 $\sqrt{2}$ + 8 $\sqrt{2}$ + 16+16 = 32+ 16 $\sqrt{2}$$

Calculate the perimeter of the above parallelogram if $BD = 30$\sqrt{6}$$ .

Calculate the perimeter of the above parallelogram if $BD = 8$\sqrt{2}$$ .

By the 45-45-90 Theorem,

$AB= CD = BD = 8$\sqrt{2}$$

$AD= BC = BD \cdot $\sqrt{2}$ = 8$\sqrt{2}$ \cdot $\sqrt{2}$ = 8 \cdot 2 = 16$

The perimeter of the parallelogram is

$AB + CD + BC + AD = 8 $\sqrt{2}$ + 8 $\sqrt{2}$ + 16+16 = 32+ 16 $\sqrt{2}$$

Calculate the perimeter of the above parallelogram if $BD = 30$\sqrt{6}$$ .

Calculate the perimeter of the above parallelogram if $BD = 8$\sqrt{2}$$ .

By the 45-45-90 Theorem,

$AB= CD = BD = 8$\sqrt{2}$$

$AD= BC = BD \cdot $\sqrt{2}$ = 8$\sqrt{2}$ \cdot $\sqrt{2}$ = 8 \cdot 2 = 16$

The perimeter of the parallelogram is

$AB + CD + BC + AD = 8 $\sqrt{2}$ + 8 $\sqrt{2}$ + 16+16 = 32+ 16 $\sqrt{2}$$

Calculate the perimeter of the above parallelogram if $BD = 30$\sqrt{6}$$ .

Calculate the perimeter of the above parallelogram if $BD = 8$\sqrt{2}$$ .

By the 45-45-90 Theorem,

$AB= CD = BD = 8$\sqrt{2}$$

$AD= BC = BD \cdot $\sqrt{2}$ = 8$\sqrt{2}$ \cdot $\sqrt{2}$ = 8 \cdot 2 = 16$

The perimeter of the parallelogram is

$AB + CD + BC + AD = 8 $\sqrt{2}$ + 8 $\sqrt{2}$ + 16+16 = 32+ 16 $\sqrt{2}$$

Calculate the perimeter of the above parallelogram if $BD = 30$\sqrt{6}$$ .