How to find the perimeter of a parallelogram

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

Questions 1 - 4

1

Parallelogram1

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

$32+ 16 \sqrt{2}$

$16+ 16 \sqrt{2}$

$32+ 32\sqrt{2}$

$48\sqrt{2}$

Explanation

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

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

2

Parallelogram2

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

$180\sqrt{2}$

$360\sqrt{2}$

$180+180\sqrt{2}$

$240+120\sqrt{2}$

Explanation

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

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

3

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

$28\text{in}$

$48\text{in}$

$36\text{in}$

$24\text{in}$

$32\text{in}$

Explanation

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

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

$P = 28\text{in}$

4

Parallelogram1

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

$2 \sqrt{X} + 2 \sqrt{2X}$

$4 \sqrt{2X}$

$4 \sqrt{X }$

$\sqrt{6X}$

$2 \sqrt{3X}$

Explanation

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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