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

The base length of a parallelogram is 10 inches and the side length is 6 inches. Give the perimeter of the parallelogram.

Explanation

Like any polygon, the perimeter of a parallelogram is the total distance around the outside, which can be found by adding together the length of each side. In case of a parallelogram, each pair of opposite sides is the same length, so the perimeter is twice the base plus twice the side length. Or as a formula we can write:

Where:

is the base length of the parallelogram and is the side length. So we can write:

The side length of a parallelogram is and the base length is three times more than side length. Give the perimeter of the parallelogram in terms of .

Explanation

The base length is three times more than the side length, so we have:

Base length

The perimeter of a parallelogram is:

Where:

is the base length of the parallelogram and is the side length. So we get:

$Perimeter=2(w+h)=2\left [ (3t^2+3)+(t^2+1) \right ]$

The base length of a parallelogram is identical to its side length. If the perimeter of the parallelogram is 40, give the base length.

Explanation

The perimeter of a parallelogram is:

Where:

is the base length of the parallelogram and is the side length. In this problem the base length and side length are identical, that means:

So we can write:

$Perimeter=40=2(w+w)\Rightarrow 20=2w\Rightarrow w=10$

The base length of a parallelogram is and the side length is . Give the perimeter of the parallelogram in terms of and calculate it for .

Explanation

The perimeter of a parallelogram is:

where:

is the base length of the parallelogram and is the side length. So we have:

$Perimeter=2(w+h)=2\left [ (2t+3)+(2t-3) \right ]=2(4t)=8t$

and:

$t=1\Rightarrow 8t=8\times 1=8$

Parallelogram2

Give the perimeter of the above parallelogram if .

$40+20\sqrt{3}$

$40+40\sqrt{3}$

$40+20\sqrt{2}$

$60\sqrt{2}$

$60\sqrt{3}$

Explanation

By the $30^{\circ}-60^{\circ}-90^{\circ}$ Theorem:

$AB = CD= BD\sqrt{3}= 10\sqrt{3}$ , and

$AD=BC= 2\cdot BD = 2\cdot 10 = 20$

The perimeter of the parallelogram is

$AB + CD + BC + AD = 20+20+ 10\sqrt{3}+ 10\sqrt{3} = 40+20\sqrt{3}$

Parallelogram1

The above parallelogram has area 100. Give its perimeter.

$20+20\sqrt{2}$

$20+20\sqrt{3}$

$40\sqrt{2}$

$40\sqrt{3}$

Explanation

The height of the parallelogram is , and the base is . By the $45^{\circ}-45^{\circ}-90^{\circ}$ Theorem, . Since the product of the height and the base of a parallelogram is its area,