McGraw-Hill Education GRE 2019

In a certain rectangle, the length is double the width. If the area of the

rectangle is 72, what is the length of the rectangle?

SOLUTION: Let w = the width of the rectangle. The length therefore equals 2w.

The area of the rectangle is lw = 2w × w = 2w^2 = 72. Thus

w^2 = 36

w = 6

2 w = 2(6) = 12

Maximizing the Area of a Polygon

Some tougher GRE questions will give you the perimeter of a parallelogram or

triangle and ask you for the maximum area. Given a fixed perimeter, the maximum

area of any polygon will be reached when all sides are equal.

Maximum Area of a Rectangle

To maximize the area of a rectangle, you should make the side lengths equal.

Notice that when you do so, you create a square!

What is the maximum area of a rectangle with a perimeter of 28?

SOLUTION: To maximize the area of the rectangle, make all the sides equal.

Let x = the length of one side. The perimeter is thus

4 x = 28

x = 7

If x = 7, the area is 7^2 = 49.

Maximizing Area with Given Side Lengths

In the preceding examples, you used the perimeter of the shapes to determine

the side lengths that would maximize the area. Sometimes you will be given the

lengths of the two sides of a parallelogram or triangle and asked to determine the

maximum area from this information. In these situations, use the following rule:

Given the lengths of two sides of a triangle or parallelogram, you can maximize the

area by making the two sides perpendicular. In the case of a parallelogram, this

means creating a rectangle. In the case of a triangle, this means creating a right

triangle.

CHAPTER 13 ■ GEOMETRY 399

04-GRE-Test-2018_313-462.indd 399 12/05/17 12:04 pm