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# Geometric Figures Practice Problems

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By McGraw-Hill Professional
Updated on Sep 26, 2011

## Geometric Figures Practice Problems

### Set 1: Introduction to Geometric Figures in Word Problems - Solving for Three Unknowns

To review algebra problems involving geometric figures, go to Geometric Figures Help

#### Practice

1. A box’s width is its length. The perimeter of the box is 40 inches. What are the box’s length and width?
2. A rectangular yard is twice as long as it is wide. The perimeter is 120 feet. What are the yard’s dimensions?

#### Solutions

1. The perimeter of the box is 40 inches, so P = 2 l + 2 w becomes 40 = 2 l + 2 w . The width is its length, and w = (2 l /3), so 40 = 2 l + 2 w becomes 40 = 2 l + 2(2 l /3) = 2 l + (4 l /3).

The length of the box is 12 inches and its width is inches.

2. The perimeter of the yard is 120 feet, so P = 2 l + 2 w becomes 120 = 2 l + 2 w . The length is twice the width, so l = 2 w , and 120 = 2 l + 2 w becomes 120 = 2(2 w ) + 2 w .

The yard’s width is 20 feet and its length is 2 l = 2(20) = 40 feet.

### Set 2: Increasing and Decreasing Dimensions - Rectangles and Squares

To review geometric problems where one or more dimensions changed, go to Geometric Figures Help.

#### Practice

1. A rectangular piece of cardboard starts out with its width being three-fourths its length. Four inches are cut off its length and two inches from its width. The area of the cardboard is 72 square inches smaller than before it was trimmed What was its original length and width?
2. A rectangle’s length is one-and-a-half times its width. The length is increased by 4 inches and its width by 3 inches. The resulting area is 97 square inches more than the original rectangle. What were the original dimensions?

#### Solutions

1. Let l represent the original length and w , the original width. The original area is A = lw . The new length is l – 4 and the new width is w – 2. The new area is then ( l – 4)( w – 2). But the new area is 72 square inches smaller than the original area, so ( l – 4)( w – 2) = A – 72 = lw – 72. So far, we have ( l – 4)( w – 2) = lw – 72.

The original width is three-fourths its length, so l . We will now replace w with

The original length was 16 inches and the original width was inches.

2. Let l represent the original length and w , the original width. The original area is then given by A = lw . The new length is l + 4 and the new width is w + 3. The new area is now ( l + 4)( w + 3). But the new area is also the old area plus 97 square inches, so A + 97 = ( l + 4)( w + 3). But A = lw , so A + 97 becomes lw + 97. We now have

lw + 97 = ( l + 4)( w + 3).

Since the original length is of the original width, . Replace each l by .

The original width is 10 inches and the original length is inches.

### Set 3: Increasing and Decreasing Dimensions - Circles

To review geometric problems where the radius of a circle changed, go to Geometric Figures Help.

#### Practice

A circle’s radius is increased by 5 inches and as a result, its area is increased by 155π square inches. What is the original radius?

#### Solution

Let r represent the original radius. Then r + 5 represents the new radius, and A = πr 2 represents the original area. The new area is 155π square inches more than the original area, so 155π + A = π( r + 5) 2 = 155π + πr 2 .

The original radius is 13 inches.

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