Chapter 4.6, Problem 56PS

### Intermediate Algebra

10th Edition
Jerome E. Kaufmann + 1 other
ISBN: 9781285195728

Chapter
Section

### Intermediate Algebra

10th Edition
Jerome E. Kaufmann + 1 other
ISBN: 9781285195728
Textbook Problem

# For Problems 45-56, set up an algebraic equation and solve each problem. (Objective 3)The perimeter of a rectangle is 114 centimeters. If the ratio of its width to its length is 7 to 12, find the dimensions of the rectangle.

To determine

To Find:

The dimensions of the rectangle.

Explanation

Approach:

The comparison of the two numbers by division, we can write in the form of the ratio aÂ toÂ b as ab. The statement of equality between two ratios is known as a proportion. Therefore if abÂ andÂ pq are two same ratios, the proportion ratio can be formed abâ€‰=â€‰pqâ€‰(bâ‰ 0,qâ‰ 0).

This property can be solved as follows.

abâ€‰=â€‰pq,Â (bâ‰ 0,qâ‰ 0)

Multiply bq on both sides.

bq(ab)â€‰=â€‰bq(pq)qa=bp

Calculation:

The perimeter of the rectangle can be expressed as.

perimeter=2(l+w)

Here, the length of the rectangle is l in cm and the width of the rectangle is w in cm

Substitute 114â€‰cm for perimeter in the above expression.

2(l+w)=114â€‰cml+w=114â€‰cm2

Further solving.

l+w=57â€‰cmÂ (1)

For the given rectangle the ratio of width to length given as 512

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