Exercise 39 Page 63

We are given the area of a rectangle and a relationship between its length and width. We are asked to find the perimeter. We can do this in two steps.

We can find the length and width using the given information
Using the length and width, we can find the perimeter.

Let's do this!

Finding the Length and Width

Let's introduce some variables and express the given relationship between length and width using these variables

Information	Expression
Length	l
Width	w
A rectangle's length is twice its width.	l= 2 w

It is also given that the area of the rectangle is 48 square inches. We can use the formula A=l w to find the length and width.

A=l w

SubstituteII

A= 48, l= 2w

48=( 2w)w

▼

Solve for w

Multiply

48=2w^2

DivEqn

.LHS /2.=.RHS /2.

24=w^2

SqrtEqn

sqrt(LHS)=sqrt(RHS)

sqrt(24)=w

RearrangeEqn

Rearrange equation

w=sqrt(24)

The width of the rectangle is sqrt(24) inches. Since we know that the rectangle's length is twice its width, this means that the rectangle is 2 sqrt(24) inches long.

Finding the Perimeter

The formula P=2l+2w relates the length l and the width w of a rectangle to the perimeter P. We now know the length and width, so this formula will help us find the perimeter.

P=2l+2w

SubstituteII

l= 2sqrt(24), w= sqrt(24)

P=2( 2sqrt(24))+2( sqrt(24))

▼

Simplify

CommutativePropMult

Commutative Property of Multiplication

P=(2* 2)sqrt(24)+2sqrt(24)

Multiply

P=4sqrt(24)+2sqrt(24)

FactorOut