Exercise 35 Page 547

Let's begin with recalling that the perimeter of a rectangle is twice the sum of the length and width. Since we know that the perimeter is 220 inches, we can write an equation. Let l be the length and w be the width of a rectangle. 2(l+w)=220 Now, we will use the fact that the ratio of the length to the width is 7: 3. Using this, we can express l using the w-term.

l/w=7/3

MultEqn

LHS * w=RHS* w

l=7/3w

With this, we can substitute the value of l into the perimeter equation and solve for w.

2(l+w)=220

Substitute

l= 7/3w

2( 7/3w+w)=220

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Solve for w

NumberToFrac

a = 3* a/3

2(7/3w+3/3w)=220

AddFrac

Add fractions

2(10/3w)=220

DivEqn

.LHS /2.=.RHS /2.

10/3w=110

MoveRightFacToNum

a/c* b = a* b/c

10w/3=110

MultEqn

LHS * 3=RHS* 3

10w=330

DivEqn

.LHS /10.=.RHS /10.

w=33

The width of this rectangle is 33 inches. With this, we can evaluate the length.

l=7/3w

Substitute

w= 33

l=7/3( 33)

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Simplify right-hand side

MoveRightFacToNum

a/c* b = a* b/c

l=7*33/3

ReduceFrac

a/b=.a /3./.b /3.

l=7*11

Multiply

Multiply

l=77

The length of this rectangle is 77 inches. Finally, we will find the area of this rectangle by multiplying its length, 77, and width, 33. 77*33=2541 The area of this rectangle is 2541 square inches.