Arectangle has a length of 4 feet and a perimeter of 14 feet. what is the perimeter of a similar rectangle with a width of 9 feet? a. 36 ft b. 108 ft c. 42 ft d. 126 ft

First Rectangle: P= 14; L= 4, W= ?
Second Rectangle: P= ?, L= ?, W= 9

STEP 1:
solve for the width in the first rectangle using perimeter formula

Perimeter= 2L + 2W
14= 2(4) + 2W

14= 8 + 2W
subtract 8 from both sides

6= 2W
divide both sides by 2

3= W

STEP 2:
Since the two rectangles are similar, they may be different sizes, but they have corresponding angles. We can set up a ratio to compare width and length.

First Rectangle: P= 14, L= 4, W= 3
Second Rectangle: L= ?, W= 9

length/width= length/width
4/3= L/9

cross multiply
4 * 9= 3 * L

36= 3L
divide both sides by 3

12= L

STEP 3:
find perimeter of second rectangle; use length from step 2

Perimeter= 2L + 2W
=2(12) + 2(9)
=24 + 18
=42 feet

ANSWER: (C) 42 feet

Hope this helps! :)

c = 9.7

Step-by-step explanation:

Using the law of sines

sin 55       sin 82

=

8                    c

Using cross products

c sin 55 = 8 sin 82

c = 8 sin 82 / sin 55

c=9.67115

To the nearest tenth

c = 9.7