SOLUTION: The 1st problem is: The perimeter of a rectangle is 860 cm. The length is 4 times the width. Find the length.
The 2nd problems is: The perimeter of a rectangle is 146 cm. The wi
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Question 757671: The 1st problem is: The perimeter of a rectangle is 860 cm. The length is 4 times the width. Find the length.
The 2nd problems is: The perimeter of a rectangle is 146 cm. The width is 10 cm longer than TWICE the length. Find the width.
I am totally stumped on these problems it is just confusing.
Thank you for your help.
Answer by nerdybill(7384) (Show Source): You can put this solution on YOUR website!
The 1st problem is: The perimeter of a rectangle is 860 cm. The length is 4 times the width. Find the length.
Let x = length of rectangle
then
4x = width of rectangle
.
from: "The perimeter of a rectangle is 860 cm" we get our equation:
2(4x + x) = 860
2(5x) = 860
10x = 860
x = 86 cm (length)
.
The 2nd problems is: The perimeter of a rectangle is 146 cm. The width is 10 cm longer than TWICE the length. Find the width.
Let x = length
then from " width is 10 cm longer than TWICE the length"
2x+10 = width
.
from:" The perimeter of a rectangle is 146 cm"
2(2x+10 + x) = 146
2(3x+10) = 146
6x+20 = 146
6x = 126
x = 21 cm (length)
width:
2x+10 = 2(21)+10 = 42+10 = 52 cm (width)
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