SOLUTION: ABCD is a rectangle. The area of the rectangle is 40 square feet.
The length is 3 more than the width.
A) Find the measure of each length and each width
B) Find the length of
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-> SOLUTION: ABCD is a rectangle. The area of the rectangle is 40 square feet.
The length is 3 more than the width.
A) Find the measure of each length and each width
B) Find the length of
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Question 715224: ABCD is a rectangle. The area of the rectangle is 40 square feet.
The length is 3 more than the width.
A) Find the measure of each length and each width
B) Find the length of the diagonal, rounded to the nearest tenth Answer by mananth(16946) (Show Source):
You can put this solution on YOUR website! Area of rectangle = l*w
Width be x
Length = x+3
x(x+3) =40
x^2+3x=40
x^2+3x-40=0
x^2+8x-5x-40=0
x(x+8)-5(x+8)=0
(x+8)(x-5)=0
Therefore x= -8 OR x=5
Ignore negative
x=5 is the width
width = 5 feet
