SOLUTION: I've been stuck on this problem since Friday. Help please.
The length of a rectangle is 4 inches shorter than twice the width. If the perimeter of the rectangle is 34 inches,
Question 753059: I've been stuck on this problem since Friday. Help please.
The length of a rectangle is 4 inches shorter than twice the width. If the perimeter of the rectangle is 34 inches, find the length and the width of the triangle.
The answer on the paper is L=10 W=7 Found 2 solutions by DrBeeee, MathTherapy:Answer by DrBeeee(684) (Show Source):
You can put this solution on YOUR website! Let W = the width of the rectangle
Let L = the length of the rectangle
The problem most of my students have is to remember the formula for the perimeter, p. It is
(1) p = 2*L + 2*W, do you have this?
Next write the expression for length as given by the problem statement, namely that the lenght is 4 less than twice the width or
(2) L = 2*W - 4. Do you have that?
Now put L of (2) into (1) and get
(3) p = 2*(2*W - 4) + 2*W
Now set (3) to the perimeter of 34 inches and get
(4) 2*(2*W - 4) + 2*W = 34
Now simplify (4) to obtain
(5) 4*W - 8 + 2*W = 34 or
(6) 6*W = 34 + 8 or
(7) 6*W = 42 or
(8) W = 42/6 or
(9) W = 7
Then from (1) we get
(10) L = 2*7 - 4 or
(11) L = 10
Let's check these values.
Is (2*10 + 2*7 = 34)?
Is (20 + 14 = 34)?
Is (34 = 34)? Yes
Answer: The lenght of the rectangle is 10 inches and the width is 7 inches.
You can put this solution on YOUR website!
I've been stuck on this problem since Friday. Help please.
The length of a rectangle is 4 inches shorter than twice the width. If the perimeter of the rectangle is 34 inches, find the length and the width of the triangle.
The answer on the paper is L=10 W=7
Let width be W
Then length = 2W - 4
Since perimeter = 34, then 2L + 2W = 34 ----- 2(L + W) = 2(17) ----- L + W = 17
Since L = 2W - 4, then L + W = 17 becomes: 2W - 4 + W = 17
3W = 17 + 4
3W = 21
W, or width = , or inches
Substituting 7 for W, L or length = 2(7) - 4, or inches
You can do the check!!
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