SOLUTION: We are stumped, I am trying to help my daughter and I have no clue help please.... The length of a rectangle is 1 inch more than twice its width. The value of the area of the r

Algebra ->  Rectangles -> SOLUTION: We are stumped, I am trying to help my daughter and I have no clue help please.... The length of a rectangle is 1 inch more than twice its width. The value of the area of the r      Log On


   



Question 1018183: We are stumped, I am trying to help my daughter and I have no clue help please....
The length of a rectangle is 1 inch more than twice its width. The value of the area of the rectangle (in square inches) is 5 more than the value of the perimeter of the rectangle (in inches). find the width.

Found 2 solutions by solver91311, josmiceli:
Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!


You are given that

The perimeter of a rectangle is . Substituting for you get the perimeter as a function of the width:

The area of a rectangle is . Substituting for you get the area as a function of the width:

Since the value of the area is 5 greater than the perimeter, you can say:



which simplifies to:



Solve the factorable quadratic for the positive root. Then calculate .

John

My calculator said it, I believe it, that settles it

Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
They are doing something a little bit illegal here
and you should be aware of it. They are saying
an area ( in square inches ) is equal to a perimeter
( in inches ).
You can only do this if you say the VALUES are equal,
but you can't make an equation and say
[ inches ] = [ square inches ]
--------------------------------
Let the length of the rectangle = +L+
Let the width of the rectangle = +W+
(1) +L+=+2W+%2B+1+
------------------
Let the area of the rectangle = +A+
Let the perimeter of the rectangle = +P+
+A+=+L%2AW+
and
+P+=+2L+%2B+2W+
----------------
They are saying:
+A+=+P+%2B+5+ ( even though the units don't match )
(2) +L%2AW+=+2L+%2B+2W+%2B+5+
Substitute (1) into (2)
(2) +%28+2W+%2B+1+%29%2AW+=+2%2A%28+2W+%2B+1+%29+%2B+2W+%2B+5+
(2) +2W%5E2+%2B+W+=+4W+%2B+2+%2B+2W+%2B+5+
(2) +2W%5E2+%2B+W+=+6W+%2B7+
(2) +2W%5E2+-+5W+-+7+=+0+
Use the quadratic formula to solve
+W+=+%28+-b+%2B-+sqrt%28+b%5E2+-+4%2Aa%2Ac+%29%29+%2F+%282%2Aa%29+
+a+=+2+
+b+=+-5+
+c+=+-7+
+W+=+%28+-%28-5%29+%2B-+sqrt%28+%28-5%29%5E2+-+4%2A2%2A%28-7%29+%29%29+%2F+%282%2A2%29+
+W+=+%28+5+%2B-+sqrt%28+25+%2B+56+%29%29+%2F4+
+W+=+%28+5+%2B-+sqrt%28+81+%29%29+%2F4+
+W+=+5%2F4+%2B+9%2F4+
+W+=+14%2F4+
+W+=+7%2F2+ ( note that I can't use the negative square root )
The width is 3.5 in
-----------------
check:
(1) +L+=+2W+%2B+1+
(1) +L+=+2%2A%287%2F2%29+%2B+1+
(1) +L+-+8+
and
(2) +L%2AW+=+2L+%2B+2W+%2B+5+
(2) +L%2A%287%2F2%29+=+2L+%2B+2%2A%287%2F2%29+%2B+5+
(2) +%287%2F2%29%2AL+=+2L+%2B+12+
(2) +%287%2F2%29%2AL+-+%284%2F2%29%2AL+=+12+
(2) +%283%2F2%29%2AL+=+12+
(2) +L+=+%282%2F3%29%2A12+
(2) +L+=+8+
OK