SOLUTION: My teacher just started doing this with me can someone help me out. A rectangle has an area of 240 {cm}^2 and a perimeter of 64 cm. What are its dimensions? Need to find t

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Question 624573: My teacher just started doing this with me can someone help me out.


A rectangle has an area of 240 {cm}^2 and a perimeter of 64 cm. What are its dimensions? Need to find the length and the with
Found 3 solutions by solver91311, ewatrrr, lwsshak3:
Answer by solver91311(24713) About Me  (Show Source):
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The perimeter of a rectangle is given by and the area of a rectangle is given by

First, we know that the perimeter is equal to 64, so we write:







And we know that



Make the subsitution from the perimeter equation:



A little algebra music, Sammy:



Solve the quadratic for , then calculate . Hint: It factors.

John

My calculator said it, I believe it, that settles it
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Answer by ewatrrr(24785) About Me  (Show Source):
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Hi,



A rectangle has an area of 240 {cm}^2 and a perimeter of 64 cm. What are its dimensions

 2%2Al+%2B+2%2Aw+=+64cm

   l+%2B+w+=+32cm  or l+=+32-w

 A = length·width = 240cm^2

     (32-w)w = 240

     32w - w^2 = 240

      w^2 - 32w + 240 = 0

    (w - 12)(w - 20) = 0

       w is 12cm or  20cm.   Dimensions of rectangle are 12cm by 20cm

Answer by lwsshak3(11628) About Me  (Show Source):
You can put this solution on YOUR website!
A rectangle has an area of 240 {cm}^2 and a perimeter of 64 cm. What are its dimensions? Need to find the length and the width
**
let x=length
let y=width
Area of a rectangle=length*width=xy
Perimeter=twice the length+twice the width=2x+2y
64=2x+2y
32=x+y
y=32-x
..
240=x*y=x(32-x)=32x-x^2
x^2-32x+240=0
(x-12)(x-20)=0
x=12(reject, x=length)
or
x=20
32-x=12
Dimensions of rectangle:
length=20 cm
width=12 cm