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The length of a rectangle exceedss its width by 8 inches. The perimeter is 80 inches. Find the length and width of the rectangle

 Oct 24, 2019
 #1
avatar+107022 
+4

Hint,

If you let the width be x units long.  Then how long will the length be?

 

If the perimeter is 80 what will length + width be?

 Oct 24, 2019
 #2
avatar+8 
+2

So would it be 

 

2a+2a=8=80

 Oct 24, 2019
 #4
avatar+107022 
+2

Notreallymath has asked if this is correct:

2a+2a+8=80

 

No, Not quite.  frown

--------

If    a    is the width then

The the length will be      a+8

 

This means that the perimeter will be     a+a+    a+8    +   a+8  = 4a+16

so

4a+16 = 80

 

But what I was getting at is that the perimeter is 2*length + 2*width

so a HALF of the perimeter is  length+ width

so

x + x+8 = 40

2x+8=40

2x     = 32

x = 16

so the width is 16   and the length is ??

 

You can do it either way though.    laugh

Melody  Oct 24, 2019
edited by Melody  Oct 24, 2019
 #3
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+2

Let length=x+8 

Width= x

 

Apply the perimeter formula for rectangle which is: 

2*(Length+Width) 

we already know that length=x+8 and width is x 

so,

2*(x+8+x)= 2*(2x+8) = 4x+16

well the perimeter as the question says is 80 inches

so set,

4x+16=80

subtract 16 from both sides

4x=64

divide by 4

x=16

Well width is x so it is 16

length= 16+8=24 

 

You can make sure by subistuting length and width in the perimeter formula

so

2*(24+16)=80

2*(40)=80

yes indeed these values work.

 Oct 24, 2019

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