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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
+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
+2

So would it be

2a+2a=8=80

Oct 24, 2019
#4
+2

Notreallymath has asked if this is correct:

2a+2a+8=80

No, Not quite. --------

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