Find the length of a rectangle lot with a perimeter of 82 meters if the length is 7 meters more then the width?

2 Answers
May 8, 2018

24 meters

Explanation:

If the short side of the rectangle = s, then the long side would be s+7.

The total perimeter then would be the sum of all 4 sides, where the sids are equal in pairs, i.e.
Perimeter = s+(s+7)+s+(s+7) = 4s+14 = 82
Therefore 4s=82-14 = 68
so s=17
As this is the short side, the long side is s+7=17+7=24
The length of the rectangle, therefore, is 24 meters

May 8, 2018

The length of the rectangle lot is #color(red)(24)# meters.

Explanation:

To start, I'll set up an equation to help solve this...

#2l+2w=p#, meaning the length twice plus the width twice equals the perimeter. We can fill this in with the known values:

#2l+2(l-7)=82#

#w# can be replaced by #(l-7)# because the problem states that the width is #7# meters shorter than the length.

Distribute the #2# to the #(l-7)# to get:

#2l+2l-14=82#

And combine like terms to get:

#4l=96#

Finally divide both sides by #4# to finish with:

#l=24#, meaning the length is #color(red)(24)# meters.

I'll check my work by plugging this info back in to the original equation to see if it works:

#2l+2w=p#
#2l+2w=82#
#2(24)+2(24-7)=82#
#2(24)+2(17)=82#
#color(green)(48+34=82)#

The equation works, meaning the length is #24# m and the width is #17# m.