Help Math problems!!! test tommrow!!?

T

Tom

New member
1) you have the units confused. Perimeter is units of length, area is units of length squared. Assuming the perimeter is 60 inches (not squared) then:
L = 18
2L+2W = 60
2(18) + 2W = 60
W = (60 - 36)/2 = 12

2) L = W+26
2L + 2W = 136
2(W+26) + 2W = 136
4W = 136 - 2*26 = 136 -52 = 84
W=84/4 = 21
 
1) you have the units confused. Perimeter is units of length, area is units of length squared. Assuming the perimeter is 60 inches (not squared) then:
L = 18
2L+2W = 60
2(18) + 2W = 60
W = (60 - 36)/2 = 12

2) L = W+26
2L + 2W = 136
2(W+26) + 2W = 136
4W = 136 - 2*26 = 136 -52 = 84
W=84/4 = 21
 
Rectangle and width problems:( Algedra 1)

A rectangle has a length of 18 inches. If the rectangle has a perimeter of 60 in^2, what is the width of the rectangle?

A rectangular plot of land is measured to have a perimeter of 136 ft^2. If the land is known to have a length that is 26 feet greater than the width, what is the width of the rectangular plot?
 
I will assume you mean 60 in. instead of 60 in^2 since perimeter is a measure of distance not ares.

two sides are 18, the other two sides are equal but we don't know what they are.

So 18 + 18 + 2 other sides = 60

36 + 2 sides = 60

2 sides = 60 - 36 = 24

1 side = half of 24 = 12

So the width is 12

Next problem

Let A = small side (width)
So big side (length) = A + 26

A rectangle has 4 sides (in this case, 2 big, 2 small)

So...

A + A + A +26 + A + 26 = 136

4 * A + 52 = 136

4 * A = 84

A = 84 / 4 = 21

So, the width is 21 feet and the length is 21 + 26 = 47 feet.

hope that helps
 
a "perimeter" CANNOT be in^2. it is a line around the outside, like a fence. Area has ^2 (or square) units, which are like square tiles on a floor.

So, I don't know if 60 is the area or the perimeter.

Check out your problem and post it again.
 
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