The length of a rectangle is
2

yd
longer than its width.
If the perimeter of the rectangle is
32

yd
, find its area.

Question

The length of a rectangle is 
2
 
yd
 longer than its width. 
If the perimeter of the rectangle is 
32
 
yd
, find its area.

anonymous · Answer

does anybody know how to solve problems like this?

dayakar · Answer

can u draw a rough figure

anonymous · Answer

i think its 16 because i divided the perimeter by 4 and times it by 2

dayakar · Answer

can u show ur work

quantummechanics · Answer

Hello, so lets let $w = width$ and $l = length$ we have that the length is 2 yards longer than its width so we can represent our length as $$l = w+2$$ and the perimeter is the distance around a 2d shape, since it's a rectangle we then have $$P = 2w+2l$$ with this you may find the length by plugging in the equation $$l=w+2$$ for length in the perimeter equation and solve for the width then you may find the width after and apply the area of a rectangle which is$$A = lw$$

anonymous · Answer

so its 62?

dayakar · Answer

no

quantummechanics · Answer

Please use the steps I've shown