The perimeter of a rectangular garden is 80 feet, and the area of the

Since perimeter = 2L + 2W, and area = WL, we can create the following two equations:

2L + 2W = 80

L + W = 40

L = 40 - W

WL = 391

We can substitute 40 - W for L in the equation WL = 391:

W(40 - W) = 391

40W - W^2 = 391

W^2 - 40W + 391 = 0

(W - 17)(W - 23) = 0

W = 17 or W = 23

Thus we see that the shorter side of the rectangle is 17.

Alternate Solution:

Since perimeter = 2L + 2W:

2L + 2W = 80

L + W = 40

Since this is the sum of the shorter and longer sides of the rectangle, the shorter side must be less than 20 (if the shorter side is greater than 20 and the longer side is even longer, they would add up to a value greater than 40). The only answer choice that is less than 20 is A.

Answer: A

You could even try process elimination for this. D & E can not be the answer as rectangle has 4 sides.
Let's look at A

If the perimeter was 80 then 80 - 34 = 46

The length of the longer side is 23

17 x 23 = 391

Therefore the answer is A

Given data
2 l + 2 w = 80
l x w= 391

to get the answer quickly

Division of 391 are 17 & 23.... both prime numbers

l and b = 17 & 23
Shorter is 17 : Ans

Perimeter: 80 and Area = 391

=> 2 (L + B) = 80
=> L + B = 40

Possible choices: 23 and 17 [23 + 17 = 40 and 23 * 17 = 391]

Shortest side: 17

Answer A

Hi,

At first I was about to set an equation but I got a headache because of 391 factorization.
We already know that the perimeter is 80 , therefore x+y= 40 (let's say x is the shorter side)
This eliminates options D&E

Then we have another trick here : because our area is an odd number, both x and y must be odd. This eliminates option C.
Now you just have to try with B: if x=23 then y =17 which is impossible x being the shorter side.
We are left with option A

Answer A

Edit: thinking again about it, we could from the first look select A because if x+y=40 and one side is shorter than the other, then x must be less than 20. No need for any further calculation, you can just perform 17*23 to confirm the area.