Question 293420
You trimmed 4 inches from the length and 5 inches from the width. (a square card)
 The area of the resulting card is 20 square inches.
:
a. what were the original dimensions
Let x = side length of the the original square card
then
(x-4) = one side of the trimmed card
and
(x-5) = the other side
:
Area of resulting card given as 20 sq/in, therefore:
(x-4) * (x-5) = 20
FOIL
x^2 - 5x - 4x + 20 = 20
:
x^2 - 9x + 20 - 20 = 0
:
x^2 - 9x = 0
Factor out x
x(x - 9) = 0
Two solutions
x = 0 (obviously not this one)
x = 9 in by 9 are the dimensions of the original square card
:
:
b. What was the perimeter of the original card?
4(9) = 36 in 
:
:
c. What is the difference in the areas of the original and trimmed cards?
9^2 - 20 = 61 sq inches
:
:
Check solution
Trimmed card dimensions
9 - 5 = 4
9 - 4 = 5
4 by 5 inches
:
trimmed card area: 4 * 5 = 20
:
:
Did I explain this well enough for you to understand it?