SOLUTION: You made a square card to send to a friend. The card did not fit in the envelope so you had to trim the card. You trimmed 4 inches from the length and 5 inches from the width. The
Algebra ->
Polynomials-and-rational-expressions
-> SOLUTION: You made a square card to send to a friend. The card did not fit in the envelope so you had to trim the card. You trimmed 4 inches from the length and 5 inches from the width. The
Log On
Question 293420: You made a square card to send to a friend. The card did not fit in the envelope so you had to trim the card. You trimmed 4 inches from the length and 5 inches from the width. The area of the resulting card is 20 square inches.
a. what were the original dimensions
b. What was the perimeter of the original card?
c. What is the difference in the areas of the original and trimmed cards?
could you please also show me how you got the answers. Thank You Found 2 solutions by richwmiller, ankor@dixie-net.com:Answer by richwmiller(17219) (Show Source):
You can put this solution on YOUR website! 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?
