Okay, so you made a square card to send to a friend. The card did not fit in the envelope, soo you had to trim the card. You trimmed 4 in. from the length and 5 in. from the width. The area of the resulting card is 20 sq. in.

-What were the original dimensions of the card?

-What was the perimeter of the original card?

-And, what's the difference in the areas of the original and the trimmed cards?

It is a square card, therefore all of the original sides were the same.....x

x = original width

x = original length

x - 4 = new length

x - 5 = new width

Area is length x width

(x - 4)(x - 5) = 20 {resulting area is 20}

x² - 9x + 20 = 20 {used foil method}

x² - 9x = 0 {subtracted 20 from both sides}

x(x - 9) = 0 {factored x out)

x = 0 or x - 9 = 0 {set each factor equal to 0}

x = 9 {x, the original sides, cannot be 0}

**Original dimensions were 9 in. by 9 in.**Perimeter is 2(length) + 2(width)

= 2(9) + 2(9) {perimeter of original card}

= 18 + 18 {multiplied}

= 36 in {added}

**Original perimeter was 36 in.**Difference in areas:

Area of original card = 9 x 9 = 81 in²

Area of new card = 5 x 4 = 20 in²

**Difference in areas = 61 in²***© Algebra House*
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