# Help :o

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## Help :o

 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?
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## Re: Help :o

 This post was updated on . 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