Help :o

classic Classic list List threaded Threaded
2 messages Options
BonafideLove95 BonafideLove95
Reply | Threaded
Open this post in threaded view
|

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?
AlgebraHouse AlgebraHouse
Reply | Threaded
Open this post in threaded view
|

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