120k views
2 votes
You made a square card. The card did not fit in the envelope so you trimmed it. You trimmed 4 inches from the length and 5 inches from the width. The area of the resulting card is 20 square inches. What were the original dimensions for the card?

User JimEvans
by
8.2k points

2 Answers

5 votes

Answer:

The original dimensions of the card were 9 inches length and 9 inches width

Explanation:

Let x be the original length and width of the card in inches (remember that it was squared originally). The exercise says that the area of the new width, with dimensions x-4 and x-5 is 20 square inches, therefore

(x-4)*(x-5) = 20

x²-9x+20 = 20

x²-9x = 0

x*(x-9) = 0

x = 0 or x = 9

Since x must be positive, then it cant be 0, thus, x has to be 9. The original dimensions of the card were 9 inches length and 9 inches width.

User Vinnie
by
8.7k points
2 votes

Answer:

original dimension for the card = 9 inches by 9 inches

Explanation:

Let the length and width of the card be x and x respectively since its a square card and they are both the same length.

Upon trimming, new length and width are (x-4) inches and (x-5) inches respectively.

Area of new card = (x-4)(x-5) = 20

Expanding the brackets to form a quadratic equation

Area of new card = x² - 5x - 4x + 20 = 20 ≡ x² -9x = 0

Hence x² = 9x

and x = 9

Therefore original dimension for the card = 9 inches by 9 inches

User Gustavo Guerra
by
7.9k points

No related questions found

Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories