Question

An envelope is 6 in. wide and 4 in. high. Will a card with an 8 in. diagonal fit in it? Identify the correct explanation. A) Because 7.2 < 8, it is possible that the card fits in the envelope. B) Because 8.9 > 8, it is possible that the card fits in the envelope. C) Because 7.2 < 8, it is NOT possible that the card fits in the envelope. D) Because 8.9 > 8, it is NOT possible that the card fits in the envelope.

Answer 1

Using the Pythagorean theorem, we calculate the envelope's maximum diagonal to be about 7.2 inches, which is less than the 8-inch diagonal of the card. Hence, the card cannot fit inside the envelope.

Step-by-step explanation:

To determine if a card with an 8 in. diagonal will fit inside an envelope that is 6 in. wide and 4 in. high, we can use the Pythagorean theorem (Diagonal of a rectangle = √(width² + height²)).

First, we calculate the maximum diagonal that can fit in the envelope:

√(6² + 4²) = √(36 + 16) = √52 ≈ 7.2 inches.

Since the card's diagonal is 8 inches, which is greater than the envelope's maximum diagonal length of approximately 7.2 inches, the card will not fit. Therefore, the correct explanation is C) Because 7.2 < 8, it is NOT possible that the card fits in the envelope.

Answer 2

To determine if a card with an 8 in. diagonal will fit in an envelope that is 6 in. wide and 4 in. high, we can use the Pythagorean theorem to calculate the diagonal. The diagonal is found to be approximately 7.21, which is less than 8. Therefore, the card will fit in the envelope.

Step-by-step explanation:

To determine if a card with an 8 in. diagonal will fit in an envelope that is 6 in. wide and 4 in. high, we can use the Pythagorean theorem. The Pythagorean theorem states that in a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides. In this case, the envelope can be considered as the rectangle and the diagonal of the card as the hypotenuse of a right triangle. So the diagonal can be calculated as √(6^2 + 4^2) = √(36 + 16) = √52.

Since the square root of 52 is approximately 7.21, which is less than 8, we can conclude that the card with an 8 in. diagonal will fit in the envelope.

Therefore, the correct explanation is: Because 7.2 < 8, it is possible that the card fits in the envelope. (A)