Use the image to answer the question.

An illustration shows a rectangle of length as whole number 6 start fraction 1 over 2 end fraction centimeters and width as whole number 4 start fraction 1 over 2 end fraction centimeters. The lengths are horizontally positioned and widths are vertically positioned.

Byron calculated the area of the rectangle as 22 square centimeters. What is the error in his reasoning and what is the correct answer?

(1 point)
Responses

Byron found the perimeter of the rectangle. The correct area is 2914
square centimeters.
Byron found the perimeter of the rectangle. The correct area is 29 Start Fraction 1 over 4 End Fraction square centimeters.

Byron found the perimeter of the rectangle. The correct area is 2414
square centimeters.
Byron found the perimeter of the rectangle. The correct area is 24 Start Fraction 1 over 4 End Fraction square centimeters.

Byron found the perimeter of the rectangle. The correct area is 22 square centimeters.
Byron found the perimeter of the rectangle. The correct area is 22 square centimeters.

Byron found the perimeter of the rectangle. The correct area is 11 square centimeters.

1 answer

To find the correct area of the rectangle, we can use the formula for the area of a rectangle, which is:

\[ \text{Area} = \text{Length} \times \text{Width} \]

The length of the rectangle is \( 6 \frac{1}{2} \) cm, which can be converted to an improper fraction:

\[ 6 \frac{1}{2} = 6 + \frac{1}{2} = \frac{12}{2} + \frac{1}{2} = \frac{13}{2} \text{ cm} \]

The width of the rectangle is \( 4 \frac{1}{2} \) cm, which can also be converted to an improper fraction:

\[ 4 \frac{1}{2} = 4 + \frac{1}{2} = \frac{8}{2} + \frac{1}{2} = \frac{9}{2} \text{ cm} \]

Now we can calculate the area:

\[ \text{Area} = \frac{13}{2} \times \frac{9}{2} = \frac{13 \times 9}{2 \times 2} = \frac{117}{4} \text{ square centimeters} \]

Now converting \(\frac{117}{4}\) to a mixed number gives:

\[ \frac{117}{4} = 29 \frac{1}{4} \]

Byron calculated the area as 22 square centimeters, which is incorrect. The error in his reasoning is that he did not correctly calculate the multiplication of the dimensions of the rectangle.

Thus, the correct answer to the problem is:

Byron found the perimeter of the rectangle. The correct area is \( 29 \frac{1}{4} \) square centimeters.