Question

1. a rectangle has an area of 34.2 square meters. if the width of the rectangle is 4.5 inches, find its length.
2. find the area of a rectangle that has a length of 16 yards and a perimeter of 54 yards.

Explanation:

Problem 13

Step1: Recall the area formula for a rectangle.

The area \( A \) of a rectangle is given by \( A = l \times w \), where \( l \) is the length and \( w \) is the width. We need to find \( l \), so we can rearrange the formula to \( l=\frac{A}{w} \).

Step2: Substitute the given values into the formula.

We know that \( A = 34.2 \) square meters and \( w = 4.5 \) inches. Wait, there is a unit inconsistency here (meters and inches). But assuming we proceed with the numbers as given (maybe a typo in units, but we'll use the numbers), so \( l=\frac{34.2}{4.5} \).

Step3: Perform the division.

\( \frac{34.2}{4.5}=7.6 \).

Step1: Recall the perimeter formula for a rectangle.

The perimeter \( P \) of a rectangle is given by \( P = 2(l + w) \), where \( l \) is the length and \( w \) is the width. We know \( P = 54 \) yards and \( l = 16 \) yards. First, we solve for \( w \).

Step2: Rearrange the perimeter formula to solve for \( w \).

From \( P = 2(l + w) \), divide both sides by 2: \( \frac{P}{2}=l + w \), then \( w=\frac{P}{2}-l \).

Step3: Substitute the values of \( P \) and \( l \) into the formula for \( w \).

\( \frac{54}{2}-16 = 27 - 16=11 \) yards. So the width \( w = 11 \) yards.

Step4: Recall the area formula for a rectangle.

The area \( A=l\times w \).

Step5: Substitute \( l = 16 \) yards and \( w = 11 \) yards into the area formula.

\( A = 16\times11 = 176 \) square yards.

Answer:

The length of the rectangle is \( 7.6 \) (units as per the inconsistent units, but mathematically the value is \( 7.6 \)).

Problem 15