Question

1. a rectangular field has an area of 2100 square feet. the leng is 50 feet.

a. how wide is the field?
b. the field is divided into 3 rectangles, as shown. write and solve an equation to find x.
c. determine the dimensions of each rectangle.

Explanation:

Part a

Step1: Recall area formula for rectangle

The area of a rectangle is given by \( A = l \times w \), where \( l \) is the length and \( w \) is the width. We know \( A = 2100 \) square feet and \( l = 50 \) feet. We need to find \( w \).

Step2: Solve for width

From \( A = l \times w \), we can rearrange to \( w=\frac{A}{l} \). Substituting the values, we get \( w = \frac{2100}{50} \).

Step3: Calculate the value

\( \frac{2100}{50}=42 \). So the width of the field is 42 feet.

Step1: Analyze the total width

The total width of the field is the sum of the widths of the three rectangles, which is \( x + 3x+2x \). We know from part (a) that the total width is 42 feet. So the equation is \( x + 3x+2x=42 \).

Step2: Simplify the left - hand side

Combining like terms, \( x + 3x+2x=(1 + 3+2)x = 6x \). So the equation becomes \( 6x = 42 \).

Step3: Solve for x

Divide both sides of the equation \( 6x = 42 \) by 6: \( x=\frac{42}{6}=7 \).

Step1: Dimensions of the first rectangle

The length of each rectangle is 50 feet (since they are all part of the same rectangular field with length 50 feet). The width of the first rectangle is \( x \) feet, and we found \( x = 7 \) feet. So the dimensions of the first rectangle are 50 feet (length) and 7 feet (width).

Step2: Dimensions of the second rectangle

The width of the second rectangle is \( 3x \) feet. Substitute \( x = 7 \): \( 3x=3\times7 = 21 \) feet. The length is 50 feet. So the dimensions of the second rectangle are 50 feet (length) and 21 feet (width).

Step3: Dimensions of the third rectangle

The width of the third rectangle is \( 2x \) feet. Substitute \( x = 7 \): \( 2x=2\times7 = 14 \) feet. The length is 50 feet. So the dimensions of the third rectangle are 50 feet (length) and 14 feet (width).

Answer:

The width of the field is 42 feet.

Part b