Question

1. the area of a rectangle is 36 square centimeters, and its length is 9 centimeters.

a. what is the width of the rectangle?
b. elsa wants to draw a second rectangle that is the same length but is 3 times as wide. draw and label elsa’s second rectangle.
c. what is the perimeter of elsa’s second rectangle?
lesson 2: solve multiplicative comparison word problems by applying the area and perimeter formulas.

Explanation:

Step1: Find the width of the first rectangle

Use the area formula $A = l\times w$ (where $A$ is area, $l$ is length, $w$ is width). Given $A = 36$ square - centimeters and $l = 9$ centimeters. We can solve for $w$ by $w=\frac{A}{l}$. So $w=\frac{36}{9}=4$ centimeters.

Step2: Find the width of the second rectangle

The second rectangle has the same length as the first one ($l = 9$ centimeters) and is 3 times as wide as the first one. So the width of the second rectangle $w_2=3\times4 = 12$ centimeters.

Step3: Find the perimeter of the second rectangle

Use the perimeter formula $P = 2(l + w)$. Substitute $l = 9$ centimeters and $w_2=12$ centimeters into the formula. $P=2(9 + 12)=2\times21 = 42$ centimeters.

Answer:

a. 4 centimeters
b. (Drawing: Draw a rectangle, label the length as 9 centimeters and the width as 12 centimeters)
c. 42 centimeters