Question

the following figure is a rectangle made up of two smaller rectangles.
(a) find the area of the following (in square units).
the light rectangle (on the top):
the dark rectangle (on the bottom):
(b) give the area of the entire figure (in square units) in two different ways.
as a sum of two areas:
as a product of the length and width:

Explanation:

Part (a)

Light Rectangle (Top)

Step1: Recall rectangle area formula

The area of a rectangle is given by \( A = \text{length} \times \text{width} \). For the light rectangle, the length is 6 and the width is \( x \).

Step2: Calculate the area

Using the formula, the area \( A_{\text{light}} = 6 \times x = 6x \).

Dark Rectangle (Bottom)

Step1: Recall rectangle area formula

The area of a rectangle is \( A = \text{length} \times \text{width} \). For the dark rectangle, the length is 6 and the width is 7.

Step2: Calculate the area

Using the formula, the area \( A_{\text{dark}} = 6 \times 7 = 42 \).

Part (b)

As a sum of two areas

Step1: Identify the two areas

We have the area of the light rectangle (\( 6x \)) and the area of the dark rectangle (42).

Step2: Sum the areas

The total area \( A_{\text{total}} = 6x + 42 \).

As a product of length and width

Answer:

Step1: Determine the total length and width

The total height of the entire rectangle is \( x + 7 \) (since it's made by stacking the two smaller rectangles) and the length is 6.

Step2: Calculate the area using the formula

Using the rectangle area formula \( A = \text{length} \times \text{width} \), the area \( A_{\text{total}} = 6 \times (x + 7) \). We can also expand this to \( 6x + 42 \) (which matches the sum from the previous method), but the product form is \( 6(x + 7) \).

Final Answers

(a) Light rectangle: \( \boldsymbol{6x} \); Dark rectangle: \( \boldsymbol{42} \)
(b) Sum of two areas: \( \boldsymbol{6x + 42} \); Product of length and width: \( \boldsymbol{6(x + 7)} \) (or \( 6x + 42 \) if expanded, but the product form is \( 6(x + 7) \))