Question

the following figure is a rectangle made up of two smaller rectangles.
(a) find the area of the following (in square units).
the dark rectangle (on the left):
the light rectangle (on the right):
(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:

Step1: Area of dark rectangle

The dark rectangle has length \( x \) and width \( 6 \). The formula for the area of a rectangle is \( \text{Area} = \text{length} \times \text{width} \). So, the area is \( 6 \times x = 6x \).

Step2: Area of light rectangle

The light rectangle has length \( 5 \) and width \( 6 \). Using the area formula for a rectangle, \( \text{Area} = 5 \times 6 = 30 \).

Step3: Sum of two areas (entire figure)

To find the area as the sum of the two smaller rectangles, we add their areas. So, it is \( 6x + 30 \).

Step4: Product of length and width (entire figure)

The entire rectangle has a length of \( x + 5 \) (since it's the sum of the lengths of the two smaller rectangles) and a width of \( 6 \). Using the area formula, \( \text{Area} = 6 \times (x + 5) = 6x + 30 \) (which matches the sum, as expected).

Part (a)

• Dark rectangle: \( 6x \)
• Light rectangle: \( 30 \)

Part (b)

• Sum of two areas: \( 6x + 30 \)
• Product of length and width: \( 6(x + 5) \) (or \( 6x + 30 \) when expanded)

Answer:

(a) Dark rectangle: \( \boldsymbol{6x} \); Light rectangle: \( \boldsymbol{30} \)
(b) Sum of two areas: \( \boldsymbol{6x + 30} \); Product of length and width: \( \boldsymbol{6(x + 5)} \) (or \( \boldsymbol{6x + 30} \))