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 left):
the dark 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: Calculate the area of the light rectangle (on the left)

The formula for the area of a rectangle is \(A = length\times width\). For the light rectangle, length \(= 7\) and width \(=x\). So, \(A_{light}=7\times x = 7x\)

Step2: Calculate the area of the dark rectangle (on the right)

For the dark rectangle, length \(=7\) and width \(=6\). Using the area formula \(A = length\times width\), we get \(A_{dark}=7\times6=42\)

Step3: Calculate the area as a sum of two areas

The total area \(A_{total}\) as a sum is \(A_{light}+A_{dark}\). Substituting the values from Step1 and Step2, we have \(A_{total}=7x + 42\)

Step4: Calculate the area as a product of length and width

The total length of the combined rectangle is \((x + 6)\) and the width is \(7\). Using the area formula \(A=length\times width\), we get \(A_{total}=7\times(x + 6)=7x+42\)

Answer:

• The light rectangle (on the left): \(7x\)
• The dark rectangle (on the right): \(42\)
• As a sum of two areas: \(7x + 42\)
• As a product of the length and width: \(7(x + 6)\)