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: Calculate area of dark rectangle

The formula for the area of a rectangle is \(A = length\times width\). For the dark rectangle, length is \(x\) and width is \(9\). So the area \(A_{dark}=9x\).

Step2: Calculate area of light rectangle

For the light rectangle, length is \(5\) and width is \(9\). Using the area formula \(A = length\times width\), we get \(A_{light}=9\times5 = 45\).

Step3: Calculate area as sum of two areas

The total area as a sum of two areas is \(A_{sum}=A_{dark}+A_{light}=9x + 45\).

Step4: Calculate area as product of length and width

The total length of the big rectangle is \((x + 5)\) and the width is \(9\). Using the area formula \(A=length\times width\), we have \(A_{product}=9(x + 5)\). Expanding \(9(x + 5)\) using the distributive property \(a(b + c)=ab+ac\) (here \(a = 9\), \(b=x\), \(c = 5\)), we get \(9x+45\).

Answer:

• The dark rectangle: \(9x\)
• The light rectangle: \(45\)
• As a sum of two areas: \(9x + 45\)
• As a product of the length and width: \(9(x + 5)\) (or \(9x+45\))