The following figure is a rectangle made up of two smaller rectangles.

Question

Top of the rectangle : 6
Side of the light rectangle: x
Side of the dark rectangle: 7
Bottom of the rectangle: 6
(a) Find the area of the following (in square units).
Thelightrectangle(onthe top):_
The dark rectangle(onthe 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: _

Answer 1

To solve the problem, let's first break down the information provided about the rectangle which consists of two smaller rectangles.

Given:

• Width of the entire rectangle (Top and Bottom): 6 units.
• Height of the light rectangle (on top): \(x\) units.
• Height of the dark rectangle (on bottom): 7 units.

(a) Find the area of each rectangle:

1. Area of the light rectangle (top):

   • Width = 6 units
   • Height = \(x\) units
   • Area = Width × Height = \(6 \times x\)

   Area of the light rectangle = \(6x\) square units.

2. Area of the dark rectangle (bottom):

   • Width = 6 units
   • Height = 7 units
   • Area = Width × Height = \(6 \times 7 = 42\)

   Area of the dark rectangle = 42 square units.

(b) Area of the entire figure:

1. As a sum of two areas:

   • Total Area = Area of the light rectangle + Area of the dark rectangle
   • Total Area = \(6x + 42\) square units.

2. As a product of the length and width:

   • The entire figure has a width of 6 units and a total height which is the sum of the heights of the two rectangles.
   • Total Height = \(x + 7\) units.
   • Total Area = Width × Total Height = \(6 \times (x + 7)\)

   Total Area as a product = \(6(x + 7)\) square units.

Summary:

(a)

• Area of the light rectangle: \(6x\) square units.
• Area of the dark rectangle: 42 square units.

(b)

• Total area as a sum: \(6x + 42\) square units.
• Total area as a product: \(6(x + 7)\) square units.