Pearson Geometry Common Core, 2011
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Pearson Geometry Common Core, 2011 View details
1. Areas of Parallelograms and Triangles
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Exercise 43 Page 621

Divide the given figure into recognizable polygons.

12 800m^2

Practice makes perfect

Before we try to find the area of the given figure, notice that the figure can be divided into a rectangle and a triangle.

First, we will find any missing lengths from these more recognizable polygons. Then we can calculate the area of the rectangle and the area of the triangle. Finally, we will add the individual areas together to obtain the total area of the figure.

Missing Lengths

In the diagram, we can see that the length and the width of the rectangle are 120m and 60m, respectively. Let b be the base of the triangle and let h be the height. We can use the Segment Addition Postulate to write two equations in terms of b and h.

Let's solve the equations in the above diagram! c|c 60+ b=200 & h+40= 120 ⇓ & ⇓ b= 140 & h=80 Now that we have the base and the height of the triangle, let's visualize these shapes separately.

Area of the Rectangle

To find the area of the rectangle, we will substitute w= 60 and l= 120 into the formula for the area of a rectangle.
A=l w
A= 120( 60)
A=7200
The area of the rectangle is 7200m^2.

Area of the Triangle

To find the area of the triangle, we will substitute b= 140 and h=80 into the formula for the area of a triangle, and simplify.
A=1/2bh
A=1/2( 140)(80)
â–Ľ
Evaluate right-hand side
A=1/2(11 200)
A=11 200/2
A=5600
The area of the triangle is 5600m^2.

Area of the Figure

We now know the area of the rectangle and the area of the triangle.

To find the area of the figure, we can add these two values. Area of the Figure 7200+5600=12 800m^2