GED Mathematical Reasoning: Combined Figures

  • A combined figure is an irregular shape made up of figures that we are familiar with. Combined figures can be two dimensional or three dimensional. For an example, the two dimensional combined figure showing on the screen is composed of a triangle, a rectangle, and a trapezoid.

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  • For a three dimensional example, here is a combined figure made up of two pyramids.

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  • The general strategy for doing this usually involves a “divide and conquer” approach whereby we divide the figure up into the basic shapes we are familiar with.

 

Example 1

Cara would like to put a wallpaper border around the room with the dimensions shown. How much border will she need?

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In example one, we are being asked to determine how much wallpaper border is needed. Since wall paper border is placed on the wall going around the entire room, we need to calculate the perimeter of the room to solve this problem.

To calculate the perimeter of a combined figure we add the lengths of all the sides.

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In this example, the room has 6 sides but we are given the side measure for only four of those sides. So we’ll need to use the measures provided to find the length of the two missing sides.

To do this, let’s divide the room up into two rectangles. And for reference purposes, let’s name the missing sides a and b.

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Side a equals the length of the opposite wall minus 5 feet. Since 12 minus 5 is equal to 7, side a measures 7 feet.

Side b equals the length of the opposite wall minus 8 feet. Since 15 minus 8 is equal to 7, side b also measures 7 feet.

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Now that we know the measure of all 6 sides, we’re ready to add up all the side measures to find the perimeter of the room and thus the amount of wallpaper border needed.

P = 15 + 5 + 7 + 7 + 8 + 12 = 54ft

Therefore, the perimeter of the room is 54 feet and the amount of wallpaper border needed is 54 feet.

 

Example 2

Find the area of the figure shown.

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The figure shown in Example 2 is a combined figure made up of a rectangle and a triangle – two shapes for which we do know how to find the area.

We can use our knowledge of finding the area of a rectangle and triangle to find the area of the individual pieces that make up this figure. Then, we’ll add those values together to find the total area.

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The area of the rectangle can be found by multiplying its length and width.

Since 20 times 18 is equal to 360, the area of the rectangular piece is 360m^2.

A = L \times W = 20 \times 18 = 360m^2

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Next, our task is to calculate the area of the triangular piece.

Recall that to find the area of a triangle, we use the formula:

\frac{1}{2} \times base \times height

We are given that the height of the triangle is 16 meters, but the length of the base is not obvious. Let’s call the missing portion of the base x.

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Notice that x forms the opposite side of the rectangular piece. And in a rectangle, opposite sides have the same length. That means that x has a length of 18 meters.

If x equals 18 meters, the base of the triangle is equal to 18 plus 8, which is 26 meters.

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Now we have all the information needed to calculate the area of the triangle.

The area of the triangle equals one-half times the base of 26 times the height of 16.

A = \frac{1}{2} \times 26 \times 16
A = 208m^2

So the area of the triangular piece is 208m^2.

Now let’s put this all together.

The area of the rectangular piece is 360m^2 and the area of the triangular piece is 208m^2. Adding these two measures together gives a total area of 568m^2.

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