Lesson

We can think of a composite shape as one that is made from a number of smaller shapes. Many complicated shapes can be made by combining simple shapes like triangles, squares, rectangles, and parallelograms in different ways.

Often of the properties of these simpler shapes can be used to understand the composite shape. For example, knowing the total area of all the smaller shapes is the same as knowing the area of the whole composite shape.

Find the area of the following composite shape.

**Think**: One way to break up this composite shape is to divide it into one triangle and one rectangle, as shown in the figure below. We can then use the formulas for the area of each of these simple shapes to find the area of the whole composite shape.

**Do**: Combining the formula for the area of a triangle and the area of a rectangle, we have

Total area | $=$= | Area of triangle $+$+ Area of rectangle | |

$=$= | $\frac{1}{2}\times\text{base }\times\text{height }+\text{length }\times\text{width }$12×base ×height +length ×width | ||

$=$= | $\frac{1}{2}\times6\times4+6\times5$12×6×4+6×5 | ||

$=$= | $12+30$12+30 | ||

$=$= | $42$42 |

So the area of the composite shape is $42$42 cm^{2}.

**Reflect**: We chose one type of approach to break up the original composite shape, but there are many others. Consider some alternative ways to break up the shape, and think about which ones could lead to simpler calculations of the area.

We may not always initially know the length of every edge of a shape, but we can use the given information to work out missing lengths. This can be useful if a missing length is needed to find the area of a composite shape.

In the figure below, the vertical side on the left has a length of $7$7 m. Since the horizontal sides are parallel, the two vertical sides on the right must have a combined length of $7$7 m as well.

Using this fact, we know that the missing vertical length will add to $4$4 m to give $7$7 m. This number is found by taking the difference between the known vertical lengths, which gives $7-4=3$7−4=3 m.

Now that we know all the side lengths of the figure, we can break it up into two rectangles to find the area. In the figure below we see that the top rectangle has an area of $5\times3=15$5×3=15 m^{2}, and the bottom rectangle has an area of $8\times4=32$8×4=32 m^{2}. So the total figure has an area of $15+32=47$15+32=47 m^{2}.

There will usually be more than one way to break up a composite shape. Some ways may be easier than others, depending on the information that we start with, and whether it is possible to determine initially unknown information.

Find the area of the following composite shape.

**Think**: The figure below shows one way we could break up this shape. We can determine the area of the bottom rectangle using the given dimensions, but we will need to find the length and width of the top rectangle before we can find its area.

**Do**: When we add the length of the top side of the shape to the lengths of the other two shorter horizontal sides, the total will be equal to the length of the bottom horizontal side. This means the width of the top rectangle is given by $12-4-5=3$12−4−5=3 mm. Similarly, the length of the top rectangle is given by $10-2=8$10−2=8 mm.

Now we have enough information to determine the total area.

Total area | $=$= | Area of top rectangle $+$+ Area of bottom rectangle |

$=$= | $8\times3+12\times2$8×3+12×2 | |

$=$= | $24+24$24+24 | |

$=$= | $48$48 |

So the area of the composite shape is $48$48 mm^{2}.

The examples above have explored how adding areas of simple shapes can help us determine the area of a complicated shape. A similar method involves subtracting the area of constituent shapes, and this is particularly useful for composite shapes that have holes and cutaways.

Imagine a rectangular sheet of paper. The area of the paper can be found easily by calculating the product of its length and width. Now picture taking a pair of scissors and cutting a small triangle off one of the corners of the sheet of paper. If we discard the small triangle, what is the area of the remaining paper?

The remaining area is what we get after having taken away the area of the small triangle from the original sheet. That is, we subtract the area of the small triangle from the area of the larger rectangle, and the result is the area of the composite shape that is the remaining paper.

Notice that we could have found the same area by breaking up the composite shape into two rectangles and one triangle, as shown in the figure below. But this method involves adding together three areas, while the method that uses negative area only involves taking the difference between two areas.

Find the area of the following composite shape.

**Think**: We can start with a rectangle that covers the whole figure, then look for the simple shapes we can cut away from this rectangle to get back to the composite shape. These will be the areas we will subtract from the area of the rectangle.

The figure below shows that we will be subtracting the areas of two triangles (triangle B and triangle C), and two parallelograms (parallelogram D and parallelogram E) from the area of one large rectangle (rectangle A).

**Do**: Combining the formulas for the area of each simple shape, we have:

Total area | $=$= | Area of rectangle $-$− Area of triangles $-$− Area of parallelograms |

$=$= | $14\times13-\frac{1}{2}\times6\times13-\frac{1}{2}\times6\times13-2\times5-8\times3$14×13−12×6×13−12×6×13−2×5−8×3 | |

$=$= | $14\times13-6\times13-2\times5-8\times3$14×13−6×13−2×5−8×3 | |

$=$= | $182-78-10-24$182−78−10−24 | |

$=$= | $70$70 |

So the area of the composite shape is $70$70 m^{2}.

**Reflect**: Labeling each constituent shape can be useful to keep track of which shape is being added and which shape is being subtracted.

Find the area of the figure shown.

Find the total area of the figure shown.

Find the shaded area in the figure shown.

Find area of right triangles, other triangles, special quadrilaterals, and polygons by composing into rectangles or decomposing into triangles and other shapes; apply these techniques in the context of solving real-world and mathematical problems.