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Lesson 2.3.3Lesson 2.3.3

Finding Complex AreasFinding Complex Areas

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Lesson

2.3.3Finding Complex AreasFinding Complex Areas

California Standards:Algebra and Functions 3.1

Use variables in expressions describing geometric quantities (e.g., P = 2w + 2l, A = ½bh, C = πd — the formulas for the perimeter of a rectangle, the area of a triangle, and the circumference of a circle, respectively).

Mathematical Reasoning 1.3

Determine when and how to break a problem into simpler parts.

What it means for you:You’ll see how you can use the formulas for the areas of rectangles and triangles to find areas of much more complex shapes too.

Key word:• complex shape

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Finding Complex AreasFinding Complex Areas

Finding the area of a rectangle or a triangle is one thing.

Lesson

2.3.3

But once you can do that, you can start to find out theareas of some really complicated shapes using those very same techniques.

6 in.

4 in.

Area = 6 × 4 = 24 in2

This is an important idea in math — using what you know about simple situations to find out about more complex ones.

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Finding Complex AreasFinding Complex Areas

Find Complex Areas by Breaking the Shape Up

There’s no easy formula for working out the area of a shape like this one.

But you can find the area by breaking the shape up into two smaller rectangles.

Lesson

2.3.3

5

b

h

Finding Complex AreasFinding Complex Areas

Example 1

Lesson

2.3.3

Area of large rectangle = 10 × 4 = 40 in2.

Solution

Divide the shape into two rectangles, as shown.

Now you need the dimensions of the small rectangle, b and h.

So the area of the small rectangle = bh = 4 × 4 = 16 in2.

So the total area of the shape is 40 + 16 = 56 in2.

b = 10 – 6 = 4 in. And h = 8 – 4 = 4 in.

Find the area of this shape.

Solution follows…

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Finding Complex AreasFinding Complex Areas

You don’t always have to break a complicated shape down into rectangles.

You just have to break it down into simple shapes that you know how to find the area of.

Lesson

2.3.3

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Finding Complex AreasFinding Complex Areas

Example 2

Lesson

2.3.3

Divide the shape into a rectangle and a triangle.

Solution

Area of rectangle = 9 × 4 = 36 in2.

So the total area of the shape is 36 + 6 = 42 in2.

Area of triangle = × 4 × 3 = 6 in2.1

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Find the area of the shape below.

Solution follows…

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Finding Complex AreasFinding Complex Areas

Guided Practice

Solution follows…

Lesson

2.3.3

1.Find the areas of the shapes below.

(4 × 6) + (12 × 6) + (4 × 6) = 120 cm2

2.

3. 4.(3 × 4) + (3 × 2) = 18 in2 (6 × 2) + (0.5 × 4 × 3) = 18 cm2

(8 × 6) + (8 × 18) + (8 × 6) = 240 in2

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Finding Complex AreasFinding Complex Areas

Complex Areas Can Involve Variables

Sometimes you have to use variables for the unknown lengths.

Lesson

2.3.3

But you can write an expression in just the same way.

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Finding Complex AreasFinding Complex Areas

Example 3

Lesson

2.3.3

Divide the shape into two rectangles.

Solution

Area of the large rectangle = xy.

Area of the small rectangle = ab.

So the total area of the shape is xy + ab.

Find the area of this shape.

Solution follows…

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Finding Complex AreasFinding Complex Areas

Guided Practice

Solution follows…

Lesson

2.3.3

5.

Find the areas of the shapes below.

bc + ab + bc = ab + 2bc

6.

7. 8.(x × 2x) + (x × x) = 3x2

ab + 3ab + ab = 5ab

xy + ab12

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Finding Complex AreasFinding Complex Areas

You Can Subtract Areas As Well

Sometimes it’s easier to find the area of a shape that’s too big, and subtract a smaller area from it.

Lesson

2.3.3

–=

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p

q

q

q

Finding Complex AreasFinding Complex Areas

Example 4

Lesson

2.3.3

This time it’s easier to work out the area of the rectangle with the red outline, and subtract the area of the gray square.

Solution

Area of red rectangle = p × 2q = 2pq.

Area of gray square = q × q = q2.

So area of original shape = 2pq – q2.

This time it’s easier to work out the area of the rectangle with the red outline…

Solution follows…

Calculate the area of the shape below.

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Finding Complex AreasFinding Complex Areas

Guided Practice

Solution follows…

Lesson

2.3.3

9.

Use subtraction to find the areas of the shapes below.

(16 × 10) – (6 × 5) = 130 in2 2ac – bc

10.

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Finding Complex AreasFinding Complex Areas

Solution follows…

Lesson

2.3.3

1.

Find the areas of the shapes below.

2.

3. 4.

42 cm2 90 cm2

31 in2 8ab

Independent Practice

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Round UpRound Up

Remember that it doesn’t matter whether your lengths are numbers or variables — you treat the problems in exactly the same way.

Lesson

2.3.3Finding Complex AreasFinding Complex Areas

That’s one of the most important things to learn in algebra.