1 Lesson Finding Complex Areas

2 Lesson Finding Complex Areas California Standards: Algebra and Functions 3.1 Use variables in expressions describing geometric quantities (e.g., P = 2 w + 2 l, A = ½ bh, 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

3 Finding Complex Areas Finding the area of a rectangle or a triangle is one thing. Lesson But once you can do that, you can start to find out the areas of some really complicated shapes using those very same techniques. 6 in. 4 in. Area = 6 × 4 = 24 in 2 This is an important idea in math — using what you know about simple situations to find out about more complex ones.

4 Finding 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 Areas Example 1 Lesson Area of large rectangle = 10 × 4 = 40 in 2. 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 in 2. So the total area of the shape is = 56 in 2. b = 10 – 6 = 4 in. And h = 8 – 4 = 4 in. Find the area of this shape. Solution follows…

6 Finding 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

7 Finding Complex Areas Example 2 Lesson Divide the shape into a rectangle and a triangle. Solution Area of rectangle = 9 × 4 = 36 in 2. So the total area of the shape is = 42 in 2. Area of triangle = × 4 × 3 = 6 in Find the area of the shape below. Solution follows…

8 Finding Complex Areas Guided Practice Solution follows… Lesson Find the areas of the shapes below. (4 × 6) + (12 × 6) + (4 × 6) = 120 cm (3 × 4) + (3 × 2) = 18 in 2 (6 × 2) + (0.5 × 4 × 3) = 18 cm 2 (8 × 6) + (8 × 18) + (8 × 6) = 240 in 2

9 Finding Complex Areas Complex Areas Can Involve Variables Sometimes you have to use variables for the unknown lengths. Lesson But you can write an expression in just the same way.

10 Finding Complex Areas Example 3 Lesson 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…

11 Finding Complex Areas Guided Practice Solution follows… Lesson Find the areas of the shapes below. bc + ab + bc = ab + 2 bc ( x × 2 x ) + ( x × x ) = 3 x 2 ab + 3 ab + ab = 5 ab xy + ab 1 2

12 Finding 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 –=

13 p q q q Finding Complex Areas Example 4 Lesson 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 × 2 q = 2 pq. Area of gray square = q × q = q 2. So area of original shape = 2 pq – q 2. 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.

14 Finding Complex Areas Guided Practice Solution follows… Lesson Use subtraction to find the areas of the shapes below. (16 × 10) – (6 × 5) = 130 in 2 2 ac – bc 10.

15 Finding Complex Areas Solution follows… Lesson Find the areas of the shapes below cm 2 90 cm 2 31 in 2 8 ab Independent Practice

16 Round Up Remember that it doesn’t matter whether your lengths are numbers or variables — you treat the problems in exactly the same way. Lesson Finding Complex Areas That’s one of the most important things to learn in algebra.