 # 1 Lesson 3.1.2 Areas of Polygons. 2 Lesson 3.1.2 Areas of Polygons California Standard: Measurement and Geometry 1.2 Use formulas routinely for finding.

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1 Lesson 3.1.2 Areas of Polygons

2 Lesson 3.1.2 Areas of Polygons California Standard: Measurement and Geometry 1.2 Use formulas routinely for finding the perimeter and area of basic two-dimensional figures, and the surface area and volume of basic three-dimensional figures, including rectangles, parallelograms, trapezoids, squares, triangles, circles, prisms, and cylinders. What it means for you: You’ll use formulas to find the areas of regular shapes. Key words: area triangle parallelogram trapezoid formula substitution

3 Areas of Polygons Lesson 3.1.2 Area is the amount of space inside a shape. Like for perimeter, there are formulas for working out the areas of some polygons. You’ll practice using some of them in this Lesson. Area Perimeter

4 Square: A = s 2 s s l w Rectangle: A = lw Parallelogram: A = bh b h Areas of Polygons Area is the Amount of Space Inside a Shape Lesson 3.1.2 Triangles and other shapes can be a little more difficult, but there are formulas for those too — which we’ll come to next. Area is the amount of surface covered by a shape. Parallelograms, rectangles, and squares all have useful formulas for finding their areas.

5 Areas of Polygons Example 1 Solution follows… Lesson 3.1.2 Use a formula to evaluate the area of this shape. Use the formula for the area of a rectangle. Solution Substitute in the values given in the question. A = 14 in 2 7 in 2 in A = lw A = 7 in × 2 in Evaluate the area.

6 Areas of Polygons Example 2 Solution follows… Lesson 3.1.2 You can also rearrange the formulas to find a missing length: Find the height of a parallelogram of area 42 cm 2 and base length 7 cm. Solution Rearrange the formula for the area of a parallelogram, and substitute. A = bh A ÷ b = bh ÷ b = h h = 42 ÷ 7 = 6 cm b h bh Divide both sides by the base ( b ) Substitute values and evaluate

7 Areas of Polygons Guided Practice Solution follows… Lesson 3.1.2 1. Find the area of a square of side 2.4 m. 2. Find the length of a rectangle if it has area 30 in 2, and width 5 in. A = s 2 = 2.4 2 = 5.76 m 2 A = lw l = A ÷ w = 30 ÷ 5 = 6 in

8 Areas of Polygons The Area of a Triangle is Half that of a Parallelogram Lesson 3.1.2 The area of a triangle is half the area of a parallelogram that has the same base length and vertical height. height ( h ) base ( b ) +height ( h ) base ( b )

9 Areas of Polygons Lesson 3.1.2 In math language, the area of a triangle is given by: Area of triangle= area of parallelogram = (base × height) = bh 1 2 1 2 1 2 A = bh 1 2

10 Areas of Polygons Example 3 Solution follows… Lesson 3.1.2 Find the base length of the triangle shown if it has an area of 20 in 2 and a height of 8 in. 8 in b Solution Rearrange the formula for the area to give an expression for the base length of the triangle. A = bh 1 2 2 A = bh 2 A ÷ h = bh ÷ h b = 2 A ÷ h Multiply both sides by 2 Write out the formula Divide both sides by the height ( h ) Now substitute in the values and evaluate to give the base length. b = 2 A ÷ h = 5 in = (2 × 20) ÷ 8 Simplify Solution continues…

11 Areas of Polygons Guided Practice Solution follows… Lesson 3.1.2 3. Find the area of a triangle of base length 3 ft and height 4.5 ft. 4. Find the base length of a triangle with height 50 m and area 400 m 2. A = 0.5(3 × 4.5) = 0.5 13.5 = 6.75 ft 2 b = 2 A ÷ h = (2 400) ÷ 50 = 16 m

12 Areas of Polygons Break a Trapezoid into Parts to Find its Area Lesson 3.1.2 The most straightforward way to find the area of a trapezoid is to split it up into two triangles. Notice that both triangles have the same height but different bases. You then have to work out the area of both triangles and add them to find the total area. 1 2 height ( h ) base of triangle 1 ( b 1 ) base of triangle 2 ( b 2 ) 1 2

13 Areas of Polygons So, the area of the trapezoid is the sum of the areas of each triangle. Lesson 3.1.2 h b1b1 1 2 b2b2 Area of trapezoid= area of triangle 1 + area of triangle 2 = b 1 h + b 2 h 1 2 1 2 Take out the common factor of h to give: 1 2 Area of trapezoid = h ( b 1 + b 2 ) 1 2 A = h ( b 1 + b 2 ) 1 2

14 Areas of Polygons Example 4 Solution follows… Lesson 3.1.2 Find the area of the trapezoid shown. 8 ft 30 ft 12 ft Solution Area of trapezoid = h ( b 1 + b 2 ) 1 2 Substitute in the values given in the question and evaluate. Area of trapezoid = × 8 × (12 + 30) = × 8 × 42 = 168 ft 2. 1 2 1 2

15 Find the areas of the trapezoids in Exercises 5–8, using the formula. 5.6. 7.8. Areas of Polygons Guided Practice Solution follows… Lesson 3.1.2 3 in 10 in 5 in 4 cm 20 cm 11 cm 1.5 m 1.1 m 0.7 m 245 ft 105 ft 80 ft 0.5 3 (5 + 10) =22.5 in 2 0.5 11 (20 + 4) = 132 cm 2 0.5 0.7 (1.1 + 1.5) = 0.91 m 2 0.5 80 (105 + 245) = 14,000 ft 2

16 Areas of Polygons Independent Practice Solution follows… Lesson 3.1.2 Find the area of each of the shapes in Exercises 1–2. 1.2. 1.2 ft 1 m 1.44 ft 2 0.5 m 2

17 Areas of Polygons Independent Practice Solution follows… Lesson 3.1.2 Find the area of each of the shapes in Exercises 3–4. 3.4. 2 in 2.3 in 7 in 2.5 in 2.3 in 2 17.5 in 2

18 Areas of Polygons Independent Practice Solution follows… Lesson 3.1.2 Find the area of each of the shapes in Exercises 5–6. 5.6. 20 cm 11 cm 12 cm 4.5 ft 3.1 ft 186 cm 2 13.95 ft 2

19 Areas of Polygons Independent Practice Solution follows… Lesson 3.1.2 7. Miguel wants to know the area of his flower bed, shown below. Find the area using the correct formula. 3.1 m 2.4 m 3.72 m 2

20 Areas of Polygons Round Up Lesson 3.1.2 Later you’ll use these formulas to find the areas of irregular shapes. Make sure you practice all this stuff so that you’re on track for the next few Lessons.

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