1 / 12

9-6

9-6. Area of Irregular Figures. Course 2. Math Minute. Vocabulary. Lesson Presentation. 9-6. Area of Irregular Figures. Course 2. Math Minute Find the area of the following figures. 1. A triangle with a base of 12.4 m and a height of 5 m

salena
Download Presentation

9-6

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. 9-6 Area of Irregular Figures Course 2 Math Minute Vocabulary Lesson Presentation

  2. 9-6 Area of Irregular Figures Course 2 Math Minute Find the area of the following figures. 1.A triangle with a base of 12.4 m and a height of 5 m 2. A parallelogram with a base of 36 in. and a height of 15 in. 3. A square with side lengths of 2. yd 31 m2 540 in2 4 yd2

  3. 9-6 Area of Irregular Figures Course 2 Learn to find the area of irregular figures.

  4. 9-6 Area of Irregular Figures Course 2 You can find the area of an irregular figure by separating it into non-overlapping familiar figures. The sum of the areas of these figures is the area of the irregular figure. You can also estimate the area of an irregular figure by using graph paper.

  5. 9-6 Area of Irregular Figures 2 The area of the figure is about 15 yd . Course 2 Example 1 Estimate the area of the figure. Each square represents one square yard. Count the number of filled or almost-filled squares: 11 red squares. Count the number of squares that are about half-full: 8 green squares. Add the number of filled squares plus ½ the number of half-filled squares: 11 + ( • 8) = 11 + 4 =15. 1 2

  6. 9-6 Area of Irregular Figures 8 yd A = 72 A = 8 •9 Step 1: Separate the figure into smaller, familiar figures. Step 2: Find the area of each smaller figure. Area of the rectangle: A =lw Course 2 Example 2: Finding the Area of an Irregular Figure Find the area of the irregular figure. Use 3.14 for p. 9 yd 9 yd Use the formula for the area of a rectangle. 3 yd Substitute 8 for l. Substitute 9 for w.

  7. 9-6 Area of Irregular Figures 1 1 1 __ __ __ 2 2 2 The area of a triangle is the b•h. 12 8 yd Area of the triangle: A = (18) A = 9 A = bh A = (2•9) Course 2 Continued Find the area of the irregular figure. Use 3.14 for p. 9 yd 9 yd Substitute 2 for b and 9 for h. 2 yd Multiply.

  8. 9-6 Area of Irregular Figures A =bh A = 16 •9 A = 144 Step 1: Separate the figure into smaller, familiar figures. Step 2: Find the area of each smaller figure. Area of the parallelogram: Course 2 Now You Try… Find the area of the irregular figure. Use 3.14 for p. 16 m 9 m 16 m Use the formula for the area of a parallelogram. Substitute 16 for b. Substitute 9 for h.

  9. 9-6 Area of Irregular Figures 1 1 1 __ __ __ 2 2 2 The area of a semicircle is the area of a circle. 12 A ≈ (200.96) A ≈ (3.14•82) A ≈ 100.48 A = (pr) Area of the semicircle: Course 2 Continued Find the area of the irregular figure. Use 3.14 for p. 16 m 9 m 16 m Substitute 3.14 for p and 8 for r. Multiply.

  10. 9-6 Area of Irregular Figures A ≈ 144 + 100.48 = 244.48 The area of the figure is about 244.48 m2. Step 3: Add the area to find the total area. Course 2 Additional Example 2 Continued Find the area of the irregular figure. Use 3.14 for p. 16 m 9 m 16 m

  11. 9-6 Area of Irregular Figures A = 72 + 9 = 81 The area of the figure is about 81 yd2. Step 3: Add the area to find the total area. Course 2 Continued Find the area of the irregular figure. Use 3.14 for p.

  12. 9-6 Area of Irregular Figures Course 2 Example 3: Finding the Shaded Region Find the area of the shaded region Use 3.14 for pi. Step 1: Write the formula. A = area of square – area of circle A = (l x w) – ( πr2) 4 cm Step 2: Substitute values. 2 cm A = (4 x 4) – (3.14 x 22) Step 3: Solve. A = (16) – (3.14 x 4) A = (16) – (12.56) 4 cm A = 3.44 cm2

More Related