1. How to find the area of a parallelogram and the area of a triangle. Chapter 10.1GeometryStandard/Goal 2.2

2. 1. Check and discuss the assignment from yesterday. 2. Work on Quiz 6.1 and 1.9 3. Read, write, and discuss how to find the area of a parallelogram. 4. Read, write, and discuss how to find the area of a triangle. 5. Work on assignment.

3. The area of a rectangle is the product of its bases and height. b h

4. The area of a parallelogram is the product of a base and its corresponding height. h b

5. Base of a parallelogram is any of its sides altitude is a segment perpendicular to the line containing that base, drawn from the side opposite the base. height is the length of an altitude.

6. The area of a triangle is half the product of a base and corresponding height. h b

7. Base of a triangle is any of its sides height is the length of an altitude to the line containing that base.

8. Find the area of the parallelogram. A = bh Area of a parallelogram A = 12(8)Substitute 12 for b and 8 for h. A = 96Simplify. The area of the parallelogram is 96 m 2. You are given two sides with lengths 12 m and 10.5 m and an altitude that measures 8 m to the side that measures 12 m. Choose the side with a corresponding height to use as a base. Lesson 10-1

9. A parallelogram has 9-in. and 18-in. sides. The height corresponding to the 9-in. base is 15 in. Find the height corresponding to the 18-in. base. Find the area of the parallelogram using the 9-in. base and its corresponding 15-in. height. A = bh Area of a parallelogram A = 9(15)Substitute 9 for b and 15 for h. A = 135Simplify. The area of the parallelogram is 135 in. 2 Lesson 10-1

10. Use the area 135 in. 2 to find the height to the 18-in. base. The height corresponding to the 18-in. base is 7.5 in. A = bh Area of a parallelogram 135 = 18 h Substitute 135 for A and 18 for b. = h Divide each side by 18. 7.5 = h Simplify. 135 18 Lesson 10-1 (continued)

11. A = 195Simplify. A = bh Area of a triangle 1212 A = (30)(13)Substitute 30 for b and 13 for h. 1212 XYZ has area 195 cm 2. Find the area of XYZ. Lesson 10-1

12. The front of a garage is a square 15 ft on each side with a triangular roof above the square. The height of the triangular roof is 10.6 ft. To the nearest hundred, how much force is exerted by an 80 mi/h wind blowing directly against the front of the garage? Use the formula F = 0.004 Av 2. Draw the front of the garage, and then use the area formulas for rectangles and triangles to find the area of the front of the garage. The total area of the front of the garage is 225 + 79.5 = 304.5 ft 2. Area of the square: bh = 15 2 = 225 ft 2 Area of the triangular roof: bh = (15)(10.6) = 79.5 ft 2 1212 1212 Lesson 10-1

13. (continued) Find the force of the wind against the front of the garage. F = 0.004 Av 2 Use the formula for force. F = 0.004(304.5)(80) 2 Substitute 304.5 for A and 80 for v. An 80 mi/h wind exerts a force of about 7800 lb against the front of the garage. Lesson 10-1 A = 7795.2Simplify. A 7800Round to the nearest hundred.

14. Kennedy, D., Charles, R., Hall, B., Bass, L., Johnson, A. (2009) Geometry Prentice Hall Mathematics. Power Point made by: Robert Orloski Jerome High School.