1. Section 10-1 Area of Parallelograms &Triangles Objectives: find area of a parallelogram and triangle Area = base x height A = bh h b Area = ½ base x height A = ½ bh h b

2. Finding Area of a Parallelogram Base of a parallelogram is any of its sides The corresponding altitude is a segment  to the line containing the base, drawn from the opposite side of the base. The height is the length of the altitude.

3. Find the area of a Parallelogram What is the area of the parallelogram? Find the area of PQRS with vertices P(1, 2), Q(6, 2), R(8, 5) and S(3, 5). 1. Graph the parallelogram. 2. Choose the base. 3. Substitute in A = bh. A = bh = 2(3.5) = 7 cm 2 b = PQ = 5 h = 3 A = bh = 5(3) = 15

4. 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 = bhArea 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 Find a missing dimension Use the area 135 in. 2 to find the height to the 18-in. base.

5. The height corresponding to the 18-in. base is 7.5 in. A = bhArea of a parallelogram 135 = 18hSubstitute 135 for A and 18 for b. = hDivide each side by 18. 7.5 = hSimplify. 135 18..... continued

6. Finding Area of a Triangle Base of a triangle is any of its sides The corresponding height is the length of the altitude to the line containing the base base height Area = ½ bh

7. A = 195Simplify. A = bhArea 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. Finding Area of a Triangle

8. 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 = 0.004Av 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 Real-world Connection to Area Problems

9. Find the force of the wind against the front of the garage. F = 0.004Av 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. A = 7795.2Simplify. A 7800Round to the nearest hundred. …. continued