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Section 10-1 Area of Parallelograms &Triangles

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

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Section 10-1 Area of Parallelograms &Triangles

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

  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 A = bh = 2(3.5) = 7 cm2 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. b = PQ = 5 h = 3 A = bh = 5(3) = 15

  4. Find a missing dimension 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 = 135 Simplify. The area of the parallelogram is 135 in.2 Use the area 135 in.2 to find the height to the 18-in. base.

  5. . . . . . continued Use the area 135 in.2 to find the height to the 18-in. base. A = bhArea of a parallelogram 135 = 18hSubstitute 135 for A and 18 for b. = hDivide each side by 18. 7.5 = hSimplify. 135 18 The height corresponding to the 18-in. base is 7.5 in.

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

  7. 1 2 A = bhArea of a triangle 1 2 A = (30)(13) Substitute 30 for b and 13 for h. XYZ has area 195 cm2. Finding Area of a Triangle Find the area of XYZ. A = 195 Simplify.

  8. 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. Real-world Connection to Area Problems Do It! 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 = 0.004Av2. Area of the square:bh = 152 = 225 ft2 Area of the triangular roof:bh = (15)(10.6) = 79.5 ft2 1 2 1 2 The total area of the front of the garage is 225 + 79.5 = 304.5 ft2.

  9. …. continued Find the force of the wind against the front of the garage. F = 0.004Av2Use the formula for force. F = 0.004(304.5)(80)2Substitute 304.5 for A and 80 for v. A = 7795.2 Simplify. A 7800 Round to the nearest hundred. An 80 mi/h wind exerts a force of about 7800 lb against the front of the garage.

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