Geometry 11.3 Areas of Trapezoids.

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Presentation transcript:

Geometry 11.3 Areas of Trapezoids

Trapezoids In a trapezoid, the bases are the parallel sides. An altitude of a trapezoid is defined in the same way as an altitude of a parallelogram. The altitude of a trapezoid is any segment perpendicular to a line containing one base from a point on the opposite base. In a trapezoid, all altitudes have the same length, called the height, h. b1 h h h b2

A = ½h(b1+ b2) A = mh Area of a Trapezoid The area of a trapezoid equals half the product of height and the sum of the bases. b1 8 A = ½h(b1+ b2) 6 h or 12 A = mh b2 m is the median, the average of the bases Area = ½(6)(8 + 12) = 60 square units Area = 10(6) = 60 square units

Exercises 1. A = ½(11)(18+8) 3. A = ½(6√3)(30+15) A = 143 A = 135√3 Try # 2 and #4! 10 3 7 30º 8 2 8 11 18 15 12 30 60 45 10 6√3 8 6 2 1. A = ½(11)(18+8) 3. A = ½(6√3)(30+15) A = 143 A = 135√3 2. A = ½(10)(7+3) 4. A = ½(8)(10+2) A = 50 A = 48

Easy method : Find m first A = ½h(b1+ b2) m is the median length of the trapezoid m = ½ (b1+ b2) So, more simply: It is the average of the bases A = h•m

Exercises 5 3 15 6x 8 70 35 14 17.5 11.5 5.5 10x = h•m edian 6. 7. 8. 9. 10. b1 15 25 8 14x b2 13 10 9 h 5 4 7x A 140 46 70x2 m 7 5 3 15 6x 8 = h•m 70 35 edian 14 17.5 11.5 5.5 10x 7. 46 = 4m m = 11.5 11.5 = ½(8 + b2) 23 = 8 + b2 b2 = 15 9. 6√3m = 33√3 m = 5.5 5.5 = ½(b1 + 8) 11 = b1 + 8 b1 = 3 5. m = ½(15 + 13) m = 14 A = 5(14) = 70 6. m = ½(25 + 10) m = 17.5 140 = 17.5 h h = 8 8. 7 = ½(b1 + 9) 14 = b1 + 9 b1 = 5 10. 70x² = 7x • m m = 10x b1 = 6x

Exercises Find the area of each isosceles trapezoid. 5-12-13 Triple x 6 12 24 60 10 30 26 6√3 x 10 10 m = ½(24 + 12) x² + 10² = 26² m = ½(30+ 10) m = 18 x² + 100 = 676 m = 20 x² = 576 A = 6√3 • 18 A = 24 • 20 x = √576 A = 108√3 A = 480 x = 24

Exercises 13. Find the area of an isosceles trapezoid with legs 25 cm and bases 16 cm and 30 cm. 16 7-24-25 Triple 25 25 24 30 - 16 7 30 7 2 m = ½(16 + 30) 552 cm² m = 23 A = 24 • 23 = 552

Exercises 14. Find the area of a trapezoid with 45˚ base angles and bases 17 and 23. 17 45-45-90 Rt. ∆ 3√2 3 45 45 23 - 17 3 23 3 2 m = ½(17 + 23) m = 20 A = 3 • 20 = 60 60 sq. units

Homework pg. 436 #1-19, 23 odd pg. 470 #1-9 For #5 see the bonus from Powerpoint 11.2