+1418
In trapezoid $ABCD,$ $\overline{AB} \parallel \overline{CD}$. Find the area of the trapezoid.
+129771 Angle BDC = 180 - 70 -80 = 30
Law of Sines
6 /sin 30 = DC /sin 80
DC= 6sin 80 / sin 30 = 6sin 80 / (1/2) =12sin 80 cm
Area of triangle BCD = (1/2) ( 6) (12 sin 80) * sin 70 = 36 sin 80 * sin 70 ≈ 33.3 cm^2
Height of DBC = Height of trapezoid
33. 3 = (1/2) (DC) * height of trapezoid
66.6 = (12 sn 80) * h
66.6 / (12 sin 80) ≈ 5.64 cm
Area of trapezoid = (1/2) h * ( sum of bases) = (1/2) (5.64) ( 12sin 80 + 3) ≈ 41.79 cm^2
