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In trapezoid $ABCD,$ $\overline{AB} \parallel \overline{CD}$.  Find the area of the trapezoid.

 Jan 13, 2024
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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

 

cool cool cool

 Jan 13, 2024

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