Given: AB ∥ DC DE ⊥ AB , AD = BC m∠ADC = 134° AD = 40, DC = 32 Find: Area of ABCD

Accepted Solution

Answer: [tex]1,719.87\ units^2[/tex]Step-by-step explanation: The missing figure is attached. The formula to calculate the area of trapezoid is: [tex]A=\frac{h}{2}(B+b)[/tex] Where "B" is the long base, "b" is the short base and "h"  is the height. 1. First we need to find the height DE: - Since: [tex]m\angle ADC + m\angle FDA=180\°[/tex] [tex]m\angle ADC=134\°[/tex] We can find [tex]m\angle FDA[/tex]. This is: [tex]m\angle FDA=180\°-134\°=46\°[/tex]  - Observe in the second figure that the triangles EAD and FDA are equal. Then: [tex]m\angle FDA=m\angle EAD=46\°[/tex] - Use the Trigonometric Identity [tex]sin\alpha=\frac{opposite}{hypotenuse}[/tex]. In this case: [tex]\alpha =46\°\\opposite=DE\\hypotenuse=40[/tex] Then: [tex]sin(46\°)=\frac{DE}{40}\\\\40*sin(46\°)=DE\\\\DE=28.77\ units[/tex] 2.Use the Trigonometric Identity [tex]cos\alpha=\frac{adjacent}{hypotenuse}[/tex] to find the lenght AE, which is equal to the lenght GB In this case: [tex]\alpha =46\°\\adjacent=AE\\hypotenuse=40[/tex] Then: [tex]cos(46\°)=\frac{AE}{40}\\\\40*cos(46\°)=AE\\\\AE=GB=27.78\ units[/tex] - Therefore, the large base AB is: [tex]AB=AE+DC+GB\\\\AB=27.78+32+27.78\\\\AB=87.56\ units[/tex] - Now we can substitute values into the formula and calculate the aera of the trapezoid ABCD. This is: [tex]A=\frac{28.77\ units}{2}(87.56\ units+32\ units)=1,719.87\ units^2[/tex]