Question

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

Answer 1

Answer: $1,719.87\ units^2$

Step-by-step explanation:

The missing figure is attached.

The formula to calculate the area of trapezoid is:

$A=\frac{h}{2}(B+b)$

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:

$m\angle ADC + m\angle FDA=180\°$

$m\angle ADC=134\°$

We can find $m\angle FDA$. This is:

$m\angle FDA=180\°-134\°=46\°$

 - Observe in the second figure that the triangles EAD and FDA are equal. Then:

$m\angle FDA=m\angle EAD=46\°$

- Use the Trigonometric Identity $sin\alpha=\frac{opposite}{hypotenuse}$.

In this case:

$\alpha =46\°\\opposite=DE\\hypotenuse=40$

Then:

$sin(46\°)=\frac{DE}{40}\\\\40*sin(46\°)=DE\\\\DE=28.77\ units$

2.Use the Trigonometric Identity $cos\alpha=\frac{adjacent}{hypotenuse}$ to find the lenght AE, which is equal to the lenght GB

In this case:

$\alpha =46\°\\adjacent=AE\\hypotenuse=40$

Then:

$cos(46\°)=\frac{AE}{40}\\\\40*cos(46\°)=AE\\\\AE=GB=27.78\ units$

- Therefore, the large base AB is:

$AB=AE+DC+GB\\\\AB=27.78+32+27.78\\\\AB=87.56\ units$

- Now we can substitute values into the formula and calculate the aera of the trapezoid ABCD. This is:

$A=\frac{28.77\ units}{2}(87.56\ units+32\ units)=1,719.87\ units^2$