Given the right trapezoid calculate angle theta and the area of triangle hat(EAD), provided EA=4, AB=BC=CD=DA=2, AB_|_EC?

Jul 11, 2016

$\angle \theta = \arctan \left(\frac{\sqrt{3} - 1}{2}\right) \approx 0.350879411 r a d$

${S}_{E A D} = 2$

Explanation:

From right triangle $\Delta A B E$, knowing hypotenuse $A E = 4$ and cathetus $A B = 2$ we can find another cathetus:
$B E = \sqrt{{4}^{2} - {2}^{2}} = \sqrt{12} = 2 \sqrt{3}$.

In the right triangle $\Delta D C E$ cathetus $C D = 2$. Second cathetus $C E = C B + B E = 2 + 2 \sqrt{3}$

Now we can determine tangent of angle $\angle \theta$:
$\tan \left(\theta\right) = \frac{C D}{C E} = \frac{2}{2 + 2 \sqrt{3}} = \frac{1}{\sqrt{3} + 1} = \frac{\sqrt{3} - 1}{2}$
Angle $\angle \theta$ can be determined using an inverse function ${\tan}^{- 1} \left(\right)$ or, as it is often expressed, $\arctan \left(\right)$:
$\angle \theta = \arctan \left(\frac{\sqrt{3} - 1}{2}\right) \approx 0.350879411 r a d$

Area of triangle $\Delta E A D$ can be calculated using the length of its base $A D$ and altitude $D C$:

${S}_{E A D} = \frac{1}{2} \cdot A D \cdot D C = \frac{1}{2} \cdot 2 \cdot 2 = 2$