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

Answer 1

#/_theta = arctan((sqrt(3)-1)/2)~~0.350879411 rad#

#S_(EAD) = 2#

From right triangle #Delta ABE#, knowing hypotenuse #AE=4# and cathetus #AB=2# we can find another cathetus: #BE=sqrt(4^2-2^2)=sqrt(12)=2sqrt(3)#.

In the right triangle #Delta DCE# cathetus #CD=2#. Second cathetus #CE = CB+BE = 2+2sqrt(3)#

Now we can determine tangent of angle #/_ theta#: #tan(theta) = (CD)/(CE) = 2/(2+2sqrt(3)) = 1/(sqrt(3)+1)=(sqrt(3)-1)/2# Angle #/_theta# can be determined using an inverse function #tan^(-1)()# or, as it is often expressed, #arctan()#: #/_theta = arctan((sqrt(3)-1)/2)~~0.350879411 rad#

Area of triangle #Delta EAD# can be calculated using the length of its base #AD# and altitude #DC#:

#S_(EAD) = 1/2*AD*DC = 1/2*2*2 = 2#

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