Which triangles in the figure above are congruent and/or similar? Find the value of x, angle #/_ACE# and the area #hat(AEC)#, and #BFDC#.

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2 Answers
Aug 11, 2016

Assuming:

#AE_|_CE# and

#BF_|_AC#

There are three similar (not congruent) triangles: #Delta ACE#, #Delta ABF# and #Delta FDE#

#x=13#

#/_ACE=45^o#

#S_(AEC)~~187.88#

#S_(BFDC)~~102.88#

Explanation:

There are three triangles:
#Delta ACE#, #Delta ABF# and #Delta FDE#.

Given that all of them are right triangles and all of them are isosceles. Therefore, all of them have acute angles of #45^o# and, therefore, all are similar to each other.

Since all three hypotenuses are different, there are no congruent pairs of them.

Length of #x#.
Since #Delta ABF# is a right triangle with catheti #x# and #x# and hypotenuse #13sqrt(2)#, Pythagorean Theorem gives:
#x^2+x^2=(13sqrt(2))^2#
#=> 2x^2=338#
#=> x^2=169#
#=> x=13#

Angle #/_ACE=45^o# because, as we stated above, triangle #Delta ACE# is right isosceles.

Area of #Delta AEC#.
#S_(AEC) = 1/2*AE*CE =#
#= 1/2*(13sqrt(2)+1)^2=1/2(338+26sqrt(2)+1)~~187.88#

Area of #BFDC#.
We have to subtract areas of #Delta ABF# and #Delta FDE# from the area of #Delta ACE# to get the area of #BFDC#.
#S_(BFDC)=S_(AEC) - S_(ABF) - S_(FDE) =#
#=1/2(338+26sqrt(2)+1) - 1/2*x^2 - 1/2*1^2 =#
#=1/2(339+26sqrt(2)-169-1) = 1/2(169+26sqrt(2))~~102.88#

Aug 11, 2016

We have a triangle #Delta FED# such that #bar (FE) = bar (ED) = 1# and
#bar (FD)=2# defining a null area triangle!

Explanation:

We have a triangle #Delta FED# such that #bar (FE) = bar (ED) = 1# and
#bar (FD)=2# defining a null area triangle!