Question

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`.

Answer 1

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`

Answer 2

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!