Puzzle-43. Find the area AED in the square ABCD when AE=4 cm and angle AEB=90 degree.

[ttokoro]

Puzzle-43. Find the area AED in the square ABCD when AE=4 cm and angle AEB=90 degree.

[Werner_E · Accepted solution]

That's not a puzzle where Mathcad or Prime could be of much help (other than using it as equation editor) 😞

Let AF be perpendicular to AD. Because ABCD is a square, AB=AD and the area of triangle AED = 1/2*AB*EF.

1) Geometric solution using similar triangles and ratios:

Triangles ABE and AEF are similar and so the ratios AB:4 and 4:EF are equal So we have

and the area of AED is half of it, -> 8

2) Without much thinking using law of cosine in right angled triangles

Let alpha = angle BAE

So we have AE=4*cos(alpha) and AD=AB=4/cos (alpha). Therefore we have 1/2*AE*AD=8.

3) Pragmatic solution, relying on the fact that the task makes sense and the solution is unique.

If the task makes sense, the solution must be independent of the exact position of E (as long as AE=4), independent of alpha.

So we must get the same result when we choose alpha=0, which means B=E and so AB=4. The triangle AED is now half of the square with side length 4  (ED is its diagonal) and so has the area 8.

 

[ppal]

X - Line from point E to AD that Intersects at 90 degrees at AD

Y -  Line from point E to AB that Intersects at 90 degrees at AB

[ppal]

[Werner_E]

Nice, but your solution assumes (without proof!) that the problem is uniquely solvable and that the solution is therefore independent of the variable size of the square.
Why then s=1.1*AE?
Better to use s=AE, then you can immediately see with the naked eye that the triangle AED is half of the 4x4 square and therefore has the area 8.
Basically, this brings us to my 'pragmatic' solution #3. 😉

[ppal]