 All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars  perplexus dot info  Another triangle in a square problem (Posted on 2023-03-29) ABCD is a square with point E inside the square.
Line segments AE, BE, and DE are drawn.
The length of AE is 8 and angle AEB is a right angle.

What is the area of triangle AED?

 See The Solution Submitted by Brian Smith No Rating Comments: ( Back to comment list | You must be logged in to post comments.) Solution | Comment 1 of 4
I'm going to guess 32 before solving.
Draw 1/2 circle from A to B centered at the midpoint of AB.  This includes E because AEB is a right angle.
Imagine any point on this half circle:  if E can be anywhere along this path, then you can consider the side length of the square as changing its size to whatever magnification factor keeps the length AE staying always 8.  So I suspect the area of AED may be the same no matter where E is located along the half circle.  So let E be identical to B.  Now triangle AED is triangle ABD: an equilateral right triangle with sides 8, 8, and 8 sqrt(2).  (1/2)*8*8 = 32.  Or let E be the exact center of the square:  both AEB and AED are right triangles of the same dimension we had when E was coincident with B.  Again 32.

OK, now an actual solution:
Let x be the side length of the square.  Angle t is ABE.  Angle EAD is also t.  The base of of triangle AED is S, which we don't know yet; but its height is 8*sin(t).  But AEB is a right triangle, so sin(t) is 8/x.
So the area of triangle AED is:
(1/2)(x)(8*sin(t)) = (1/2)(x)(8*8/x) = 32.
As expected, this area is independent of angle t.

 Posted by Larry on 2023-03-29 12:02:49 Please log in:

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