Triangle area dilemna

[BMAD]

Why can't a right triangle with an hypotenuse of sqrt(24) have an area of 13 squared inches? 

[araver]

Spoiler

I'm assuming 24 is also measured in inches. Otherwise, the question is ambiguous and it has no solutions ifand only if the actual measurement unit for the hypotenuse is less than sqrt(13/6) of an inch.

1. Let's denote a,b,c the sides of the triangle with c being the hypotenuse, c^2 = 24. Pitagora yields a^2+b^2 = 24.

Assume the area is 13 = a*b/2. Therefore a*b = 26. However (a-b)^2 >=0 so a^2+b^2 >= 2*a*b hence 24 >= 2*26, contradiction. Thus, the area cannot be 13 square inches.

2. Moving an arbitrary point on a circle with diameter sqrt(24), we can see that the area of the right triangle ranges from 0 (degenerate triangle) to the area of the isosceles right triangle with an hypotenuse of sqrt(24) which is c/sqrt(2)*c/sqrt(2)/2=6.

 

 

Edited August 12, 2016 by araver

[DejMar]

It very well may be possible that a right triangle with an hypotenuse of sqrt(24) inches to have an area of 13 squared inches, yet not a Euclidean right triangle. One might consider verifying the assertion as it relates to right triangles that are hyperbolic. 

[araver]

Nice catch. 

I disregarded those geometries since the word hypothenuse used in the OP is ambiguous in an elliptic geometry where two angles can be "right" at the same time yielding two "hypothenuses". But "square units" kinda need a rectangle to be defined which is not true in hyperbolic geometry.

If I recall the definition of area correctly, in a hyperbolic geometry one could theoretically get any area size up to an upper bound by the product of a constant and the defect of the triangle. Presumably, if one chooses the "right" hyperbolic curvature one might get any number for the area including the elusive "square inches" measurement. 

But since the Euclidean case fails to produce such a "big" area, I'd look for an elliptic geometry.