The problem is that right triangles have to obey a constraint. The angle opposite the hypotenuse is 90 degrees.
Thus, once you've fixed the two endpoints of the hypotenuse, not all points are eligible to be the final point of the triangle. All other points in space can form a triangle with those two points, but it may not be a right triangle.
If you interpret the hypotenuse as the diameter of a circle, all -- and only -- the points on the circle, except the hypotenuse's endpoints, will form a right triangle with the hypotenuse. If the diameter's length is, as specified in the problem, 10 inches, this tells us that the circle has radius 5 inches. This is the maximum distance between the hypotenuse and the third corner of the triangle. The problem tells us that the distance from the hypotenuse to the third corner is 6 inches, which is impossible.
> If you interpret the hypotenuse as the diameter of a circle, all -- and only -- the points on the circle, except the hypotenuse's endpoints, will form a right triangle with the hypotenuse.
I've known that since I was a kid. What I didn't know until ~40 years later is that there is a generalization of that. If AB is a chord of a circle, and C and D are any points on the circle that are on the same side of AB, then angles ACB and ADB are the same. I have no idea how I never came across that before. It's called the Inscribed Angle Theorem, and is in Euclid.
When I read that I tried to prove it. First try was geometrically. I just could not get it. (Yes, I know that in fact it is easy...I've always sucked at geometry).
Second try was with vectors. What it is saying is that the dot product of AC and CB should be the same no matter where C is if you move C around on the same side of AB. That led to some ugly expression that would need to be constant. Mathematica said it was constant, but Mathematica doesn't show its work and I could not figure out how to show it.
Next try was with physics. Imagine that the circle is a very large circular train track, and there is a train on the track whose front is at A and back is at B. Imagine you have two cameras at the center of the circle, one pointed at A and the other, mounted directly on top of the first, pointed at B.
If the train starts moving, you'd have to turn the cameras to keep them pointing at the front and back of the train. With the cameras at the center of the circle, you'd have to turn them at the same rate. That's because from your point of view at the center of the circle, the angular velocities of any two points on the train are the same.
What the the Inscribed Angle theorem implies is that this also works if the cameras are on the circle. I.e., from the point of view of someone standing on the circular track, looking at a train moving elsewhere on the track, all parts of the train have the same angular velocity.
Dropping the train, what we have then is that the Inscribed Angle theorem is equivalent to claiming that a point moving around the circle at constant angular velocity as seen from the center of the circle also had constant (but not the same constant!) angular velocity as seen from an observer on the circle.
It was then easy to set up a point moving around a circle at constant angular velocity in polar coordinates (r = 1, θ = t), convert to Cartesian coordinates, shift the viewpoint to somewhere on the circle, go back to polar coordinates, and differentiate θ(t) with respect to t. That gave an expression that was easy to see was a constant. QED. Whew...
...and then I had another go at doing it with elementary geometry, and it turned out to be easy after all. Something you might reasonably see on a high school geometry homework assignment.
Thus, once you've fixed the two endpoints of the hypotenuse, not all points are eligible to be the final point of the triangle. All other points in space can form a triangle with those two points, but it may not be a right triangle.
If you interpret the hypotenuse as the diameter of a circle, all -- and only -- the points on the circle, except the hypotenuse's endpoints, will form a right triangle with the hypotenuse. If the diameter's length is, as specified in the problem, 10 inches, this tells us that the circle has radius 5 inches. This is the maximum distance between the hypotenuse and the third corner of the triangle. The problem tells us that the distance from the hypotenuse to the third corner is 6 inches, which is impossible.