A rectangle ABCD and right-angled triangle AEF share a common right-angled vertex FAE and BAD.

If the area of both shapes is 50cm2, what are the upper and lower bounds of the length FC?

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It seems to me that, by shrinking the height of the rectangle (length AD) to something infinitesimal, one could move point C indefinitely far away from segment AF, and therefore from point F, as well. Given that, I don’t think there is an upper bound for this distance, FC.

correct.

Yay!

There is a lower one though!

True, and I’ve been thinking about it, but have yet to figure it out!

infinity and zero is my guess.

One of those is bang on

Without any symbolic math let the green rectangle get longer and thinner indefinitely, then length FC approaches infinity.

For the other extreme let the green rectangle have a width one quarter of the length of the base of the triangle. Then its height is the same as the triangle height. Length FC is now the width of the rectangle. Now stretch and narrow both objects until the width of the rectangle shrinks to zero. the objects keep their heights equal, so the lower bound of FC is zero.

Aha! I just had FC = DC scribbled down. But yes, that would go towards zero.