Another Geometry Question?

Question

The legs of an isosceles right triangle have length 12. Find the lengths of the hypotenuse and the altitude to the hypotenuse.

Answer 1

The hypoteneuse has length 12 sqrt(2), or 16.9706... ,

and the altitude to the hypoteneuse half that, 6 sqrt(2) or 8.4853... .

There are a couple of easy ways to think about this:

1. By the simplest application of Pythagoras's Theorem, (1, 1, sqrt(2)) is an isoceles right triangle with "legs" of length 1 and a hypoteneuse of length sqrt(2). Your triangle is simply this basic one scaled up in the lengths of its sides by a factor of 12, that is 12 sqrt(2).*** Then if you think of drawing in the altitude to the hypoteneuse, you now have two smaller but similar triangles. In those triangles, the length 12 is now the length of the new hypoteneuse in EACH of those two, new smaller triangles. The "altitude" that you want is then 12/[sqrt(2)] = 6 sqrt(2).

2. Another way, using geometrical properties of the square for the "altitude" part. To begin with, you can consider the original triangle as two adjacent sides of a square and the diagonal joining the free ends of that square. Again, by Pythagoras and/or scaling up the (1, 1, sqrt(2)) triangle, you get the length of the hypoteneuse as 12 sqrt(2). Now, if you consider the OTHER diagonal of that same square, the "altitude" that you want is clearly, by symmetry, exactly half of the original hypoteneuse/diagonal, i.e. 6 sqrt(2).

Live long and prosper.

*** P.S. Of course, you can always plug 12^2 into Pythagoras's Theorem twice and work out the hypoteneuse from that directly. However, I'm a great believer in using scaling where appropriate to save myself unnecessary labour. (It seems redundant and simply "make work" to calculate each special case when it's clear by scaling that a right triangle with "legs" of length 'a' must necessarily be (a, a, a sqrt(2)). Realizations like this are part of learning to work math problems efficiently.

Answer 2

Use the Pythagorean Theorem. For a right triangle:

a² + b² = c²
12² + 12² = c²
144 +144 = c²
288 = c²
c = 12√2

The altitude to the hypotenuse is once again an isosceles right triangle. The altitude is equal to the length of half the hypotenuse of the original triangle.

The altitude is 6√2.