Find the length of the shortest altitude of a triangle with side lengths 10,24,26

Look at the following pic, Mellie

Angle ABC can be found thusly :

sin CAB/ 26 = sin ABC / 24

And sin CAB = sin90 = 1

So

Sin ABC = 24/26

arcsin (24/26)  = ABC

And the altitude  AX drawn from A to to side BC will be one leg ot the right triangle AXB........so this altitude will be shorter than the altitude drawn from B  to side AC, since this altitude is the hypotenuse of triangle AXB. And AX will clearly be shorter than the remaining altitude drawn from C  to side AB.

And we can find AX thusly

AX / sinABC = AB / sin AXB

AX / sin[arcsin(24/26)]  = AB / sin AXB

But AXB = 90 so the sine of this angle  = 1   and   AB = 10.....so we have

AX = 10 * sin[arcsin(24/26)]  = 120/13 units  = about 9.23 units

And this is the shortest altitude

There is an easier way to solve the problem. You know that the triangle is a right triangle because it's sides are a Pythagorean triple. Saying this we can call the point perpendicular to B as X. Then we will call AX = y and CX = 26 - y. BX we will call a. Then using the Pythagorean Theorem y ^ 2 + x ^ 2 = 10^2, y^2 + x^2 = 100. Then we can make another equation using the other triangle. (26 - y)^2 + x ^ 2 = 24^2. This simplifies to x^2 + y^2 - 52y = -100. Plugging in information from the other equation 100 - 52y = -100. y = 50/13. Then plugging this in x = 120/13