Calculate the length of the altitude through A of an isosceles triangle in which AB=AC=26 cm and BC=20 cm.
Bisecting triangle BAC (drawing a line from A to the middle of line segment BC) creates 2 right triangles. For any right triangle, the sum of the squares of each side equals the square of the hypotenuse, or a^2 + b^2 = (hypotenuse)^2. Call the point of intersection on line segment BC point D. So what is the length of segment AD? The length of BA is 26, the length of BD is 10, so for this right triangle
1) (BD)^2 + (AD)^2 = (BA)^2
2) 10^2 + (AD)^2 = 26^2
3) 100 + (AD)^2 = 676
4) (AD)^2 = 576
5) AD = sqrt(576)
6) AD = 24
Let D be the midpt. of the side BC, and let AD be the altitude of the triangle ABC
Thus, BD=DC= 20/2 = 10cm
Using Pythogaras' theorem
BD^2 + AD^2= BA^2
(10)^2 + AD^2= (26)^2
AD^2= 26^2-10^2= 576 cm
The straight line down from the tip to the middle of the base makes a right triangle with the little bottom side measuring 10 and the hypotenuse, the angley side, measuring still what it is, 26. Bummer it's not a 3, 4, 5 triangle, but anyway, all you have to do is plug the numbers into the Pythagorean theorem and solve. But 10 and 26 are pretty big numbers, so I'd first divide both by two, making them 5 and 13 and then solve the problem 13 squared = 5 squared + what squared?
169 = 25 + x squared
169 -25 = x squared
144 = x squared (o thank gosh, an easy square!)
so x = the square root of 144, 12.
But then remember to multiply it back up (because you halved the sides to make getting the answer easier).
So 12 x 2 = 24.
So the altitude is 24.
Now I'm going to see if 24's what the other answerer got, just a sec . . .
Yes! cool. not to shabby for an English teacher, eh?