Right Triangles and Altitudes I am once again stuck on a question about geometry, this problem is about altitudes that crate right triangles:
Let there be a triangle that has side lengths of 13, 20, and 21. Given this, find the length of the altitude drawn to the side of length 21.
I have drawn the following picture to make my understanding clearer:

However, I am still not sure how to get the length of this altitude.  My definition of an altitude is a segment drawn from one vertex to a point on the line opposite the point such that this segment is perpendicular to the line opposite the vertex.
I am very sure that to find the length, I will have to use the pythagorean theorem at some point, but I am not sure how to start this. 
If anybody could give a starting point or hint that would be great :)
Note: I cannot use the cosine rule.
Thank you in advance.
 A: Let $a$ be altitude, and x be the base of the right-most right triangle.
$$20^2 = a^2 + x^2$$ $$13^2 = a^2 + (21 -x)^2$$
$$a^2 = \color{blue}{\bf 20^2 - x^2 = 13^2 - (21-x)^2 }$$
$$400 - x^2 = 169 - (441 - 42x + x^2) = -272 + 42x - x^2$$
$$672 = 42x \iff x = 16.$$
$$x = 16 \implies a = \sqrt{400 - x^2} = \sqrt{400 - 256} = 12$$
A: Hint
You can also find altitude from area of triangle, recall that if $a,b,c$ are the sides of $ABC$ then the area of $ABC$ can be calculated by
$$\sqrt{u.(u-a).(u-b).(u-c)}$$
where $2u=a+b+c$.
So from the following you can find the height
$$\sqrt{u.(u-a).(u-b).(u-c)}=\frac{h.21}{2}$$
A: Use Heron's formula to find the area of the triangle:
$$A = \sqrt{s(s-a)(s-b)(s-c)},$$
where $s = \frac{a+b+c}{2}$ is the semiperimeter of the triangle.
Then find the altitude using the usual formula for the triangle area.
A: The simplest way, which will not work all the time, is that any time you see right triangles you should think about Pythagorean triples.  One is $3,4,5$, which we can scale up to $12,16,20$ (note the hypotenuse of $20$ in your figure).  Another is $5,12,13$.  Clearly the altitude is the common figure, $12$ and the base is $5+16=21$.
A: Use the Law of Cosines to find $\angle C$.  Then use $\sin(\angle C)=\frac{opposite}{hypotenuse}$. 
