geometry - 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.

Answer

Let $a$ be altitude, and x be the base of the right-most right triangle.

$$20^2 = a^2 + x^2$$ $$13^2 = a^2 + (21 -x)^2$$

$$a^2 = \color{blue}{\bf 20^2 - x^2 = 13^2 - (21-x)^2 }$$

$$400 - x^2 = 169 - (441 - 42x + x^2) = -272 + 42x - x^2$$

$$672 = 42x \iff x = 16.$$

$$x = 16 \implies a = \sqrt{400 - x^2} = \sqrt{400 - 256} = 12$$