# 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.

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

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

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

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

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

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

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Thank you very much, this is exactly the solution I was after. I would upvote this if I had enough reputation :) –  Michael Ferashire Silva Oct 28 '13 at 13:52
But isn't a^2 = 13^2 - (21 - x)^2? –  Michael Ferashire Silva Oct 28 '13 at 13:58
Michael, yes indeed! That simplifies matters! See the clarification above. –  amWhy Oct 28 '13 at 14:05
@amWhy: Very nice use of color +1 –  Amzoti Oct 29 '13 at 1:35
@Michael Using Herons formula would have been a good trick as well, if you allowed it. –  Sawarnik Oct 29 '13 at 10:44

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 $$\sqrt{u.(u-a).(u-b).(u-c)}=\frac{h.21}{2}$$

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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 $5+16=21$.

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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.

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I don't think I'm allowed to use this formula, but it is interesting, thanks :) –  Michael Ferashire Silva Oct 28 '13 at 13:53

Use the Law of Cosines to find $\angle C$. Then use $\sin(\angle C)=\frac{opposite}{hypotenuse}$.

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