Question 90334
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I need to find the altitude of the triangle 
13 by 10 by 13 
I have no idea how to do this.

Here's the triangle:
 
{{{drawing(300,300,-7,7,-1,13, locate(-3.7,6,"13"),
locate(3.1,6,"13"), locate(-.3,-.3,"10"),

triangle(-5,0,5,0,0,12) ) }}}

Draw in the altitude, a, which divides the base (which
measures 10) into two parts measuring 5 each:

{{{drawing(300,300,-7,7,-1,13, locate(-3.7,6,"13"),
locate(3.1,6,"13"), triangle(-5,0,0,0,0,12), 
locate (-2.5,-.3,"5"), locate(2.5,-.3,"5"), locate(0.3,6,"a"),
triangle(-5,0,5,0,0,12) ) }}}

So now you have two congruent right triangles.  I'll split 
them apart for clarity:

{{{drawing(300,300,-7,7,-1,13, locate(-4.7,6,"c=13"),
locate(3.1,6,"13"), triangle(-5,0,0,0,0,12), 
locate (-3.5,-.3,"b=5"), locate(4.5,-.3,"5"), locate(0.3,6,"a"),
triangle(2,0,7,0,2,12) ) }}}

With the right triangle on the left we can
use the Pythagorean theorem to find a:

{{{c^2 = a^2+b^2}}}

{{{13^2 = a^2+5^2}}}

{{{169 = a^2+25}}}

Subtract 25 from both sides

{{{144 = a^2}}}

{{{12 = a}}}

So the altitude of the original triangle is 12.

Edwin</pre>