Question 975491
<font face="Times New Roman" size="+2">


Construct your triangle and the desired perpendicular.  Note that the intersection of the perpendicular and the side measuring 13cm divides the 13cm side into two segments.  Let *[tex \Large x] represent the measure of the shorter segment and then the longer segment must measure *[tex \Large 13\ -\ x].  While we are at it, let *[tex \Large h] represent the height we are trying to find.  Since the height segment is perpendicular to the 13cm side of the triangle, we have divided the triangle into two right triangles, one with a hypotenuse that measures 5cm and the other with a hypotenuse that measures 12 cm.


We now invoke the magic of Mr. Pythagoras to say:



*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ +\ h^2\ =\ 25]


and


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (13\ -\ x)^2\ +\ h^2\ =\ 144]


which we can expand to read


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 169\ -\ 26x\ +\ x^2\ =\ 144]


Taking the first equation, multiplying by -1, and adding to the second equation gives:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 169\ -\ 26x\ =\ 119]


Solving for *[tex \Large x] we get


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ \frac{25}{13}]


Calling upon Mr. Pythagoras again we substitute into the first equation to get


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ h^2\ =\ 25\ -\ \left(\frac{25}{13}\right)^2]


After an abundance of arithmetic which I will let you verify for yourself, you get


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ h\ =\ \frac{60}{13}]


Which is the desired measurement.


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it

*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \