Question 991242
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Two right triangles are formed by the perpendiculars, namely triangle ADL and triangle ABM.  I will leave it as an exercise for the student to prove that these two triangles are similar.  By similarity, AL is in proportion to AM as AD is to AB, so:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{12}{18}\ =\ \frac{6.4}{x}]


Solve for *[tex \Large x]


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

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