Question 1080661
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In an acute triangle ABC, the altitudes AD and CE are drawn. 
Find the length of the line segment DE if AB = 15 cm, BC = 18 cm, and AD = 10 cm.
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<pre>
0.  Make a sketch to follow my arguments.


1.  The area of the triangle ABC is

    S = {{{(1/2)*abs(AB)*abs(CE)}}} = {{{(1/2)*abs(BC)*abs(AD)}}}, 


    which gives you an equation 

    15*|CE| = 18*10.

    It implies |CE| = {{{(18*10)/15}}} = 12 cm.


2.  Then from the right-angled triangle BCE  |BE| = {{{sqrt(abs(BC)^2-abs(CE)^2)}}} = {{{sqrt(18^2-12^2)}}} = {{{sqrt(180)}}}.


3.  From the right-angled triangle ABD  |BD| = {{{sqrt(abs(AB)^2-abs(AD)^2)}}} = {{{sqrt(15^2-12^2)}}} = {{{sqrt(81)}}} = 9.


4.  From the right-angled triangle ABD  cos(B) = {{{abs(BD)/abs(AB)}}} = {{{9/15}}} = {{{3/5}}}.


5.  Now, to answer the problem's question, apply the cosine law:


    {{{abs(DE)^2}}} = {{{abs(BE)^2 + abs(BD)^2 - 2*abs(BE)*abs(BD)*cos(B)}}}
 = {{{(sqrt(180))^2 + 9^2 - 2*sqrt(180)*9*(3/5)}}} = {{{180 + 81 - 2*sqrt(180)*9*(3/5)}}} = {{{261-(2*2*3*9*3*sqrt(5))/5}}} = {{{261-(324*sqrt(5))/5}}}.


    Hense,  |DE| = {{{sqrt(261-(324*sqrt(5))/5)}}} = 10.78 (approximately).
</pre>

Solved.