Question 557966
<pre>

The other tutor's answer is correct, but maybe you're studying the 
law of cosines instead of Heron's formula.  And maybe you haven't
proved that the shortest altitude is the one to the longest side.
(His "inverse proportional" statement is incorrect).

Maybe this is the approach you should take.
 
{{{drawing(200,175,-1,15,-1,13, triangle(0,0,14,0,5,12),
 
locate(-1,7,b=13), locate(6,0,c=14),
locate(0,0,A),locate(14,0,B), locate(4.7,13,C), locate(10,7,a=15)
 

 )}}}
 
Let's draw the three altitudes:
  
{{{drawing(200,175,-1,15,-1,13, triangle(0,0,14,0,5,12),
 
locate(-1,7,b=13), locate(6,0,c=14),
locate(0,0,A),locate(14,0,B), locate(4.7,13,C), locate(10,7,a=15),
green(line(5,0,5,12)),locate(5.2,6,h[1])
 
 )}}}{{{drawing(200,175,-1,15,-1,13, triangle(0,0,14,0,5,12),
 
 
 
locate(-1,7,b=13), locate(6,0,c=14),
locate(0,0,A),locate(14,0,B), locate(4.7,13,C), locate(10,7,a=15),
green(line(0,0,8.96,6.72)), locate(5.3,4.5,h[2])
 

 )}}}{{{drawing(200,175,-1,15,-1,13, triangle(0,0,14,0,5,12),
 
 
 
locate(-1,7,b=13), locate(6,0,c=14),
locate(0,0,A),locate(14,0,B), locate(4.7,13,C), locate(10,7,a=15),
green(line(14,0,350/169,840/169)), locate(5.7,5.2,h[3])
 

 )}}}
 
The law of cosines for the cos(A) is

{{{cos(A)}}} = {{{(b^2+c^2-a^2)/(2bc)}}} = {{{(13^2+14^2-15^2)/(2*13*14)}}} = {{{140/264}}} = {{{5/13}}} 
 
{{{h[1]/b}}} = {{{sin(A)}}} = {{{sqrt(1-cos^2(A))}}} = {{{sqrt(1-(5/13)^2)}}} = {{{sqrt(1-25/169)}}} = {{{sqrt(169/169-25/169)}}} = {{{sqrt(144/169)}}} = {{{12/13}}}
 
{{{h[1]/b}}} = {{{12/13}}}
{{{h[1]/13}}} = {{{12/13}}}

Multiply both sides by 13:

{{{h[1] = 12}}}
 
The law of cosines for cos(B)

{{{cos(B)}}} = {{{(a^2+c^2-b^2)/(2ac)}}} = {{{(15^2+14^2-13^2)/(2*15*14)}}} = {{{252/420}}} = {{{3/5}}}

{{{h[2]/c}}} = {{{sin(B)}}} = {{{sqrt(1-cos^2(B))}}} = {{{sqrt(1-(3/5)^2)}}} = {{{sqrt(1-9/25)}}} = {{{sqrt(25/25-9/25)}}} = {{{sqrt(16/25)}}} = {{{4/5}}}
 
{{{h[2]/c}}} = {{{4/5}}}
{{{h[2]/14}}} = {{{4/5}}}
Cross multiply
{{{5h[2] = 56}}}
{{{h[2] =56/5}}}
{{{h[2]=11.2}}}

{{{h[3]/c}}} = {{{sin(A)}}} which we have already calculated as {{{12/13}}}

{{{h[3]/c}}} = {{{12/13}}}

{{{h[3]/14}}} = {{{12/13}}}
Cross multiply
{{{13h[3] = 168}}}
{{{h[3] =168/13}}}
{{{h[3]=12.92307692}}}


So {{{h[2]=11.2}}} is the shortest altitude.

Incidentally, if you draw all three altitudes, they all three
intersect at the same point, called the orthocenter.

{{{drawing(200,175,-1,15,-1,13, triangle(0,0,14,0,5,12),
 
 
 
locate(-1,7,b=13), locate(6,0,c=14),
locate(0,0,A),locate(14,0,B), locate(4.7,13,C), locate(10,7,a=15),
green(line(14,0,350/169,840/169),line(0,0,8.96,6.72),line(5,0,5,12))
 

 )}}}
Edwin</pre>