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Segments $BE$ and $CF$ are the altitudes in $\triangle ABC$.
$E$ is on line $AC$ and $F$ is on line $AB$.
$BC = 65$, $BE = 60$ and $CF = 56$.
Find $A(\triangle ABC)/100$.

figure

By the Pythagorean theorem , $CE=25$ , and $BF= 33$.
If the length of altitude from $A$ to B$C$ can be calculated then the area of $\triangle ABC$ can be calculated since the length of $BC$ is known.
But I'm stuck here , so any hints are apreciated .

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  • $\begingroup$ One can grind it out. Let $x=EA$ and $y=FA$. By areas, $60(x+25)=56(y+33)$. Then Pythagorean Theorem gives us another equation (or two). Solve. There is undoubtedly also a clever way. $\endgroup$ – André Nicolas Mar 31 '14 at 13:41
  • $\begingroup$ @AndréNicolas Got it , but solving those equations would be pretty time consuming , so what's the clever way? $\endgroup$ – A Googler Mar 31 '14 at 14:15
  • $\begingroup$ @AGoogler: The solution by Sawarnik qualifies! $\endgroup$ – André Nicolas Mar 31 '14 at 14:24
  • $\begingroup$ I have a related question for you. If $m_a=12$ and $m_b=9$ are medians, and the other side is $c=10$. Then the area would be? $\endgroup$ – Sawarnik Apr 2 '14 at 17:30
  • $\begingroup$ @Sawarnik 12 is the length of median from A to BC , right? $\endgroup$ – A Googler Apr 2 '14 at 17:42
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We have that $\cot C=25/60$ and $\cot B=33/56$.

Now use the area formula: $$\text{Area}=\frac{a^2}{2(\cot B + \cot C)}$$

Thus: $$\frac{(ABC)}{100}=\frac{65^2}{200(\frac{25}{60} + \frac{33}{56})}=\frac{13^2}{8(\frac{25}{60} + \frac{33}{56})}=\frac{13^2}{\frac{10}{3} + \frac{33}{7}}=\frac{13^2}{\frac{70+99}{21}}=21$$

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  • $\begingroup$ I have never seen that formula before. How is this derived? $\endgroup$ – Laars Helenius Mar 31 '14 at 14:37
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    $\begingroup$ @AGoogler No, sorry. I just confused up with the two formulas. Now its correct :) $\endgroup$ – Sawarnik Mar 31 '14 at 18:02
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    $\begingroup$ @Sawarnik I didn't try , but I'll try to prove it now. Thanks for the tip , I'll remember it now. Also I couldn't find it on the internet , from where did you find it? $\endgroup$ – A Googler Apr 1 '14 at 13:06
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    $\begingroup$ Very fine and clear cut solution. $\endgroup$ – Ajay Apr 1 '14 at 14:36
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    $\begingroup$ It' derivation is simple. Drop a perpendicular from A on BC. (Area of triangle = 1/2 .b.h). Now use cot B and cot C to get the formula. (Derivation is mailed to you.) $\endgroup$ – Ajay Apr 2 '14 at 3:01
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MPA entrance 2011. Can you tell me the answer of Q.55 from Maths 2010 of MPA. Best of luck for 6th April.I am there too.

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  • $\begingroup$ Can't type with latex. $\endgroup$ – Ajay Apr 1 '14 at 13:26
  • $\begingroup$ Hi! Thanks for the answer . I'll try to solve Q.55 . Best of luck for you too! $\endgroup$ – A Googler Apr 1 '14 at 14:23
  • $\begingroup$ Thanks but what's your real name? How many were solved by you? $\endgroup$ – Ajay Apr 1 '14 at 14:23
  • $\begingroup$ I'm uncomfortable giving away my real name online. Can you email me instead ? (agoogler@live.com) I've solved about 30 questions randomly . Although some of them were solved by me months ago. How many have you solved? $\endgroup$ – A Googler Apr 1 '14 at 14:29
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    $\begingroup$ @Sawarnik Here - mprakashacademy.co.in/XI_SampleQuestions/Papersupto2013.pdf $\endgroup$ – A Googler Apr 1 '14 at 15:04

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