Mymensingh higher secondary 2018 #9

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samiul_samin
Posts:1007
Joined:Sat Dec 09, 2017 1:32 pm
Mymensingh higher secondary 2018 #9

Unread post by samiul_samin » Sun Feb 17, 2019 9:30 am

In triangle $ABC$,$\angle ABC=90^{\circ}$, $D$ is the midpoint of line $BC$. Point $P$ is on $AD$ line. $PM$ & $PN$ are respectively perpendicular on $AB$ & $AC$. $PM = 2PN$, $AB = 5$, $BC =a\sqrt b$ , where $a, b$ are positive integers. $a-b=$?

samiul_samin
Posts:1007
Joined:Sat Dec 09, 2017 1:32 pm

Re: Mymensingh higher secondary 2018 #9

Unread post by samiul_samin » Sun Feb 17, 2019 10:19 am

Hint
Use similar triangle
Answer
$a=6$
$b=3$
$a-b=3$

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