KHULNA 2013,Junior,P10

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tanmoy
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KHULNA 2013,Junior,P10

Unread post by tanmoy » Fri Dec 19, 2014 10:35 pm

In trapezium $ABCD$,$AD||BC$,$AD < BC$ and nonparallel sides are equal.Perpendicular drawn at $A$ on $AB$ meets $BC$ at $F$ where $BF:FC=3:2$.Perpendicular from $A$on $BC$ meets $BC$ at $E$.$BC = 10AE$,$BE < EF$,the ratio $EF:BC$ can be expressed as $\frac{a\sqrt{a}+b}{c}$.Find the LCM of $a,b,c$.
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Raiyan Jamil
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Re: KHULNA 2013,Junior,P10

Unread post by Raiyan Jamil » Mon Dec 22, 2014 1:53 pm

I think the ans is 240.
Let AE be x. So, BC= 10x , BF=6x , FC=4x . We take any point Z as the midpoint of BF. So, BZ=FZ=3x. We join A and Z. So, AZ=3x. And therefore, $EZ^2=(3x)^2-(x)^2$. And EZ= $\sqrt{8x^2}$ or, $\sqrt{8}x$.
Now, EF=EZ+FZ or, $\sqrt{8}x + 3x$. So, the ratio EF:BC can be expressed as $\frac{x\sqrt{8}+3x}{10x}$
Here, we can discover that the value of x is 8. Or, the value of a is 8, b is 3*8=24 and c is 10*8=80.So, the lcm of a,b,c will be the lcm of 8,24 and 80.That is 240.:D
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tanmoy
Posts:312
Joined:Fri Oct 18, 2013 11:56 pm
Location:Rangpur,Bangladesh

Re: KHULNA 2013,Junior,P10

Unread post by tanmoy » Mon Dec 22, 2014 10:49 pm

$\frac{x\sqrt{8}+3x}{10x}=\frac{x(\sqrt{8}+3)}{10x}=\frac{\sqrt{8}+3}{10}$=$\frac{2\sqrt{2}+3}{10}$.$\therefore$ LCM of $a,b,c$ is $30$. ;)
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Raiyan Jamil
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Re: KHULNA 2013,Junior,P10

Unread post by Raiyan Jamil » Tue Dec 23, 2014 10:32 am

Sorry, I actually did it by not taking it in the least form. This solution of yours is correct. Thanks for that.:D
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tanmoy
Posts:312
Joined:Fri Oct 18, 2013 11:56 pm
Location:Rangpur,Bangladesh

Re: KHULNA 2013,Junior,P10

Unread post by tanmoy » Tue Dec 23, 2014 1:57 pm

Also thanks to you for posting the solution :D
"Questions we can't answer are far better than answers we can't question"

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