Chittagong Junior 2013 / 9

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Labib
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Location:Dhaka, Bangladesh.
Chittagong Junior 2013 / 9

Unread post by Labib » Sun Jan 19, 2014 11:08 am

In the figure, O is the centre of the circle and $OA = 10$ is perpendicular on $OB$. Perimeter of the $OGFE$ rectangle is $24$. The perimeter of area $AEGBF$ can be written as $n+m\pi$. Find $n\times m$.
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Thanic Nur Samin
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Re: Chittagong Junior 2013 / 9

Unread post by Thanic Nur Samin » Sun Jan 19, 2014 11:42 am

O,F যোগ করি। OF=ব্যাসার্ধ=OA=10।
এখন,
$OE^2+OG^2=OF^2=10^2=100.......(1)$
$2(OE+OG)=24$বা$OE+OG=12$বা$OE=12-OG..............(2)$
(1)নং এ বসিয়ে পাই,
$(12-OG)^2+OG^2=100=> 2OG^2-24OG+144=100=> OG^2-12OG+22=0=> OG=6-\sqrt{14}$
অর্থাৎ,
$OE=6+\sqrt{14}$
$OG=12-OE=6-\sqrt{14}$
কিন্তু বৃত্তকলা $(AOB)=\frac{\pi 10^2}{4}=25\pi$
সুতরাং,
$(OGE)=\frac{1}{2}\times(6+\sqrt{14})(6-\sqrt{14})=11=> (AEGBF)=(AOB)-(GOE)=-11+25\pi$
তাই উত্তর হবে $-275$
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