Cox's Bazar+ Patuakhali BdMO 2014: Junior 8

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atiab jobayer
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Cox's Bazar+ Patuakhali BdMO 2014: Junior 8

Unread post by atiab jobayer » Sun Dec 07, 2014 10:34 am

∠BAC is a right angle, ADB, AEC, ABC are semicircles with diameters AB, AC and BC. The areas of the black portions in semicircles ADB and AEC are 13 and 16 respectively. Find the area of triangle ABC.
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tanmoy
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Re: Cox's Bazar+ Patuakhali BdMO 2014: Junior 8

Unread post by tanmoy » Sun Dec 07, 2014 6:25 pm

Assume that the area of the white portion of the semicircle $ADB$ is $x$ and the area of the white portion of semicircle $ACE$ is $y$.So,$\Pi \frac{AB^{2}}{8}-x=13$ and $\Pi \frac{AC^{2}}{8}-y=16$.Adding them,we get that
$\Pi \frac{AB^{2}}{8}-x+\Pi \frac{AC^{2}}{8}-y=29$
Or,$\Pi \frac{(AB^{2}+AC^{2})}{8}-(x+y)=29$
Or,$\Pi \frac{BC^{2}}{8}-(\Pi \frac{BC^{2}}{8}-\frac{AB\cdot AC}{2})=29$
Or,$\frac{AB\cdot AC}{2}=29$
$\therefore$ $\Delta ABC=29$ :D :) :P
"Questions we can't answer are far better than answers we can't question"

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