BDMO 2017 National round Secondary 4

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ahmedittihad
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BDMO 2017 National round Secondary 4

Unread post by ahmedittihad » Fri Feb 10, 2017 9:05 pm

$ABCD$ is a trapezoid with $AD\perp BC$ and $\angle ADC = 90^{\circ}$. $M$ is the midpoint $AB$ and $OM = 6.5$, and $ BC+CD+DA = 17$. Find the area of $ABCD$.
Frankly, my dear, I don't give a damn.

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ahmedittihad
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Re: BDMO 2017 National round Secondary 4

Unread post by ahmedittihad » Wed Jan 24, 2018 10:09 pm

ahmedittihad wrote:
Fri Feb 10, 2017 9:05 pm
$OM = 6.5$
This is a typing mistake, it should be $CM=6.5$.
Frankly, my dear, I don't give a damn.

Absur Khan Siam
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Re: BDMO 2017 National round Secondary 4

Unread post by Absur Khan Siam » Wed Jan 24, 2018 11:09 pm

How can it be $AD \perp BC$ and $\angle ADC = 90^o$ ? Is it a valid condition?
"(To Ptolemy I) There is no 'royal road' to geometry." - Euclid

samiul_samin
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Re: BDMO 2017 National round Secondary 4

Unread post by samiul_samin » Thu Jan 25, 2018 12:15 am

I think it not possible to draw a figure according to this condition?

aritra barua
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Re: BDMO 2017 National round Secondary 4

Unread post by aritra barua » Thu Jan 25, 2018 8:44 pm

ahmedittihad meant $AD$ is perpendicular to $CD$.Let $C'$ be the reflection of $C$ in $M$.So,$CC'$=$13$.$C',A,D$ are collinear.Again,[$ABCD$]=[$AMCD$]+[$BMC$]=[$AMCD$]+[$AC'M$]=[$CC'D$].Suppose $BC$=$a$,$CD$=$b$,$AD$=$c$.Since $\bigtriangleup CC'D$ is a right triangle,it suffices to find $b(a+c)$.From Pythagoras,$(a+c)^2+b^2=169$ or $(a+b+c)^2-2(ab+bc)=169$ or $ab+bc=b(a+c)=60$.So,[$CC'D$]=[$ABCD$]=$60/2$=$30$.

samiul_samin
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Re: BDMO 2017 National round Secondary 4

Unread post by samiul_samin » Thu Jan 25, 2018 10:17 pm

aritra barua wrote:
Thu Jan 25, 2018 8:44 pm
ahmedittihad meant $AD$ is perpendicular to $CD$.Let $C'$ be the reflection of $C$ in $M$.So,$CC'$=$13$.$C',A,D$ are collinear.Again,[$ABCD$]=[$AMCD$]+[$BMC$]=[$AMCD$]+[$AC'M$]=[$CC'D$].Suppose $BC$=$a$,$CD$=$b$,$AD$=$c$.Since $\bigtriangleup CC'D$ is a right triangle,it suffices to find $b(a+c)$.From Pythagoras,$(a+c)^2+b^2=169$ or $(a+b+c)^2-2(ab+bc)=169$ or $ab+bc=b(a+c)=60$.So,[$CC'D$]=[$ABCD$]=$60/2$=$30$.
Why$AC'$and $BC$ is equal?I didn't get it.

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