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BDMO 2017 National round Secondary 4
Posted: Fri Feb 10, 2017 9:05 pm
by ahmedittihad
$ABCD$ 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 $ABCD$.
Re: BDMO 2017 National round Secondary 4
Posted: Wed Jan 24, 2018 10:09 pm
by ahmedittihad
This is a typing mistake, it should be $CM=6.5$.
Re: BDMO 2017 National round Secondary 4
Posted: Wed Jan 24, 2018 11:09 pm
by Absur Khan Siam
How can it be $AD \perp BC$ and $\angle ADC = 90^o$ ? Is it a valid condition?
Re: BDMO 2017 National round Secondary 4
Posted: Thu Jan 25, 2018 12:15 am
by samiul_samin
I think it not possible to draw a figure according to this condition?
Re: BDMO 2017 National round Secondary 4
Posted: Thu Jan 25, 2018 8:44 pm
by aritra barua
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,[$CC'D$]=[$ABCD$]=$60/2$=$30$.
Re: BDMO 2017 National round Secondary 4
Posted: Thu Jan 25, 2018 10:17 pm
by samiul_samin
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.