BDMO 2017 National round Secondary 4
- ahmedittihad
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$ABCD$ 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.
- ahmedittihad
- Posts:181
- Joined:Mon Mar 28, 2016 6:21 pm
Re: BDMO 2017 National round Secondary 4
This is a typing mistake, it should be $CM=6.5$.
Frankly, my dear, I don't give a damn.
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Re: BDMO 2017 National round Secondary 4
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
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Re: BDMO 2017 National round Secondary 4
I think it not possible to draw a figure according to this condition?
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Re: BDMO 2017 National round Secondary 4
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,[$CC'D$]=[$ABCD$]=$60/2$=$30$.
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Re: BDMO 2017 National round Secondary 4
Why$AC'$and $BC$ is equal?I didn't get it.aritra barua wrote: ↑Thu Jan 25, 2018 8:44 pmahmedittihad 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,[$CC'D$]=[$ABCD$]=$60/2$=$30$.