A geometry problem

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badass0
Posts:11
Joined:Wed Feb 25, 2015 10:07 pm
A geometry problem

Unread post by badass0 » Sat Feb 28, 2015 9:48 pm

Screenshot from 2015-02-28 21-30-46.jpg
Screenshot from 2015-02-28 21-30-46.jpg (15.23KiB)Viewed 2443 times
$\triangle ABC$ এ $D$ ও $E$ বিন্দুদ্বয় $BC$ বাহূকে সমত্রিখন্ডিত করে। $F$ ও $G$ বিন্দুদ্বয় $AB$ ও $AC$ বাহূকে সমদ্বিখন্ডিত করে।যদি $MNDE$ চতুরভূজের ক্ষেত্রফল $12$ হয় তাহলে $ABC$ ত্রিভুজের ক্ষেত্রফল কত?

tanmoy
Posts:312
Joined:Fri Oct 18, 2013 11:56 pm
Location:Rangpur,Bangladesh

Re: A geometry problem

Unread post by tanmoy » Sun Mar 01, 2015 12:02 pm

Let the length of the side $BC$ is $x$.$\therefore DE=\frac{x} {3}$ and $MN=\frac{x} {6}$.Suppose the height of the triangle $ABC$ is $h$.Now,see that $MNED$ is a trapezoid.$\therefore \frac{1} {2} \times (MN+DE) \times \frac{h} {2}=\frac{1} {2} \times \frac{x} {2} \times \frac{h} {2}=12$.$\therefore \frac{1} {2} \times x \times h =48$.
$\therefore$ The area of triangle $ABC$ is $48$. :)
"Questions we can't answer are far better than answers we can't question"

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