This is not cool actually!

For discussing Olympiad level Geometry Problems
Naheed
Posts:20
Joined:Sun Dec 16, 2012 11:10 pm
This is not cool actually!

Unread post by Naheed » Wed Jan 08, 2014 1:10 pm

চিত্রে একটি বর্গের কর্ণকে ভূমি ধরে একটি সমবাহু ত্রিভুজ আঁকা হলো। শুধু বর্গটির ক্ষেত্রফল ২ হলে, ABCD ক্ষেত্রের ক্ষেত্রফলকে a+b√c আকারে প্রকাশ করা যায়। (a+b+c) এর মান কত?

In the figure, an equilateral triangle has been drawn
taking the diagonal as base and its area is 2. The area
of ABCD can be written as a+b√c. Find out the value
of (a+b+c)?

চিত্রটি এ্যাটাচ করে দেওয়া হলো।
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Tahmid
Posts:110
Joined:Wed Mar 20, 2013 10:50 pm

Re: This is not cool actually!

Unread post by Tahmid » Thu Jan 09, 2014 7:35 pm

area of square is 2 ....... so, AC=2.
$\Delta ABC= 2^{2} \times \frac{\sqrt{3}}{4}= \sqrt{3}$ [becaus ABC is equilateral]
$\Delta ADC=\frac{2}{2}=1$
$ABCD= \sqrt{3}+1= 1+(1 \times \sqrt{3})= a+b\sqrt{c}$
$a+b+c= 1+1+3=5$ .......so ans:5 :)

Naheed
Posts:20
Joined:Sun Dec 16, 2012 11:10 pm

Re: This is not cool actually!

Unread post by Naheed » Thu Jan 30, 2014 3:07 pm

How did you find the value of ABC= 2X2X √3/4 ?

Neblina
Posts:18
Joined:Sun Feb 06, 2011 8:38 pm

Re: This is not cool actually!

Unread post by Neblina » Thu Jan 30, 2014 6:18 pm

Naheed wrote:How did you find the value of ABC= 2X2X √3/4 ?
Area of triangle= 0.5*ab*sin x
As the triangle is equilateral all 3 angles are equal so =60
sin 60=$$\frac{\sqrt{3}}{2}$$
AC=2, so ABC=$$\frac{1}{2}$$*2*2*$$\frac{\sqrt{3}}{2}$$

Naheed
Posts:20
Joined:Sun Dec 16, 2012 11:10 pm

Re: This is not cool actually!

Unread post by Naheed » Thu Jan 30, 2014 7:13 pm

Thanks for the solution.

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