1solution daldo phir answer ki janch partal karenge.
A point moves inside a triangle formed by A(0,0), B(1,3^0.5), C(2,0) such that min{(PA,PB,PC)}=1,then area bounded by the curve traced by p is:
(a)3^1.5-3pie/2
(b)3^0.5+pie/2
(c)3^0.5-pie/2
(d)3^1.5+3pie/2
source:http://www.goiit.com/posts/list/integration-find-the-area-traced-by-the-curve-4173.htm
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9 Answers
62good one sky..
actually ankit it is an equilateral triangle.
we remove the circular paths from each vertex
so the area is area of triangle - area of a circle/2
62Area of triangle =
√3/4 . (2)^2
√3
Area of circular sector removed
= pi. r2 . (180/360)
= pi. r2/2
r=1 so pi/2
So the area remianing is √3-pi/2
21did that 180 factr cum frm da fact that angle of sector = 60 degree....
we hav 3 sectrs => 60 *3???
1i guess thats nt inetrgrable @ our level :P
xcept 4 agar limitz is frm 0 to pi/2