integrate 24

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

9 Answers

1
skygirl ·

ans c ??

1
ankit mahapatra ·

solution daldo phir answer ki janch partal karenge.

62
Lokesh Verma ·

good 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

62
Lokesh Verma ·

Area 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

1
ankit mahapatra ·

Thanks.

21
tapanmast Vora ·

did that 180 factr cum frm da fact that angle of sector = 60 degree....
we hav 3 sectrs => 60 *3???

1
ankit mahapatra ·

Adding one more question

integration of

1/1 +√tanx

1
°ღ•๓яυΠ·

i guess thats nt inetrgrable @ our level :P

xcept 4 agar limitz is frm 0 to pi/2

1
ankit mahapatra ·

can we take tanx =1/t2 and solve?

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