area

[abhishek.93]

consider a triangle OAB formed by O≡(0,0), A≡(2,0)AND B≡(1,√3). P(x,y) is an arbitrary interior point of the triangle moving in such a way that the sum of its distances from the three sides of the triangle is √3 units . find the area of the region representing possible positions of the point P .

[tiny-tim]

Hi abhishek! :wink:

(this is an equilateral triangle, of course, and one of the possible positions of P is its centroid)

Show us what you've tried, and where you're stuck, and then we'll know how to help! :smile:

[tauon]

the triangle is an equilateral triangle, so the distance from any arbitrary point P inside the triangle to the sides of the triangle is constant (and equal to the height of the triangle, so √3 units in this case).

proof:

consider P, an arbitrary point inside the triangle.

http://img148.imageshack.us/img148/5082/triangleo.png

the area of the triangle OAB is √3 units and it is also the sum of the areas of the triangles POA= (d1*0A)/2, POB=(d2*OB)/2 and PAB=(d3*AB)/2.
but OA=OB=AB=2 units.

So the sum of the areas of the smaller triangles is d1+d2+d3=√3. Hence, the sum of the distances from P to the sides of the triangle is constant and equal to √3.

q.e.d.

so the area of the region representing possible positions of the point P that satisfy the given positions is the area of the triangle = √3 square units...