Consider the traingle having vertices `O(0,0),A(2,0)`, and `B(1,sqrt3)`. Also `b le"min" {a_1,a_2,a_3....a_n}` means ` b le a_1` when `a_1` is least, `b le a_2` when `a_2` is least, and so on. Form this, we can say `b le a_1,b le a_2,.....b le a_n`.
Let R be the region consisting of all those points P inside `DeltaOAB` which satisfy `d(P,OA)le "min"[d(P,OB),d(P,AB)]`, where d denotes the distance from the point to the corresponding line. then the area of the region R is

Options

A. `sqrt(3)`sq,units
B. `(2+sqrt3)` sq.units
C. `sqrt(3)//2` sq.units
D. `1//sqrt(3)` sq.units

Answer

D

Explanation

`d(P,OA)le min[d(P,OB),d(P,AB)]`
or `d(P,OA)le(P,OB)`
and `d(P,OA0)le d(P,AB)`
When `d(P,OA)P` is equidistant from OA and OB, or P lies on the bisector of lines OA and OB.
Hence, when `d(P,OA)le d(P,OB)`, point P is nearer to OA than to OB or lies below the angle bisector of `angleAOB`. Similarly, when `d(P,OA)led(P,AB)P` is nearer to OA than to AB, or P lies below the bisector of `angle OAB and AB`. Therefore, the required area is equal to the area of `DeltaOIA`. Now,
`tanangleBOA=(sqrt3)/(1)=sqrt3`
or `angle BOA=60^@`
Hence, the triangle is equilateral. then I coincides. with the centroid which is `(1,1//sqrt3)`.
Therefore,the area of `Delta OIA` is
`(1)/(2) OAxxIM=(1)/(2)xx2xx(1)/(sqrt3)=(1)/(sqrt3)`sq.units