InterviewSolution
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InterviewSolution
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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} meansb 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 |
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Answer» `SQRT(3)`sq,units  `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 |
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