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Given 3 lines in the plane such that the point of intersection form a triangle with sides of lenght 20,20 and 18,the number of points equidistance from all the 3 lines is

a.)1
b.)4
c.)0
d.)3

Read Solution (Total 4)

Point to remember is that incenter and excenter are the points which are equidistant from the three sides of a triangle. Incenter always lie inside the triangle as it is the center of incircle and the excenters always lie outside the circle as they are center of three excircles.
b)4 (ans)
12 years agoHelpfull: Yes(43) No(3)

in a triangle sum of any 2 sides should be greater than the 3rd side. so 20,20,18 do not constitute a triangle
12 years agoHelpfull: Yes(2) No(23)
i think ans is C) becoz all lines are making a triangle
12 years agoHelpfull: Yes(0) No(16)
I THINK THE ANS TO BE c BEC INTERSECTION OF LINES FORM A TRIANGLE THATS WHY NO POINT IS EQUIDISTANCE FOR ALL 3
12 years agoHelpfull: Yes(0) No(9)

TCS Other Question

A lady has some fine gloves and hats in her closet.19 blue,46 red and 35 yellow.The lights are out and it is totally dark.Inspite of the darkness she can make the diff. b/w a hat and glove.She takes out a item out of the closet only if she is sure that it is a glove.
How many glove must she take out to make sure she has a pair of each colour.

a.)65
b.)83
c.)54
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a.)8
b.)704
c.)11
d.)88