Q. 64.3( 16 Votes )

Let me place √5, √6, √7, -√6, -√8, -√11 on same Number Line.

Answer :

Let the point O represent 0 on the number line.


We place a point A on the number line such that OA = 2 units.


Now, we draw AB perpendicular to OA at point A such that AB = 1unit.


By Pythagoras’ Theorem, we know that,


OB2 = OA2 + AB2


OB2 = 4 + 1 = 5


OB = √5


Now, we draw BC perpendicular to OB at point B such that BC = 1unit.


Again, by Pythagoras’ Theorem, we know that,


OC2 = OB2 + BC2


OC2 = 5 + 1 = 6


OC = √6



We draw CD perpendicular to OC at point C such that CD = 1unit.


Again, by Pythagoras’ Theorem, OD = √7


Again, we draw ED perpendicular to OD at point D such that ED = 1unit.


Again, by Pythagoras’ Theorem, OE = √8


If we extend ED till F such that EF = 2 units and applying Pythagoras’ theorem on ΔODF, we get OF = √11


Now, taking O as a centre and OB, OC, OD, OE and OF as the radius, the arcs are drawn which intersects the number line at G, H, I, K and L.


OG = √5 units, OH = √6 units, OI = √7 units, OJ = -√6 units, OK = -√8 units and OL = -√11 units


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Let me place on Number Line.

West Bengal Mathematics

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