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# Question 5 Locate √5,√10 and √17 on the number line.

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Solution

## (i) Here 5=22+12 So, draw a right angled ∆OAB, in which OA = 2 units and AB=1 unit and ∠OAB=90∘ By using Pythagoras theorem, we get OB=√OA2+AB2=√22+12=√4+1=√5 Taking OB=√5 as radius and point O as centre, draw an arc which meets the number line at point P on the positive side of it. Hence, it is clear that point P represents √5 on the number line. (ii) Here, 10=32+1 So, draw a right angled ΔOAB in which OA=3 units and AB = 1 unit and ∠OAB=90∘. By using Pythagoras theorem, we get OB=√OA2+AB2=√32+12=√9+1=√10 Taking OB=√10 as radius and point O as centre, draw an arc which meets the number line at a point P on the positive side of it. The point P represents √10 on the number line. (iii)Here,17=42+1 So, draw a right angled ΔOAB, in which OA = 4 units and AB = 1 unit and ∠OAB=90∘ By using Pythagoras theorem, we get OB=√OA2+AB2=√42+12=√16+1=√17 Taking OB=√17 as radius and point O as centre, draw an arc which meets the number line at point P on the positive side of it. The point P represents √17 on the number line.  Suggest Corrections  9      Similar questions  Related Videos   Irrational Numbers
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