Question

Locate √5 and √10 on a number line.

Answer 1

Locating √5 on number line :



Here 5 = 2² + 1¹

..............[ See the attached image (I) √5 ]..........

So, draw a line a right angled OAB, in which OA = 2 units and angle OAB = 90°

By using Pythagoras theorem, we get

$OB = \sqrt{ {(OA) }^{2} + {( AB)}^{2} } \\ \\ OB = \sqrt{ {2}^{2} + {1}^{2} } = \sqrt{4 + 1} = \sqrt{5}$

Taking OB = √5 as radius and point Pad centre, draw an arc which meets the number line point P on the positive side of it.

Hence, it is clear that point P represents √5 on the number line.

Now,

Locating √10 on a number line :

Here, 10 = 3² + 1

.............[ See the attached image (II) √10 ]..........

So, draw a right angled OAB, in which OA = 3 units and AB = 1 unit and angle OAB = 90° .

By using Pythagoras theorem, we get

$OB = \sqrt{ {(OA) }^{2} + {( AB)}^{2} } \\ \\ OB = \sqrt{ {3}^{2} + {1}^{2} } = \sqrt{9 + 1} = \sqrt{10}$

Taking OB = √10 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 √10 on the number line.

Answer 2

Answer:

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