#### Locate √10 and √17 on the number line.

**Solution:**

We have to locate √10 on the number line.

Expressing 5 as a sum of two perfect square numbers.

10 = (1)² + (3)²

1 + 9 = 10

Consider OA = 3 units

Now draw BA perpendicular to OA

Let BA = 1 units

Now join OB.

Using a compass with centre O and radius OB draw an arc that intersects the number line at the point C.

Now, C = √10

Considering triangle OAB,

By pythagoras theorem,

OB² = OA² + AB²

OB² = (3)² + (1)²

OB² = 9 + 1

OB² = 10

Taking square root,

Therefore, OB = √10

We have to locate √17 on the number line.

Expressing 5 as a sum of two perfect square numbers.

17 = (4)² + (1)²

16 + 1 = 17

Consider OA = 4 units

Now draw BA perpendicular to OA

Let BA = 1 units

Now join OB.

Using a compass with centre O and radius OB draw an arc that intersects the number line at the point C.

Now, C = √5

Considering triangle OAB,

By pythagoras theorem,

OB² = OA² + AB²

OB² = (4)² + (1)²

OB² = 16 + 1

OB² = 17

Taking square root,

Therefore, OB = √17

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