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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