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CBSE Class 9 Answered

<p>how to solve root 10 and 7.3 on number line</p> <p>&nbsp;</p>
Asked by gaurav72001 | 17 May, 2014, 03:22: PM
Expert Answer
Consider a number line. Let the point O represent 0 and point A represent 3.
Now draw a perpendicular AP at A on the number line and cut off arc AB = 1 unit.
Using Pythagoras Theorem, we have
O B squared equals O A squared plus A B squared rightwards double arrow O B squared equals open parentheses 3 close parentheses squared plus 1 squared equals 10 rightwards double arrow O B equals square root of 10
With O as the centre and O B equals square root of 10 as radius draw an arc cutting real line at C. Clearly, O C equals O B equals square root of 10.
Thus, C represents square root of 10on the number line.
Now let us find the square root of 7.3
Make a list of perfect squares, 0, 4, 9, 16, 25, 36, ....
Identify perfect squares closest to 7.3
Thus we have, 4 < 7.3 < 9
Take a positive square root of each number.
Thus, we have square root of 4 less than square root of 7.3 end root less than square root of 9
Now evaluate the square roots.
Therefore, we have, 2 less than square root of 7.3 end root less than 3
Now make a list of perfect squares between 2 and 3.
Numbers Squares
2.1 4.41
2.2 4.84
2.3 5.29
2.4 5.76
2.5 6.25
2.6 6.76
2.7 7.29
2.8 7.84
2.9 8.41
Thus, from the list it is clear that, 7.29 is closer to 7.3.
Therefore, we have, square root of 7.3 end root almost equal to square root of 7.29 end root equals 2.7
Thus, we can mark 2.7 on the number line as follows:

Thus, OA represents the square root of 7.3
Answered by Vimala Ramamurthy | 21 May, 2014, 10:02: AM
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