Question

Represent √9.3 on the number line and justify your construction.​

Answer 1

Step-by-step explanation:

Steps of Construction

1. On a numberline mark AB = 9.3 unit & BC= 1 unit.

2: Mark O the mid point of AC

3: Draw a semicircle with O as centre & OA as radius

4: At B draw a perpendicular BD.

5: BD = √9.3 unit

6: Now, B as centre, BD as radius, draw an arc, meeting the numberline at E.

Now, with BD or BE = √9.3 as radius , with 0 ( origin) of the number line as centre, draw an arc on the the number line, intersecting at point ‘k’. And this point ‘k' lies between integers 3 & 4, & represents √9.3.

JUSTIFICATION:

$bd = \sqrt{ (\frac{10.3}{2}) {}^{2} - (\frac{8.3}{2}) {}^{2} }$

$= \sqrt{ \frac{(10.3) {}^{2} - (8.3) {}^{2} }{2} }$

$= \sqrt{ \frac{(10.3 + 8.3)(10.3 - 8.3)}{4} }$

$= \sqrt{ \frac{18.6 \times 2}{4} }$

$= \sqrt{ \frac{37.2}{4} }$

$= \sqrt{9.3}$

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