Question

locate on number line
$\sqrt{29}$
$\sqrt{9.3}$

Answer 1

: 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 = √ {(10.3/2)² - (8.3/2)²}

=> BD = √{10.3²- 8.3²)/4 }

=> BD = √{(10.3+8.3)(10.3–8.3)/4}

=> BD = √{18.6*2/4}

=> BF = √{37.2/4}

=> BD = √9.3 = BE

This is√9•3 answer