Question

represent √9.3 on number line​

Answer 1

Answer:

STEPS: 1: On a numberline mark AB = 9.3 unit & BC= 1 unit. ( & why 1 unit, I've given the reason in comment section)

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:

$\tt⟼BD = \sqrt{ \bigg[ \bigg( \frac{10.3}{2} { \bigg)}^{2} - \bigg( \frac{8.3}{2} { \bigg)}^{2} \bigg ] }$

$\tt⟼BD = \sqrt{\bigg[ \frac{ {10.3 }^{2} - {8.3}^{2} }{4} \bigg] }$

$\tt⟼BD = \sqrt{ \bigg[ \frac{(10.3 + 8.3)(10.3 - 8.3)}{4} \bigg ] }$

$\tt⟼BD \: = \sqrt{\bigg[ \frac{18.6 \times 2}{4} \bigg ]}$

$\tt⟼BF = \sqrt{ \bigg[ \frac{37.2}{4} \bigg] }$

$\tt⟼BD = \sqrt{9.3} = BE$

Answer 2

Answer:

STEPS: 1: On a numberline mark AB = 9.3 unit & BC= 1 unit. ( & why 1 unit, I've given the reason in comment section)

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