Question

Represent root 9.3 on the number line
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Answer 1

Answer:

Here are the stepwise explanation for the plotting of 9.3 on the number line

Draw a line AB by measuring 9.3cm

From the point, B add 1cm and mark it as C

Mark the point of bisection by a compass and say it as ‘O’

Measure AO which is the radius and draws a semi-circle.

From B draw a perpendicular AB touching the semi-circle and mark as D

Draw an arc on the number line by taking compass pointer on B and pencil on D

The point which intersects the number line is the square root of 9.3

Answer 2

Step-by-step explanation:

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:

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........

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hope it helps you.......