Represent√9.3 on the number line
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Posted by Poonam Singh 5 years, 1 month ago
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Gaurav Seth 5 years, 1 month ago
STEPS: 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
0Thank You