represent
on the number line
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⇒ Kindly refer to the given attachment file for the figure.
⇒Construct a line AB = 9.3 cm. Now, from point B, draw a line such that BC = 1cm.
⇒Now, find the midpoint of AC = AB + BC = 10.3/2. Now, call that midpoint O.
⇒Now, taking O as centre and 0A as radius, draw a semicircle.
⇒Draw a line perpendicular to B cutting AC at a new point D. Then, BD = √9.3
⇒Now, taking B as centre and BD as radius, draw an arc making a new point on the line.
⇒Then BE = √9.3
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Noah11:
badiya handwriting!
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here is your answer OK dude
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
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
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