Hlw guys
Answer the question no. 4
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Answers
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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To Represent √9.3 on number line :
Step 1 : Draw a line and mark a point A on it
Step 2 : Mark a point B on the line drawn in Step 1 such that AB = 9.3 cm
Step 3 : Mark a point C on AB produced such that BC = 1 unit.
Step 4 : Find mid-point of AC. Let the mid-point be O.
Step 5 : Taking O as centre and OC = OA as radius draw s semi-circle. Also,draw a line passing through B perpendicular to OB. Suppose it cuts the semi-circle at D
Step 6 : Taking B as the centre and BD as radius drawn an arc cutting OC produced at E.
Here,length BE represents √9.3
[Refer to the attachment!]
