Math, asked by sanjaisanthosh33, 1 year ago

represent
  \sqrt{9.3}
on the number line

Answers

Answered by Anonymous
18
\textbf{Hey mate!}

\textbf{Answer:}

⇒ Kindly refer to the given attachment file for the figure.

\textbf{Steps of Construction}

⇒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

\boxed{Hope\:it\:helps\:you!}

✪ Be Brainly ✪
Attachments:

Noah11: badiya handwriting!
Anonymous: :)) ty bhaiya.. xD
Noah11: bhaiya =_=
Answered by vikram991
1
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
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