Question

Locate √3 on number line with a number line and suitable steps also.

Note:
• Steps should be written in simple words.
• Number line should be shown correctly


NO SPAMS
QUALITY CONTENT REQUIRED​

Answer 1

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We need to locate √3 on the number line. For the number line refer to attachment. The steps of locating would be ,

$\large\red{\dashrightarrow}\underline{\sf Steps :- }$

• Draw a line , and with suitable scale ( here I took 2cm = 1 unit ) , mark the numbers 1 , 2 , 3 ...
• Taking 1 unit from 0 , as centre , construct 90° on it. Again taking 1 unit as radius and A as centre , cut an arc on the 90° line,mark the point as B .
• Join O to B . Now we get a right angled triangle , ∆OAB .

• In ∆ OAB :-

$\sf\dashrightarrow OA^2+ AB^2 = OB^2 \\\\\sf\dashrightarrow (1u)^2 + (1u)^2 = OB^2 \\\\\sf\dashrightarrow 1u^2+1u^2=OB^2 \\\\\sf\dashrightarrow OB^2 = 2u^2 \\\\\sf\dashrightarrow \boxed{\sf \orange{OB =\sqrt2 \ units. }}$

Hence we got , OB = √2 .

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• Now construct a 90° on the line OB taking B as centre.
• Taking B as centre and radius 1u , cut an arc on the 90°. Mark the point as C .
• Join out the points O and C . By using Pythagoras Theorem in ∆OBC we will get OC = √3 u .
• Now taking O as centre , and OC as radius , cut an arc on the number line . Mark the point as M .
• Hence OM represent √3 .

• In ∆ OCB :-

$\sf\dashrightarrow OB^2+ BC^2 = OC^2 \\\\\sf\dashrightarrow (\sqrt2u)^2 + (1u)^2 = OC^2 \\\\\sf\dashrightarrow 2u^2+1u^2=OB^2 \\\\\sf\dashrightarrow OC^2 = 3u^2 \\\\\sf\dashrightarrow \boxed{\sf\orange{ OC =\sqrt3 \ units. }}$

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$\qquad\qquad\qquad\tiny\blacksquare\:\:\boxed{\pink{\sf Hence \ OM \ represents \ \sqrt{3} \:\:units .}}$

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