Question

locate √3 on the number line


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Answer 1

$</p><p> </p><p> <div style="</p><p> padding:50px 5px;</p><p> text-align: center;</p><p> font-family:Monoscape;</p><p> background-image: radial-gradient(circle at center center, rgba(108, 78, 11,0.1) 0%, rgba(108, 78, 11,0.1) 58%,transparent 58%, transparent 81%,rgba(144, 96, 58,0.1) 81%, rgba(144, 96, 58,0.1) 100%),linear-gradient(90deg, hsl(68,81%,84%),hsl(68,81%,84%)); background-size: 22px 22px;</p><p> "></p><p> </p><p> <span style="text-align: center; border-bottom:2px double #460eff; color: #460effff; ">Solution To Your Question &#x2193;</span> </p><p> </p><p> <p style="color: black;</p><p> text-shadow:0px 0px 10px rgb(215,240,204);</p><p> text-align: center;</p><p> font-weight: 500; "></p><p> </p><p> </p><p>First draw a number line having points (at least) [0,3]. If we denote the point 0 as “O” and point 1 as “A” then OA will be equal to 1 unit. Then at point A draw a perpendicular of length AB=1 unit ( equal to the distance from 0 to 1 in the number line i.e. OA). And join the points O and A, so that OAB is a right angled triangle. Then by pythagoras theorem,</p><p> </p><p>OB^2=OA^2+AB^2</p><p> </p><p>=> OB=sqrt (1^2+1^2)</p><p> </p><p>=>OB = root2</p><p> </p><p>Now, again draw a perpendicular at point B of length BC= 1 unit and join the points O and C. Again by pythagoras theorem we get OC = root 3. Then by compus taking radius =OC, draw an arc so that it cuts the number line at D. Then the distance OD will be the square root of 3.</p><p></p><p></p><p> </p><p> &#128513; Hope You Liked It &#128513;</p><p> </p></p><p> </p><p> <marquee style="color:#000;"> <b>Mark Me as the brainliest and Follow Me &#128151;</p><p> </p><p> </b></marquee></p><p> </p><p> </p><p> </p><p> </div></p><p> </p><p> </p><p>$

Answer 2

⦁First construct the BD of unit length perpendicular to OB.

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⦁Then apply the Pythagoras theorem $\rm{OD = \sqrt{(\sqrt{2})^{2} + 1^{2} = \sqrt{3}}}$

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⦁We can locate $\rm{\sqrt{3}}$ on number line by using a compass.

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⦁We know that, O is the center and radius OD. Now let's draw an arc which intersects the number line at the point Q.

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⦁Therefore, now the point Q = $\rm{\sqrt{3}}$