Question

locate √3 on the number line

Answer 1

Draw a line. At a point A on the line construct a line which is perpendicular to the first line. Mark a point C on this line a distance of one unit from A.

Using the compass, draw an arc of a circle with radius 2 and center C so that the arc intersects the first line at B. Draw the line segment CB.

Now you have a right triangle with hypotenuse of length 2 units and one side of length 1 unit. Hence, by Pythagoras' Theorem

|CA| 2 + |AB| 2 = |BC| 2That is1 + |AB| 2 = 4and thus |AB| 2 = 3 and |AB| is the square root of 3.Draw a line. At a point A on the line construct a line which is perpendicular to the first line. Mark a point C on this line a distance of one unit from A.

Using the compass, draw an arc of a circle with radius 2 and center C so that the arc intersects the first line at B. Draw the line segment CB.

Now you have a right triangle with hypotenuse of length 2 units and one side of length 1 unit. Hence, by Pythagoras' Theorem

|CA| 2 + |AB| 2 = |BC| 2That is1 + |AB| 2 = 4and thus |AB| 2 = 3 and |AB| is the square root of 3.

Answer 2

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

ㅤ

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

ㅤ

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

⦁ 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.

ㅤ

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