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Question

The number $\sqrt{2}$ is shown on a number line. Steps are given to show $\sqrt{3}$ on the number line using $\sqrt{2}$. Fill in the boxes properly and complete the activity.

Activity:

∙ The point Q on the number line shows the number ........

∙ A line perpendicular to the number line is drawn through the point Q. Point R is at unit distance from Q on the line.

∙ Right angled ∆ ORQ is obtained by drawing seg OR.

∙ l(OQ) = $\sqrt{2}$, l(QR) = 1

Activity:

∙ The point Q on the number line shows the number ........

∙ A line perpendicular to the number line is drawn through the point Q. Point R is at unit distance from Q on the line.

∙ Right angled ∆ ORQ is obtained by drawing seg OR.

∙ l(OQ) = $\sqrt{2}$, l(QR) = 1

∴ by Pythagoras theorem,

[l(OR)]^{2} = [l(OQ)]^{2} + [l(QR)]^{2}

[l(OR)]

= $\overline{){0}}$^{2} + $\overline{){0}}$^{2 }= $\overline{){0}}$ + $\overline{){0}}$

= $\overline{){0}}$

∴ l(OR) = $\overline{){0}}$

∴ l(OR) = $\overline{){0}}$

Draw an arc with centre O and radius OR. Mark the point of intersection of the line and the arc as C. The point C shows the number line $\sqrt{3}$.

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Solution

∙ The point Q on the number line shows the number $\overline{)\sqrt{2}}$.

∙ A line perpendicular to the number line is drawn through the point Q. Point R is at unit distance from Q on the line.

∙ Right angled ∆ ORQ is obtained by drawing seg OR.

∙ l(OQ) = $\sqrt{2}$, l(QR) = 1

∴ by Pythagoras theorem,

[l(OR)]

= ${\overline{)\sqrt{2}}}^{2}+{\overline{)1}}^{2}=\overline{)2}+\overline{)1}$

= 3

∴ l(OR) = $\overline{)\sqrt{3}}$

Draw an arc with centre O and radius OR. Mark the point of intersection of the line and the arc as C. The point C shows the number line $\sqrt{3}$.

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