Question

Represent $\sqrt{10.5}$ on the number line.

Answer 1

To represent $\sqrt{10.5}$ on the number line, follow the following steps of construction:

(i) Mark two points A and B on a given line such that AB = 10.5 units.

(ii) From B, mark a point C on the same given line such that BC = 1 unit.

(iii) Find the mid point of AC and mark it as O.

(iv) With O as centre and radius OC, draw a semi-circle touching the given line at points A and C.

(v) At point B, draw a line perpendicular to AC intersecting the semi-circle at point D.

(vi) With B as centre and radius BD, draw an arc intersecting the given line at point E.

Thus, let us treat the given line as the number line, with B as 0, C as 1, and so on, then point E represents $\sqrt{10.5}$.


Justification:

Here, in semi-circle, radii OA = OC = OD = $\frac{10.5+1}{2}=\frac{{\displaystyle 11.5}}{2}=5.75\mathrm{units}$

And, OB = AB $-$ AO = 10.5 $-$ 5.75 = 4.75 units

In a right angled triangle OBD,

$$\begin{gathered} \mathrm{BD}=\sqrt{{\mathrm{OD}}^2-{\mathrm{OB}}^2} \\ =\sqrt{5.{75}^2-4.{75}^2} \\ =\sqrt{\left(5.75+4.75\right)\left(5.75-4.75\right)}\left[{\mathrm{a}}^2-{\mathrm{b}}^2=\left(\mathrm{a}+\mathrm{b}\right)\left(\mathrm{a}-\mathrm{b}\right)\right] \\ =\sqrt{10.5\times 1} \\ =\sqrt{10.5} \end{gathered}$$