# Represent 4.7−−−√ geometrically on the number line.

Question:

Represent $\sqrt{4 . \overline{4}}$ geometrically on the number line.

Solution:

To represent $\sqrt{4.7}$ on the number line, follow the following steps of construction:

(i) Mark two points A and B on a given line such that AB = 4.7 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 $\sqrt{4.7}$.

Justification:

Here, in semi-circle, radii $\mathrm{OA}=\mathrm{OC}=\mathrm{OD}=\frac{4.7+1}{2}=\frac{5.7}{2}=2.85$ units

And, $\mathrm{OB}=\mathrm{AB}-\mathrm{AO}=4.7-2.85=1.85$ units

In a right angled triangle OBD,

$\mathrm{BD}=\sqrt{\mathrm{OD}^{2}-\mathrm{OB}^{2}}$

$=\sqrt{2.85^{2}-1.85^{2}}$

$=\sqrt{(2.85+1.85)(2.85-1.85)} \quad\left[a^{2}-b^{2}=(a+b)(a-b)\right]$

$=\sqrt{4.7 \times 1}$

$=\sqrt{4.7}$