# Represent 7.28−−−−√ geometrically on the number line.

Question:

Represent $\sqrt{7.128}$ geometrically on the number line.

Solution:

To represent $\sqrt{7.28}$ on the number line, follow the following steps of construction:

(i) Mark two points A and B on a given line such that AB = 7.28 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 $\sqrt{7.28}$.

Justification:

Here, in semi-circle, radii $\mathrm{OA}=\mathrm{OC}=\mathrm{OD}=\frac{7.28+1}{2}=\frac{8.28}{2}=4.14$ units

And, $\mathrm{OB}=\mathrm{AB}-\mathrm{AO}=7.28-4.14=3.14$ units

In a right angled triangle $\mathrm{OBD}$,

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

$=\sqrt{4.14^{2}-3.14^{2}}$

$=\sqrt{(4.14+3.14)(4.14-3.14)} \quad\left[a^{2}-b^{2}=(a+b)(a-b)\right]$

$=\sqrt{7.28 \times 1}$

$=\sqrt{7.28}$