Question

A circular park with radius $20m$ is situated in a colony. Three boys Ankur, Syed and David are sitting at equal distance on its boundary each having a toy telephone in this hands to talk each other. The length of the string of each phone is

• A. $20m$
• B. $17.32m$
• C. $34.64m$
• D. $30m$

Answer 1

Answer: (C)

$34.64m$

Let us denote the positions of Ankur as $A$, that of Syed as $S$ and that of David as $D$

So, $AS=SD=DA.$

$\therefore \Delta ASD$ is an equilateral triangle.

Let the centre of the park be O.

Then, we have a circle circumscribing an equilateral $△ASD$, the radius being $OA=20m.$

The length of the telephone wire= $AS=SD=DA.$

AO is joined and is produced to meet DS at N.

Now, we know that the centriod of an equilateral triangle coincides with the centre of its circumcircle and the median of the triangle contains the radius.

$\therefore AN$ is the median as well as the height of $\Delta ASD$ and O is the centroid of $\Delta ASD.$

So, $AN=\frac{3}{2}OA=\frac{3}{2}\times 20m=30m$

($OA:ON=2:1$, when O is the centroid and $AN=OA+ON$ is the median of any triangle).

Now, side a of an equilateral triangle $=\frac{2\times height}{\sqrt{3}}.$

Here, $a=AS=SD=DA=\frac{2\times 30}{\sqrt{3}}m=20\sqrt{3}m=34.64m.$