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# A carrom board

Question:

A carrom board ( $4 \mathrm{ft}^{\times} 4 \mathrm{ft}$ square) has the queen at the center. The queen, hit by the striker moves to the front edge, rebounds and goes in the hole behind the striking line. Find the magnitude of displacement of the queen (a) from the center to the front edge, (b) from the front edge to the hole and (c) from the center to the hole.

Solution:

In $\otimes \mathrm{ABC}, \tan (=x / 2$ and in $\otimes \mathrm{DCF}, \tan (=(2-\mathrm{x}) / 4$, So, $(\mathrm{x} / 2)=(2-\mathrm{x}) / 4$. Solving,

$4-2 x=4 x$

Or, $6 x=4$

Or, $x=2 / 3 f t$

(a) In $\otimes \mathrm{ABC}, \mathrm{AC}=\left(\mathrm{AB}^{2}+\mathrm{BC}^{2}\right)^{1 / 2}=2 \sqrt{10 / 3} \mathrm{ft}$

(b) $\ln \otimes C F D, D F=1-(2 / 3)=4 / 3 \mathrm{ft}$

So, $C F=4$ ft. Now, $C D=\left(C^{2}+F D^{2}\right)^{1 / 2}=4 \sqrt{10 / 3} \mathrm{ft}$

(c) In $\otimes A D E, A E=\left(A E^{2}+E D^{2}\right)^{1 / 2}=2 \sqrt{2} \mathrm{ft}$.