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Thus, take the ratio. $\frac{\pi (xR)^2}{\pi R^2 \frac{\pi}{12 \pi}}$ where $R$ is the radius of the hour hand.

$\implies \frac{S_m}{S_h} =12x^2$

Since the solution to the problem does not depend on the interval of time, we may choose a particularly easy interval. Also, since the minute hand is $x$ times bigger and we're looking at a ratio, we may assume the hour hand to be of length $1$ and the minute hand to be of length $x$ .

An amount of time that seems particularly easy, is that of 12h. The hour hand made exactly one turn, so the area swept by it, is $S_h=\pi$ . In the same interval, the minute hand has gone round exactly $12$ times, so the area swept is $S_m=12\cdot\pi x^2$ . The quotient of these is $12 x^2$ , so $k=12$ .

Call the length of the hour hand $a$ . Then the length of the minute hand is $ax$ . For any given fixed interval of time (Also assuming that the clock is circular), we have that the hour hand revolves at $\frac{1}{12}$ the rate which the minute hand moves, i.e. if the minute hand revolves an angle $\theta$ over a time interval, then the hour hand revolves $\frac{\theta}{12}$ . As a a result the area covered by the minute hand for a given fixed interval of time is:

$\large \frac{S_m}{S_h}=\frac{\frac{\theta}{2 \pi} \pi a^2 x^2} {\frac{\theta}{24 \pi} \pi a^2 } =12x^2$ .

Therefore $\large k = \boxed{12}$ .

Imagine that an hour has passed. Then, let's call the hour hand $a$ . Then, the minute hand is equal to $ax$ . If an hour has passed, then the area swept by the minute hand is equal to $\pi(ax)^{2}$ . This because we know that in an hour the minute hand forms a circle with radii $ax$ . But while the minute hand turns to it's original position, the hour hand just moves one number to the right. As there are 12 hours in a clock, then the hour hand is giving $\frac {1}{12}$ of a circle of radii $a$ (remember that $a$ is the hour hand). So the area swept by the hour hand is $\frac {\pi(a^{2})}{12}$ .

Hence, $\frac {S_{m}}{S_{h}} = \frac {\pi(ax)^{2}}{\frac {\pi(a^{2})}{12}} = 12x^{2}$ so $k = \boxed {12}$ .

1 hour is equal to 60 minutes, that means if minute hand move 2 pi rad, hour hand will move 1/6 pi rad. $S_{m}=\pi(r_{m})^2$ $S_{h}=\frac{\pi(r_{h})^2}{2\pi}\times \frac{\pi}{6}$

$\frac{S_{m}}{S_{h}}=\frac{\pi(r_{m})^{2}}{\pi(r_{h})^{2}}\times 12$ $\frac{S_{m}}{S_{h}}=(\frac{r_{m}}{r_{h}})^{2}\times 12$ $\frac{S_{m}}{S_{h}}=(\frac{xr_{h}}{r_{h}})^{2}\times 12$ $\frac{S_{m}}{S_{h}}=x^{2}\times 12$ $x^{2}\times 12=kx^{2}$ $\boxed{12}=k$

For 12hr. period: Sm = 12 X Pi X (x^2) Sn = Pi X 1 (Where 1 is the length of the hour hand - Assumed) Sm/Sn = 12 X (x^2) = k(x^2) Therefore k = 12

As an Example, consider the time interval between 12:00 noon and 6:00 PM. Let the length of hour hand be "l". Therefore, length of minute hand is "xl". During the above time interval, area covered by hour hand= 0.5(pi)(l)^2 . And, area covered by minute hand= 6 (pi)(xl)^2 Thus ratio of these two values is: 12x^2, hence ans is 12.

Let the length of hour hand is "h". Therefore length of the minute hand is hx. So say the time we are considering is one hour. So area swept by minute hand in one hour,S {m}is (22/7)hx^{2}.[This is because it will make a complete 360 degrees angle in one hour]. but in one hour the hour hand will move only 30 degrees . So area covered by the hour hand in one hour,S {h} is (30/360)(22/7)h^{2}=[(22/7)h^{2}]12 Therefore S {m}/S {h}=[(22/7)hx^{2}]/[(22/7)h^{2}]=12x^{2}. But by the problem S {m}/S {h}=kx^{2} Therefore kx^{2}=12x^{2}, so k=12

Let radius of minute hand be m & radius of hour hand be h and according to the question h = \ frac{m}{k} for some constant k. Now area {1} swept by minute hand in x minutes = \ frac{22}{7} m^{2} \ times \ frac{x}{360} & area {2} swept by hour hand = \ frac{22}{7} \ frac{m}{k}^{2} \ times \ frac{\ frac{x}{12}}{360}. Now \ frac{a {1}}{a {2}} = 12 \ times k^{2} . So, constant =12