The rate of change in the area of a circle sector

$\textbf{The hour hand moves too}$

Just for completeness I'll include a solution along that line of reasoning.

Set at noon, the initial angle is $0^c$, which makes the computation quite easy.

After $t$ minutes the angle covered by the minute hand is $t \cdot \dfrac{2\pi}{60}=\dfrac{t\pi}{30}$

Note that this is the displacement from the 12 mark. This is necessary to be able to compute the required angle.

After $t$ minutes which is $\dfrac{t}{60}$ hours which is $\dfrac{t}{60} \cdot \dfrac1{12}$ of that whole clock face meaning the hour hand is meant to displace by an angle of:

$\dfrac{t}{60} \cdot \dfrac1{12} \cdot 2\pi=\dfrac{t\pi}{360}$

So the displacement of the minute hand from the hour hand given by a time t:

$=\dfrac{t\pi}{30}-\dfrac{t\pi}{360}=\dfrac{11t\pi}{360}$

The area function to be used, as you correctly gave is:

$\dfrac{r^2}2 \times \theta=\dfrac{5^2}{2} \times \dfrac{11t\pi}{360}=\dfrac{55t\pi}{144}$

Hope I haven't spoiled the ending too much.

You also have to account for the movement of the hour hand, which would decrease the rate of change.

In one minute, the minute hand moves forward $1$ unit and the hour hand moves forward $\frac{t}{12}$ units. Hence, $$\frac{dh}{dt}=\frac{1}{12} \implies h(t)= \frac{t}{12} \\ \frac{dm}{dt} = 1 \implies m(t)= t$$

Now, $60$ units correspond to an arc length of $2\pi \times 5$ in, so $1$ unit corresponds to an arc length of $\frac{\pi}{6}$ in.

The arc length of the sector, $$a(t)= \big(m(t)-h(t)\big)\frac{\pi}{6} = \frac{11\pi}{72} \ t$$ and so

$$\theta(t) = \frac{a(t)}{r} = \frac{11\pi}{72\times 5} \ t$$

Therefore, $$\frac{dA}{dt} = \frac{r^2}{2} \frac{d\theta}{dt} = \frac{55\pi}{144} $$