The problem asks for the radius of a circle given the arc length and the central angle subtended by that arc. We are given:
The formula relating arc length, radius, and central angle requires the angle to be in radians. To convert degrees to radians, we use the conversion factor $\frac{\pi}{180^\circ}$.
$ \theta_{\text{rad}} = 30^\circ \times \frac{\pi}{180^\circ} = \frac{30\pi}{180} = \frac{\pi}{6} \text{ radians} $
The formula for the arc length ($s$) of a circle is:
$ s = r \theta_{\text{rad}} $
Where '$r$' is the radius and '$\theta_{\text{rad}}$' is the central angle in radians.
Now, we substitute the known values into the formula and solve for the radius ($r$):
$ 2 \text{ cm} = r \times \frac{\pi}{6} $
Rearranging the formula to solve for $r$:
$ r = \frac{2 \text{ cm}}{\frac{\pi}{6}} = \frac{2 \times 6}{\pi} \text{ cm} = \frac{12}{\pi} \text{ cm} $
Using the given value $\pi = \frac{22}{7}$:
$ r = \frac{12}{\frac{22}{7}} \text{ cm} = 12 \times \frac{7}{22} \text{ cm} = \frac{84}{22} \text{ cm} $
Simplify the fraction and convert it to a decimal:
$ r = \frac{84 \div 2}{22 \div 2} \text{ cm} = \frac{42}{11} \text{ cm} $
$ r \approx 3.818181... \text{ cm} $
Rounding to two decimal places, the radius is approximately 3.82 cm.
When does a parallelogram become a rectangle?
In the following figure, RS ∥ TU. Find the value of ∠PQR − ∠PRQ, if ∠QTU + ∠PRQ = 130° and ∠RPQ = 80°.

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Two circles of radii 16 cm and 4 cm, respectively, touch each other externally at Point A. PQ is the direct common tangent of these circles with centres C1 and C2, respectively. What is the length of PQ?
Let C be a circle with center O and AB be a chord of C such that the length of AB is equal to the radius of C. Let D be any point on the major arc of AB. Find ∠AOB and ∠ADB, respectively.
The centres of two circles are 84 cm apart. If the radii of these two circles are 38 cm and 26 cm, respectively, then which of the following options gives the length (in cm) of a direct common tangent of these two circles?