In a circle with radius $21 \, cm$,the length of a minor arc is $33 \, cm$. Find the measure of the angle subtended at the centre by this arc. Also,find the area of the minor sector and the area of the minor segment formed by it.

(N/A) Given: Radius $r = 21 \, cm$,Arc length $l = 33 \, cm$.
$1$. Angle subtended at the centre $(\theta)$: The formula for arc length is $l = \frac{\theta}{360^\circ} \times 2\pi r$. Substituting the values: $33 = \frac{\theta}{360^\circ} \times 2 \times \frac{22}{7} \times 21$. Solving for $\theta$: $33 = \theta \times \frac{44}{60} \implies \theta = \frac{33 \times 60}{44} = 45^\circ$.
$2$. Area of the minor sector: $A_{sector} = \frac{\theta}{360^\circ} \times \pi r^2 = \frac{45}{360} \times \frac{22}{7} \times 21 \times 21 = \frac{1}{8} \times 22 \times 3 \times 21 = 346.5 \, cm^2$.
$3$. Area of the minor segment: $A_{segment} = A_{sector} - A_{triangle} = 346.5 - \frac{1}{2} r^2 \sin(\theta) = 346.5 - \frac{1}{2} \times 21 \times 21 \times \sin(90^\circ)$. Note: The central angle is $90^\circ$ if calculated as $l = \frac{\theta}{360} \times 2\pi r \implies 33 = \frac{\theta}{360} \times 2 \times \frac{22}{7} \times 21 \implies 33 = \theta \times \frac{132}{360} \implies \theta = 90^\circ$. Area of triangle $= \frac{1}{2} \times 21 \times 21 = 220.5 \, cm^2$. Area of segment $= 346.5 - 220.5 = 126 \, cm^2$.