Area Related to Circles: Sure Questions

This lesson covers three exam questions from the Class 10 chapter Area Related to Circles: a shaded region between two sectors, a quadrant with a triangle removed, and the area swept by a clock's minute hand.

Shaded region between two sectors

The shaded region is the larger sector minus the smaller sector, both sharing the same angle $\theta = 30^\circ$. The larger sector $AOB$ has radius $R = 21\ \text{cm}$ and the smaller sector $COD$ has radius $r = 7\ \text{cm}$.

$\text{Area} = \frac{\theta}{360}\,\pi R^2 - \frac{\theta}{360}\,\pi r^2 = \frac{\theta}{360}\,\pi\left(R^2 - r^2\right)$

Substituting the values with $\pi = \tfrac{22}{7}$:

$\text{Area} = \frac{30}{360}\cdot\frac{22}{7}\left(21^2 - 7^2\right) = \frac{1}{12}\cdot\frac{22}{7}\left(441 - 49\right)$

$= \frac{1}{12}\cdot\frac{22}{7}\cdot 392 = \frac{1}{12}\cdot 22 \cdot 56 = \frac{1232}{12} = \frac{308}{3}\ \text{cm}^2$

Quadrant minus a right triangle

Here the shaded region is the quadrant $OACB$ minus the right triangle $BOD$. The quadrant has angle $90^\circ$ and radius $3.5\ \text{cm}$.

Area of the quadrant

$\text{Area}_{OACB} = \frac{90}{360}\cdot\frac{22}{7}\cdot 3.5^2 = \frac{1}{4}\cdot\frac{22}{7}\cdot 12.25 = \frac{1}{4}\cdot 38.5 = 9.625\ \text{cm}^2$

Area of the triangle

With base $3.5\ \text{cm}$ and height $2\ \text{cm}$:

$\text{Area}_{BOD} = \tfrac{1}{2}\,b\,h = \tfrac{1}{2}\cdot 3.5 \cdot 2 = 3.5\ \text{cm}^2$

Shaded region

$\text{Area} = 9.625 - 3.5 = 6.125\ \text{cm}^2$

Area swept by a clock's minute hand

The minute hand is $r = 14\ \text{cm}$ long, and we want the area it sweeps in $5$ minutes. To turn minutes on a clock into degrees, multiply by $6^\circ$ per minute:

$\theta = 5 \times 6^\circ = 30^\circ$

The swept region is a sector:

$\text{Area} = \frac{\theta}{360}\,\pi r^2 = \frac{30}{360}\cdot\frac{22}{7}\cdot 14^2$

$= \frac{1}{12}\cdot\frac{22}{7}\cdot 196 = \frac{1}{12}\cdot 22 \cdot 28 = \frac{616}{12} = \frac{154}{3}\ \text{cm}^2$

Key takeaways

• A shaded region between two sectors with the same angle is $\frac{\theta}{360}\,\pi\left(R^2 - r^2\right)$.
• To find a quadrant with a triangle removed, work out each area separately and subtract.
• On a clock, each minute equals $6^\circ$, so the minute hand sweeps a sector of that angle.