Exercise 11.1: Areas Related to Circles — NCERT Solutions

The area of a sector with central angle $\theta$ is: $\text{Area} = \frac{\theta}{360^\circ} \times \pi r^2$ Here $\theta = 60^\circ$ and $r = 7$ cm. $= \frac{60}{360} \times \frac{22}{7} \times 7 \times 7$ $= \frac{1}{6} \times 22 \times 7 = \frac{154}{6} = 25.67 \text{ cm}^2$ So the area of the sector is approximately $25.67 \text{ cm}^2$.

The length of an arc subtending angle $\theta$ at the centre is: $\text{Arc length} = \frac{\theta}{360^\circ} \times 2\pi r$ Here $\theta = 120^\circ$, $r = 21$ cm. $= \frac{120}{360} \times 2 \times \frac{22}{7} \times 21$ $= \frac{1}{3} \times 2 \times 22 \times 3 = \frac{1}{3} \times 132 = 44 \text{ cm}$ So the arc length is $44$ cm.

Here $r = 10$ cm, $\theta = 90^\circ$.

Area of sector: $= \frac{90}{360} \times 3.14 \times 10^2 = \frac{1}{4} \times 314 = 78.5 \text{ cm}^2$

Area of triangle (with the two radii as sides, angle $90^\circ$): $= \frac{1}{2} \times r \times r \times \sin 90^\circ = \frac{1}{2} \times 10 \times 10 \times 1 = 50 \text{ cm}^2$

Area of minor segment: $= \text{sector} - \text{triangle} = 78.5 - 50 = 28.5 \text{ cm}^2$

The major sector corresponds to angle $360^\circ - 90^\circ = 270^\circ$.

$\text{Area of major sector} = \frac{270}{360} \times \pi r^2$ $= \frac{3}{4} \times \frac{22}{7} \times 14 \times 14$ $= \frac{3}{4} \times 22 \times 28 = \frac{3}{4} \times 616 = 462 \text{ cm}^2$ So the area of the major sector is $462 \text{ cm}^2$.

Here $r = 12$ cm, $\theta = 120^\circ$.

Area of sector: $= \frac{120}{360} \times 3.14 \times 12^2 = \frac{1}{3} \times 3.14 \times 144 = \frac{452.16}{3} = 150.72 \text{ cm}^2$

Area of triangle: $= \frac{1}{2} r^2 \sin 120^\circ = \frac{1}{2} \times 144 \times \frac{\sqrt{3}}{2} = 36\sqrt{3}$ $= 36 \times 1.73 = 62.28 \text{ cm}^2$

Area of minor segment: $= 150.72 - 62.28 = 88.44 \text{ cm}^2$

In $60$ minutes the minute hand sweeps $360^\circ$. So in $10$ minutes it sweeps: $\theta = \frac{360}{60} \times 10 = 60^\circ$ Area swept = area of a sector with $r = 14$ cm, $\theta = 60^\circ$: $= \frac{60}{360} \times \frac{22}{7} \times 14 \times 14$ $= \frac{1}{6} \times 22 \times 28 = \frac{616}{6} = 102.67 \text{ cm}^2$ So the area swept is approximately $102.67 \text{ cm}^2$.

Side of square $= 14$ cm, so area of square: $= 14 \times 14 = 196 \text{ cm}^2$

Each quadrant has radius $7$ cm and angle $90^\circ$. Four quadrants together form one full circle of radius $7$ cm: $\text{Area of 4 quadrants} = \pi r^2 = \frac{22}{7} \times 7 \times 7 = 154 \text{ cm}^2$

Area of shaded region: $= \text{square} - \text{four quadrants} = 196 - 154 = 42 \text{ cm}^2$

The hexagon divides the circle into $6$ equal sectors, each with angle: $\theta = \frac{360^\circ}{6} = 60^\circ$ Each design is the segment = sector − triangle.

Area of one sector ($r = 28$, $\theta = 60^\circ$): $= \frac{60}{360} \times \frac{22}{7} \times 28 \times 28 = \frac{1}{6} \times 22 \times 112 = \frac{2464}{6} = 410.67 \text{ cm}^2$

Area of one triangle (equilateral, side $= r = 28$): $= \frac{\sqrt3}{4}\times 28^2 = \frac{1.73}{4}\times 784 = 1.73 \times 196 = 339.08 \text{ cm}^2$

Area of one segment: $= 410.67 - 339.08 = 71.59 \text{ cm}^2$

Total area of six designs: $= 6 \times 71.59 = 429.54 \text{ cm}^2$

The largest circle inside the square has diameter equal to the side, so: $\text{diameter} = 28 \text{ cm} \Rightarrow r = 14 \text{ cm}$

Area of square: $= 28 \times 28 = 784 \text{ cm}^2$

Area of circle: $= \frac{22}{7} \times 14 \times 14 = 616 \text{ cm}^2$

Area of remaining part: $= 784 - 616 = 168 \text{ cm}^2$

Finding the angle: Sector area $= \frac{\theta}{360}\times \pi r^2$. $9\pi = \frac{\theta}{360} \times \pi \times 6^2$ $9\pi = \frac{\theta}{360} \times 36\pi$ $9 = \frac{36\theta}{360} = \frac{\theta}{10}$ $\theta = 90^\circ$

Length of arc: $= \frac{\theta}{360}\times 2\pi r = \frac{90}{360}\times 2\pi \times 6 = \frac{1}{4}\times 12\pi = 3\pi \text{ cm}$ $\approx 3 \times 3.14 = 9.42 \text{ cm}$

Each wiper cleans a sector with $r = 25$ cm and $\theta = 115^\circ$.

Area cleaned by one wiper: $= \frac{115}{360} \times \frac{22}{7} \times 25 \times 25$ $= \frac{115}{360} \times \frac{22}{7} \times 625$ $= \frac{115 \times 22 \times 625}{360 \times 7} = \frac{1581250}{2520} \approx 627.48 \text{ cm}^2$

Total area for two wipers: $= 2 \times 627.48 = 1254.96 \text{ cm}^2$ So the total area cleaned is approximately $1254.96 \text{ cm}^2$.

Diameter $= 35$ mm, so radius $r = \frac{35}{2} = 17.5$ mm.

Area of full circle: $= \frac{22}{7} \times \frac{35}{2} \times \frac{35}{2} = \frac{22}{7} \times \frac{1225}{4} = \frac{26950}{28} = 962.5 \text{ mm}^2$

Since the circle is divided into $10$ equal sectors: $\text{Area of each sector} = \frac{962.5}{10} = 96.25 \text{ mm}^2$