Areas Related to Circles — Important Questions

Answer: (a) $231\ \text{cm}^2$

Area $=\dfrac{\theta}{360}\times\pi r^2=\dfrac{60}{360}\times\dfrac{22}{7}\times 21\times 21=\dfrac{1}{6}\times 22\times 3\times 21=231\ \text{cm}^2.$

Answer: (a) $2$ units

$2\pi r=\pi r^2\Rightarrow r=2.$

Answer: (a) $10\ \text{cm}$

$\pi R^2=\pi(8)^2+\pi(6)^2\Rightarrow R^2=64+36=100\Rightarrow R=10\ \text{cm}.$

Answer: (a) $9.625\ \text{cm}^2$

The diameter cannot exceed the shorter side, so diameter $=3.5\ \text{cm}$, radius $=1.75\ \text{cm}$. Area $=\dfrac{22}{7}\times(1.75)^2=\dfrac{22}{7}\times 3.0625=9.625\ \text{cm}^2.$

Answer: (a) Both A and R are true and R is the correct explanation of A.

Area $=\dfrac{90}{360}\times 3.14\times 10^2=\dfrac{1}{4}\times 314=78.5\ \text{cm}^2$, exactly what the formula in R gives.

$2\pi r=22\Rightarrow r=\dfrac{22\times 7}{2\times 22}=3.5\ \text{cm}.$

Area of quadrant $=\dfrac{1}{4}\pi r^2=\dfrac{1}{4}\times\dfrac{22}{7}\times 3.5\times 3.5=\dfrac{1}{4}\times 38.5=9.625\ \text{cm}^2.$

In $60$ min the hand turns $360^\circ$, so in $35$ min it turns $\dfrac{35}{60}\times 360^\circ=210^\circ.$

Area $=\dfrac{210}{360}\times\dfrac{22}{7}\times 12\times 12=\dfrac{7}{12}\times\dfrac{22}{7}\times 144=22\times 12=264\ \text{cm}^2.$

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

Area of triangle $OAB=\dfrac{1}{2}r^2\sin 120^\circ=\dfrac{1}{2}\times 144\times\dfrac{\sqrt3}{2}=36\sqrt3=36\times 1.73=62.28\ \text{cm}^2.$

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

Each blade sweeps a sector of radius $25\ \text{cm}$ and angle $115^\circ$.

Area of one sector $=\dfrac{115}{360}\times\dfrac{22}{7}\times 25\times 25=\dfrac{115\times 22\times 625}{360\times 7}=\dfrac{1581250}{2520}=627.48\ \text{cm}^2.$

Total area for two wipers $=2\times 627.48=\dfrac{158125}{126}\approx 1254.96\ \text{cm}^2.$

The centres form an equilateral triangle of side $=3.5+3.5=7\ \text{cm}.$

Area of triangle $=\dfrac{\sqrt3}{4}\times 7^2=\dfrac{1.732}{4}\times 49=21.22\ \text{cm}^2.$

At each vertex the angle is $60^\circ$, so the three sectors together make an angle of $180^\circ$; total sector area $=\dfrac{180}{360}\times\dfrac{22}{7}\times(3.5)^2=\dfrac{1}{2}\times 38.5=19.25\ \text{cm}^2.$

Enclosed area $=21.22-19.25=1.97\ \text{cm}^2.$

Each design is a segment with $\theta=60^\circ,\ r=28\ \text{cm}.$

Area of one sector $=\dfrac{60}{360}\times\dfrac{22}{7}\times 28\times 28=\dfrac{1}{6}\times 2464=410.67\ \text{cm}^2.$

Area of triangle $=\dfrac{\sqrt3}{4}\times 28^2=\dfrac{1.7}{4}\times 784=333.2\ \text{cm}^2.$

Area of one segment $=410.67-333.2=77.47\ \text{cm}^2.$

Six designs $=6\times 77.47=464.8\ \text{cm}^2.$

Cost $=464.8\times 0.35=\text{Rs }162.68.$

Tied at a corner of a square, the horse grazes a quarter circle (angle $90^\circ$).

(i) With rope $=5\ \text{m}$: area $=\dfrac{90}{360}\times 3.14\times 5^2=\dfrac{1}{4}\times 3.14\times 25=19.625\ \text{m}^2.$

(ii) With rope $=10\ \text{m}$ (still less than the side $15\ \text{m}$, so it stays a quarter circle): area $=\dfrac{1}{4}\times 3.14\times 100=78.5\ \text{m}^2.$

Increase $=78.5-19.625=58.875\ \text{m}^2.$

(i) Angle of each slice $=\dfrac{360^\circ}{6}=60^\circ.$

(ii) Area of one slice (sector) $=\dfrac{60}{360}\times\dfrac{22}{7}\times 14\times 14=\dfrac{1}{6}\times 616=102.67\ \text{cm}^2.$

(iii) Arc (crust) length $=\dfrac{60}{360}\times 2\times\dfrac{22}{7}\times 14=\dfrac{1}{6}\times 88=14.67\ \text{cm}.$

Perimeter of one slice $=\text{arc}+2r=14.67+28=42.67\ \text{cm}.$