Areas Related to Circles

Using area of a sector,

$\text{Area} = \dfrac{\theta}{360}\times \pi r^2$

Here, $r=6$ cm and $\theta=60^\circ$.

$\text{Area} = \dfrac{60}{360}\times \dfrac{22}{7}\times 6\times 6$

$= \dfrac{1}{6}\times \dfrac{22}{7}\times 36$

$= \dfrac{132}{7} \approx 18.84\text{ cm}^2$

Circumference $=22$ cm, so

$2\pi r = 22$

Using $\pi=\dfrac{22}{7}$,

$r = \dfrac{22}{2\times 22/7} = \dfrac{7}{2} = 3.5$ cm.

Area of circle:

$\pi r^2 = \dfrac{22}{7}\times 3.5\times 3.5 = 38.5\text{ cm}^2$

A quadrant is one-fourth of the circle:

$\text{Area of quadrant} = \dfrac{1}{4}\times 38.5 = 9.625\text{ cm}^2$

But the chapter’s question asks for the area of a quadrant. The correct value is $9.625\text{ cm}^2$, approximately $9.63\text{ cm}^2$.

In 5 minutes, the minute hand covers

$\theta = \dfrac{5}{60}\times 360^\circ = 30^\circ$.

Area swept = area of sector:

$\text{Area} = \dfrac{\theta}{360}\times \pi r^2$

$= \dfrac{30}{360}\times \dfrac{22}{7}\times 14\times 14$

$= \dfrac{1}{12}\times \dfrac{22}{7}\times 196$

$= \dfrac{616}{12} = 51.33\text{ cm}^2$

So the area swept is $51.33\text{ cm}^2$.

Given $r=10$ cm and the angle is $90^\circ$.

### (i) Minor segment
Area of minor sector:

$\dfrac{90}{360}\times 3.14\times 10^2 = \dfrac{1}{4}\times 3.14\times 100 = 78.5\text{ cm}^2$

Area of triangle formed by the two radii:

Since the included angle is $90^\circ$,

$\text{Area of }\triangle = \dfrac{1}{2}\times 10\times 10 = 50\text{ cm}^2$

So, minor segment:

$78.5-50=28.5\text{ cm}^2$

### (ii) Major sector
Major sector angle = $360^\circ-90^\circ=270^\circ$

$\text{Area} = \dfrac{270}{360}\times 3.14\times 10^2 = \dfrac{3}{4}\times 314 = 235.5\text{ cm}^2$

So, the required values are $28.5\text{ cm}^2$ and $235.5\text{ cm}^2$.

Given $r=21$ cm and $\theta=60^\circ$.

### (i) Length of arc
$\text{Arc length} = \dfrac{\theta}{360}\times 2\pi r$

$= \dfrac{60}{360}\times 2\times \dfrac{22}{7}\times 21$

$= \dfrac{1}{6}\times 132 = 22$ cm

### (ii) Area of sector
$\text{Area} = \dfrac{60}{360}\times \dfrac{22}{7}\times 21\times 21$

$= \dfrac{1}{6}\times 1386 = 231\text{ cm}^2$

### (iii) Area of segment
For the triangle with sides 21 cm and included angle $60^\circ$:

$\text{Area of }\triangle = \dfrac{1}{2}r^2\sin 60^\circ$

$= \dfrac{1}{2}\times 21\times 21\times \dfrac{\sqrt{3}}{2}$

Using $\sqrt{3}\approx 1.732$,

$\text{Area} \approx \dfrac{441\times 1.732}{4} \approx 190.94\text{ cm}^2$

So segment area $\approx 231-190.94 = 40.06\text{ cm}^2$.

The textbook’s Example 2 uses $r=21$ cm and $\theta=120^\circ$ for a different segment. The present question asks only for the three results above.

Given $r=15$ cm and $\theta=60^\circ$.

### Minor segment
Area of sector:

$\dfrac{60}{360}\times 3.14\times 15^2 = \dfrac{1}{6}\times 3.14\times 225 = 117.75\text{ cm}^2$

Area of triangle:

$\dfrac{1}{2}r^2\sin 60^\circ = \dfrac{1}{2}\times 225\times \dfrac{1.73}{2}$

$= 112.5\times 0.865 = 97.3125\text{ cm}^2$

Minor segment:

$117.75-97.3125 \approx 20.44\text{ cm}^2$

### Major segment
Major segment = area of circle − minor segment

$= 3.14\times 225 - 20.44 = 706.5 - 20.44 = 686.06\text{ cm}^2$

So the values are approximately $20.44\text{ cm}^2$ and $686.06\text{ cm}^2$.

Given $r=12$ cm and $\theta=120^\circ$.

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:

$\dfrac{1}{2}r^2\sin 120^\circ = \dfrac{1}{2}\times 144\times \dfrac{1.73}{2}$

$= 72\times 0.865 = 62.28\text{ cm}^2$

Area of segment:

$150.72-62.28 = 88.44\text{ cm}^2$

So the area of the corresponding segment is $88.44\text{ cm}^2$.