Chapter 10: Areas Related to Circles

Given: radius $$r = 4 \text{ cm}$$, central angle $$\theta = 30^\circ$$; take $$\pi = 3.14$$.

1. Area of the minor sector (angle $$30^\circ$$)

Sector-area formula: $$\text{Area} = \frac{\theta}{360^\circ}\times \pi r^2$$

$$\text{Area} = \frac{30^\circ}{360^\circ}\times \pi\times(4\,\text{cm})^2$$

$$= \frac{30}{360}\times\pi\times16\,\text{cm}^2 = \frac{1}{12}\times16\pi\,\text{cm}^2$$

$$= \frac{4\pi}{3}\,\text{cm}^2$$

Using $$\pi = 3.14$$, $$\text{Area} = \frac{4\times3.14}{3}\,\text{cm}^2 = 4.19\,\text{cm}^2\;\text{(approx)}$$

2. Area of the corresponding major sector

Angle of major sector $$= 360^\circ-30^\circ = 330^\circ$$

$$\text{Area} = \frac{330^\circ}{360^\circ}\times\pi\times(4\,\text{cm})^2 = \frac{11}{12}\times16\pi\,\text{cm}^2$$

$$= \frac{44\pi}{3}\,\text{cm}^2$$

Numerically, $$\text{Area} = \frac{44\times3.14}{3}\,\text{cm}^2 = 46.05\,\text{cm}^2\;\text{(approx)}$$

Hence, area of the $$30^\circ$$ sector is about $$4.19\,\text{cm}^2$$ and area of the corresponding major sector is about $$46.05\,\text{cm}^2$$.

Given data

• Radius of the circle : $$r = 21\text{ cm}$$
• Central angle : $$\angle AOB = 120^{\circ}$$
• Take $$\pi = \dfrac{22}{7}$$ and $$\sqrt3 \approx 1.732$$ (NCERT convention).

The shaded part AYB is a segment of the circle. For any segment,

$$\text{Area of segment} \;=\;\text{Area of the corresponding sector}\;\;-\n\;\text{Area of the triangle formed by the two radii.}$$

We therefore compute the two areas separately.

1. Area of sector $$OAB$$

For a sector subtending angle $$\theta^{\circ}$$ at the centre,

$$\text{Area of sector}=\frac{\theta}{360^{\circ}}\;\pi r^{2}$$

Here $$\theta = 120^{\circ}$$, hence

$$\text{Area of sector}=\frac{120^{\circ}}{360^{\circ}}\times\frac{22}{7}\times(21)^{2}$$

$$=\frac{1}{3}\times\frac{22}{7}\times441$$

First reduce $$441\div7 = 63$$, so

$$=\frac{1}{3}\times22\times63=22\times21=462\text{ cm}^{2}$$

2. Area of triangle $$\Delta OAB$$

The triangle is isosceles with two sides $$OA=OB=r=21\text{ cm}$$ and included angle $$120^{\circ}$$. Using the formula $$\text{Area}=\dfrac12 ab\sin C$$,

$$\text{Area of }\Delta OAB = \dfrac12\times21\times21\times\sin120^{\circ}$$

$$=\dfrac{441}{2}\times\dfrac{\sqrt3}{2}=\dfrac{441\sqrt3}{4}$$

Putting $$\sqrt3\approx1.732$$:

$$\dfrac{441\times1.732}{4}=\dfrac{763.812}{4}\approx190.953\text{ cm}^{2}$$

3. Area of segment AYB

$$\text{Area of segment}=462\text{ cm}^{2}-190.953\text{ cm}^{2}\approx271.047\text{ cm}^{2}$$

Rounded to one decimal place,

Area of the segment AYB ≈ $$271.1\text{ cm}^{2}$$

Given data

• Radius of the circle: $$r = 6\,\text{cm}$$
• Central angle of the sector: $$\theta = 60^{\circ}$$

Step 1 — Recall the area formula for a sector

For a circle of radius $$r$$ and a sector subtending a central angle $$\theta$$ (in degrees), the area of that sector is

$$A_{\text{sector}} = \frac{\theta}{360^{\circ}}\,\pi r^{2}$$

Step 2 — Substitute the given values

Plug $$\theta = 60^{\circ}$$ and $$r = 6\,\text{cm}$$ into the formula:

$$A_{\text{sector}} = \frac{60^{\circ}}{360^{\circ}} \times \pi \times (6\,\text{cm})^{2}$$

Step 3 — Simplify the numerical factor

$$\frac{60^{\circ}}{360^{\circ}} = \frac{1}{6}$$

Step 4 — Square the radius

$$(6\,\text{cm})^{2} = 36\,\text{cm}^{2}$$

Step 5 — Combine the factors

$$A_{\text{sector}} = \frac{1}{6} \times \pi \times 36\,\text{cm}^{2}$$

$$A_{\text{sector}} = 6\pi\,\text{cm}^{2}$$

Step 6 — Write the approximate numerical value (optional)

Using $$\pi \approx 3.14$$,

$$A_{\text{sector}} \approx 6 \times 3.14\,\text{cm}^{2} = 18.84\,\text{cm}^{2}$$

Hence, the area of the sector is $$6\pi\,\text{cm}^{2}$$ (approximately $$18.84\,\text{cm}^{2}$$).

Given: The circumference of the circle is 22 cm.

Let the radius of the circle be $$r$$ cm.

The formula for circumference is $$2\pi r$$, so

$$2\pi r = 22$$

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

$$2 \times \dfrac{22}{7} \times r = 22$$

$$\dfrac{44}{7}\,r = 22$$

$$r = \dfrac{22 \times 7}{44} = \dfrac{7}{2} = 3.5 \text{ cm}$$

Area of the whole circle

$$\pi r^2 = \dfrac{22}{7} \times (3.5)^2$$

$$= \dfrac{22}{7} \times 12.25$$

$$= \dfrac{269.5}{7} = 38.5 \text{ cm}^2$$

Area of one quadrant

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

Hence, the area of the required quadrant is $$9.625\;\mathrm{cm^2}$$.

Given: Length of the minute hand (radius of the circle) $$r = 14 \text{ cm}$$.

The minute hand completes one full revolution, i.e. an angle of $$360^{\circ}$$, in $$60$$ minutes.

In $$5$$ minutes the angle swept is

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

The area traced is therefore the area of a sector of a circle with radius $$r$$ and central angle $$\theta$$.

The area of a sector is given by $$A = \frac{\theta}{360^{\circ}}\times\pi r^{2}$$.

Substituting $$\theta = 30^{\circ}$$ and $$r = 14 \text{ cm}$$,

$$A = \frac{30^{\circ}}{360^{\circ}}\times\pi\times(14\,\text{cm})^{2}$$
$$= \frac{1}{12}\times\pi\times196\,\text{cm}^{2}$$
$$= \frac{196}{12}\pi\,\text{cm}^{2}$$
$$= \frac{49}{3}\pi\,\text{cm}^{2}.$$

Using $$\pi\approx3.14$$,

$$A\approx\frac{49}{3}\times3.14\approx51.3\,\text{cm}^{2}.$$

Thus, the area swept by the minute hand in 5 minutes is $$\dfrac{49\pi}{3}\,\text{cm}^{2}$$ (approximately $$51.3\,\text{cm}^{2}$$).

Given data

• Radius of the circle: $$r = 10 \text{ cm}$$
• Angle subtended at the centre by the chord: $$\theta = 90^\circ$$ (a right angle)
• Take $$\pi = 3.14$$ as instructed.

1. Area of the minor segment

The minor segment is the shaded region bounded by the chord and the arc with central angle $$90^\circ$$. Its area equals

$$\text{(area of the sector)} \; - \; \text{(area of the triangle OAB)}.$$

Area of the sector OAB

For a sector, $$\text{Area} = \dfrac{\theta}{360^\circ}\, \pi r^2$$. Hence

$$\text{Sector area} = \frac{90^\circ}{360^\circ}\, \pi (10)^2 = \frac14 \times 3.14 \times 100 = 3.14 \times 25 = 78.5 \text{ cm}^2.$$

Area of the triangle OAB

Triangle OAB is isosceles with sides $$OA = OB = 10 \text{ cm}$$ and included angle $$90^\circ$$ at O. Using the formula

$$\text{Area} = \dfrac12 \, a b \sin C,$$

we get

$$\text{Triangle area} = \frac12 \times 10 \times 10 \times \sin 90^\circ = 50 \text{ cm}^2.$$

Hence

$$\text{Minor segment area} = 78.5 \text{ cm}^2 - 50 \text{ cm}^2 = 28.5 \text{ cm}^2.$$

2. Area of the major sector

The major sector is subtended by the reflex angle $$360^\circ - 90^\circ = 270^\circ$$.

Again using $$\text{Area} = \dfrac{\theta}{360^\circ}\, \pi r^2$$:

$$\text{Major sector area} = \frac{270^\circ}{360^\circ} \times 3.14 \times 100 = \frac34 \times 3.14 \times 100 = 75 \times 3.14 = 235.5 \text{ cm}^2.$$

Final results

• (i) $$\boxed{28.5 \text{ cm}^2}$$ — area of the minor segment.
• (ii) $$\boxed{235.5 \text{ cm}^2}$$ — area of the major sector.

Given 

• Radius of the circle  $$r = 15\,\text{cm}$$
• Angle subtended by the chord at the centre  $$\theta = 60^\circ$$

Draw the circle with centre O, let the chord be $$\overline{AB}$$ so that $$\angle AOB = 60^\circ$$. The required regions are:

• Minor segment: the region bounded by the chord $$AB$$ and the smaller arc $$\widehat{AB}$$.
• Major segment: the region bounded by the chord $$AB$$ and the larger arc.

1. Area of the sector O A B

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

$$= \dfrac{60^{\circ}}{360^{\circ}} \times 3.14 \times 15^{2} = \dfrac16 \times 3.14 \times 225 = 3.14 \times 37.5 = 117.75\,\text{cm}^2$$

2. Area of triangle O A B

The triangle is isosceles with sides $$OA = OB = 15\,\text{cm}$$ and included angle $$60^{\circ}$$.

$$\text{Area of }\triangle OAB = \dfrac12 r^{2} \sin\theta$$

First, $$\sin 60^{\circ} = \dfrac{\sqrt3}{2} = \dfrac{1.73}{2} = 0.865$$.

$$\therefore \text{Area of }\triangle OAB = \dfrac12 \times 15^{2} \times 0.865 = \dfrac12 \times 225 \times 0.865 = 112.5 \times 0.865 = 97.3125 \approx 97.31\,\text{cm}^2$$

3. Area of the minor segment

$$\text{Minor segment} = \text{Sector} - \triangle OAB$$

$$= 117.75 - 97.31 = 20.44\,\text{cm}^2$$

4. Area of the major segment

Total area of circle:

$$\pi r^{2} = 3.14 \times 225 = 706.5\,\text{cm}^2$$

$$\text{Major segment} = \text{Area of circle} - \text{Minor segment}$$

$$= 706.5 - 20.44 = 686.06\,\text{cm}^2$$

Hence,

• Area of the minor segment  $$\approx 20.44\,\text{cm}^2$$
• Area of the major segment  $$\approx 686.06\,\text{cm}^2$$

Given data

• Radius of the circle: $$r = 12 \text{ cm}$$
• Angle subtended by the chord at the centre: $$\theta = 120^{\circ}$$
• Take $$\pi = 3.14$$ and $$\sqrt{3} = 1.73$$.

We have to find the area of the segment cut off by the chord.

The required segment is the part of the sector of angle $$120^{\circ}$$ that remains after removing the isosceles triangle formed by the two radii and the chord.

1. Area of the sector $$OAB$$ (angle $$120^{\circ}$$)

For a sector of angle $$\theta$$ in a circle of radius $$r$$,

$$\text{Area of sector} = \frac{\theta}{360^{\circ}} \times \pi r^{2}$$

Substituting the given values,

$$\frac{120^{\circ}}{360^{\circ}} \times 3.14 \times (12)^{2} = \frac{1}{3} \times 3.14 \times 144$$

$$= \frac{452.16}{3}$$

$$= 150.72 \text{ cm}^{2}$$

2. Area of the isosceles triangle $$\triangle OAB$$

The two sides $$OA$$ and $$OB$$ are radii, so $$OA = OB = r = 12 \text{ cm}$$, and the included angle is $$120^{\circ}$$.

Using the formula $$\text{Area} = \tfrac12 ab \sin C$$ for $$\triangle OAB$$,

$$\text{Area of } \triangle OAB = \tfrac12 \times 12 \times 12 \times \sin 120^{\circ}$$

We know $$\sin 120^{\circ} = \sin(180^{\circ}-60^{\circ}) = \sin 60^{\circ}$$ and $$\sin 60^{\circ} = \frac{\sqrt{3}}{2}$$.

Hence $$\sin 120^{\circ} = \frac{1.73}{2} = 0.865$$.

Therefore

$$\tfrac12 \times 12 \times 12 \times 0.865 = 72 \times 0.865$$

$$= 62.28 \text{ cm}^{2}$$

3. Area of the corresponding segment

$$\text{Area of segment} = \text{Area of sector} - \text{Area of triangle}$$

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

$$= 88.44 \text{ cm}^{2}$$

Hence, the area of the segment is $$88.44 \text{ cm}^{2}$$.

Given: The umbrella is assumed to be a flat circle of radius $$r = 45\text{ cm}$$ with $$8$$ equally-spaced ribs.

Step 1 – Central angle between two consecutive ribs
The ribs divide the full circle ( $$360^{\circ}$$ ) into $$8$$ equal sectors, so each sector subtends an angle
$$\theta = \frac{360^{\circ}}{8} = 45^{\circ}$$ at the centre.

Step 2 – Area of one sector
The area $$A$$ of a sector of angle $$\theta$$ in a circle of radius $$r$$ is
$$A = \frac{\theta}{360^{\circ}} \times \pi r^{2}$$.
Substituting $$\theta = 45^{\circ}$$ and $$r = 45\text{ cm}$$:
$$A = \frac{45^{\circ}}{360^{\circ}} \times \pi (45\text{ cm})^{2}$$
$$\;\; = \frac{1}{8}\times \pi \times 45^{2}\text{ cm}^2$$
$$\;\; = \frac{1}{8}\times \pi \times 2025\text{ cm}^2$$
$$\;\; = \frac{2025\pi}{8}\text{ cm}^2$$.

Step 3 – Numerical value (optional)
Using $$\pi \approx 3.14$$,
$$A \approx \frac{2025 \times 3.14}{8}\text{ cm}^2$$
$$\;\; = 253.125 \times 3.14\text{ cm}^2$$
$$\;\; \approx 794.8\text{ cm}^2 \approx 7.95\times10^{2}\text{ cm}^2$$.

Hence, the area between two consecutive ribs is $$\dfrac{2025\pi}{8}\text{ cm}^2$$ (approximately $$795\text{ cm}^2$$).

Given data

• Length of each wiper blade (radius of the sweep)  $$r = 25\,\text{cm}$$
• Angular sweep of each blade  $$\theta = 115^{\circ}$$
• The two wipers do not overlap.

1. Area swept by one wiper

The path traced by one blade is a sector of a circle.
For a sector of angle $$\theta$$ (in degrees) and radius $$r$$, the area is

$$A_{\text{sector}} = \frac{\theta}{360^{\circ}} \times \pi r^{2}$$

Substituting the given values,

$$A_{1} = \frac{115}{360} \times \pi \times (25)^{2}$$

$$A_{1} = \frac{115}{360} \times \pi \times 625$$

$$A_{1} = \frac{115 \times 625}{360}\,\pi = \frac{71875}{360}\,\pi \;\text{cm}^{2}$$

Using $$\pi = \dfrac{22}{7}$$ (the value usually prescribed in NCERT):

$$A_{1} = \frac{71875}{360} \times \frac{22}{7}\;\text{cm}^{2}$$

Simplifying,

$$A_{1} = \frac{1\,581\,250}{2\,520}\;\text{cm}^{2} \approx 627.5\;\text{cm}^{2}$$

2. Total area cleaned by the two wipers

The wipers do not overlap, so their swept areas simply add:

$$A_{\text{total}} = 2A_{1}$$

$$A_{\text{total}} = 2 \times 627.5\;\text{cm}^{2} \approx 1\,255\;\text{cm}^{2}$$

3. Conclusion

The two wiper blades together clean about $$1.26 \times 10^{3}\;\text{cm}^{2}$$ (that is, roughly $$1\,255\;\text{cm}^{2}$$) of the windscreen in one complete sweep.

Let the lighthouse be at the centre O of the circular region that receives the warning light.

Given:

• Radius of the illuminated sector, $$r = 16.5\,\text{km}$$.
• Angle of the sector, $$\theta = 80^\circ$$.
• Take $$\pi = 3.14$$.

The area of a sector of a circle is obtained from

$$\text{Area of sector} = \frac{\theta}{360^\circ}\,\pi r^2$$.

Substituting the given values,

$$\text{Area} = \frac{80^\circ}{360^\circ}\times 3.14 \times (16.5\,\text{km})^2$$

Simplify the fraction:

$$\frac{80}{360} = \frac{8}{36} = \frac{2}{9}$$.

Next, compute the square of the radius:

$$r^2 = (16.5)^2 = 272.25\,\text{km}^2$$.

Now evaluate the product:

$$\text{Area} = \frac{2}{9} \times 3.14 \times 272.25$$

First multiply $$3.14$$ and $$272.25$$:

$$3.14 \times 272.25 = 854.865$$.

Then multiply by $$\tfrac{2}{9}$$:

$$\text{Area} = \frac{2 \times 854.865}{9} = \frac{1709.73}{9} = 189.97\,\text{km}^2 \,(\text{approx.})$$

Therefore, the red light from the lighthouse warns ships over an area of about $$190\,\text{km}^2$$ of sea.

Step 1 : Understand the figure
The six identical designs join the centre of the round table cover to six equally–spaced points on the rim. Hence these six points form a regular hexagon whose every side equals the radius of the cover, $$r = 28\text{ cm}$$.

Step 2 : Area of one design
Each design is the triangular region between two consecutive radii; therefore it is an equilateral triangle of side $$a = 28\text{ cm}$$.
For an equilateral triangle, $$\text{Area} = \dfrac{\sqrt3}{4}\,a^{2}$$.

$$\text{Area of one design} = \frac{\sqrt3}{4}\,(28)^2 = \frac{\sqrt3}{4}\times 784 = 196\sqrt3\;\text{cm}^2$$

Step 3 : Area of six designs

$$\text{Total area} = 6 \times 196\sqrt3 = 1176\sqrt3\;\text{cm}^2$$

Given $$\sqrt3 = 1.7$$,

$$\text{Total area} = 1176 \times 1.7 = 1999.2\;\text{cm}^2$$

Step 4 : Cost of making the designs
Rate of making = Rs 0.35 per cm²

$$\text{Cost} = 1999.2 \times 0.35 = 699.72$$

Conclusion
The cost of making the six designs on the round table cover is approximately Rs 700 (exact value Rs 699.72).

Let a circle of radius $$R$$ be drawn.

Step 1 : Area of the whole circle
For the entire circle the angle at the centre is $$360^{\circ}$$ and the area is $$\pi R^{2}$$.

Step 2 : Proportionality of area and angle
For sectors of the same circle, the area is directly proportional to the central angle. Hence, if a sector subtends an angle $$p^{\circ}$$ at the centre, we write

$$\frac{\text{Area of sector}}{\text{Area of full circle}} = \frac{\text{Sector angle}}{\text{Full angle}}$$

Substituting the known values:

$$\frac{\text{Area of sector}}{\pi R^{2}} = \frac{p}{360}$$

Step 3 : Solve for the area of the sector

$$\text{Area of sector} = \frac{p}{360}\,\pi R^{2}$$

Step 4 : Match with the given options
The obtained expression can be rewritten as

$$\frac{p}{360}\,\pi R^{2}=\frac{p}{720}\,(2\pi R^{2}),$$

which is exactly the expression listed in option (D).

Therefore, option (D) is correct.