Areas Related to Circles

Find the area of a sector of a circle with radius 6 cm if angle of the sector is 60°.

Answer

We know that,

Area of sector of angle θ = $\dfrac{θ}{360°} \times πr^2$

Substituting values we get :

$\Rightarrow \text{Area } = \dfrac{60°}{360°} \times \dfrac{22}{7} \times 6^2 \\[1em] = \dfrac{1}{6} \times \dfrac{22}{7} \times 6^2 \\[1em] = \dfrac{132}{7} \text{ cm}^2.$

Hence, area of sector of circle with radius 6 cm and angle of sector = 60° is $\dfrac{132}{7}$ cm².

Find the area of a quadrant of a circle whose circumference is 22 cm.

Answer

Given,

Circumference = 22 cm

⇒ 2πr = 22

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

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

A quadrant of a circle means one of the four equal parts.

∴ The area of a quadrant = $\dfrac{1}{4} \times$ Area of circle

$= \dfrac{1}{4} \times πr^2 \\[1em] = \dfrac{1}{4} \times \dfrac{22}{7} \times (3.5)^2 \\[1em] = \dfrac{1}{4} \times 22 \times 0.5 \times 3.5 \\[1em] = \dfrac{38.5}{4} \\[1em] = \dfrac{385}{40} \\[1em] = \dfrac{77}{8} \text{ cm}^2.$

Hence, area of quadrant = $\dfrac{77}{8}$ cm².

The length of the minute hand of a clock is 14 cm. Find the area swept by the minute hand in 5 minutes.

Answer

Length of minute hand can be considered as radius (r) = 14 cm.

In 60 minutes the minute hand rotates 360°.

So, in 5 minutes, it will rotate = $\dfrac{360°}{60} \times 5 = 30°$.

We know that,

Area of sector of angle θ and radius r = $\dfrac{θ}{360°} \times πr^2$

Substituting values we get :

$\Rightarrow \text{Area } = \dfrac{30°}{360°} \times \dfrac{22}{7} \times 14^2 \\[1em] = \dfrac{1}{12} \times 22 \times 2 \times 14 \\[1em] = \dfrac{154}{3} \text{ cm}^2.$

Hence, the area swept by the minute hand in 5 minutes = $\dfrac{154}{3}$ cm².

A chord of a circle of radius 10 cm subtends a right angle at the centre. Find the area of the corresponding :

(i) minor segment

(ii) major sector.

(Use π = 3.14)

Answer

(i) Let AB be the chord subtending right angle at the center.

Given,

Radius (r) = 10 cm

From figure,

APB is the minor segment.

We know that,

Area of sector of angle θ and radius r = $\dfrac{θ}{360°} \times πr^2$

Area of right angle triangle = $\dfrac{1}{2}$ × base × height

Area of minor segment APB = Area of sector AOBP - Area of right angle triangle AOB

Substituting values we get :

$= \dfrac{90°}{360°} \times 3.14 \times 10^2 - \dfrac{1}{2} \times AO \times BO \\[1em] = \dfrac{1}{4} \times 3.14 \times 100 - \dfrac{1}{2} \times r \times r \\[1em] = \dfrac{314}{4} - \dfrac{1}{2} \times 10 \times 10 \\[1em] = 78.5 - 50 \\[1em] = 28.5 \text{ cm}^2.$

Hence, area of corresponding minor segment = 28.5 cm².

(ii) Angle subtended by major sector = 360° - 90° = 270°.

We know that,

Area of sector of angle θ and radius r = $\dfrac{θ}{360°} \times πr^2$

Substituting values we get :

$\Rightarrow \text{Area of major sector } = \dfrac{270°}{360°} \times 3.14 \times 10^2 \\[1em] = \dfrac{3}{4} \times 314 \\[1em] = 3 \times 78.5 \\[1em] = 235.5 \text{ cm}^2.$

Hence, area of major sector = 235.5 cm².

In a circle of radius 21 cm, an arc subtends an angle of 60° at the centre. Find:

(i) the length of the arc

(ii) area of the sector formed by the arc

(iii) area of the segment formed by the corresponding chord

Answer

(i) We know that,

Length of an arc of sector of angle θ = $\dfrac{θ}{360°} \times$ 2πr

Substituting values we get :

$\text{Length of arc APB} = \dfrac{60°}{360°} \times 2 \times \dfrac{22}{7} \times 21 \\[1em] = \dfrac{1}{6} \times 2 \times 22 \times 3 \\[1em] = 22 \text{ cm}.$

Hence, length of arc = 22 cm.

(ii) We know that,

Area of sector of angle θ and radius r = $\dfrac{θ}{360°} \times πr^2$

Substituting values we get :

$\Rightarrow \text{Area of sector AOBP} = \dfrac{60°}{360°} \times \dfrac{22}{7} \times 21^2 \\[1em] = \dfrac{1}{6} \times 22 \times 3 \times 21 \\[1em] = 11 \times 21 \\[1em] = 231 \text{ cm}^2.$

Hence, area of sector = 231 cm².

(iii) In triangle OAB,

OM ⊥ AB

In triangle OAM and OMB,

∠OMA = ∠OMB = 90°

OA = OB (Radius of same circle)

OM = OM (Common)

∴ △OAM ≅ △OMB (By RHS axiom)

∴ ∠MOB = ∠MOA = $\dfrac{60°}{2}$ = 30°. [By C.P.C.T.]

In △MOB,

⇒ sin 30° = $\dfrac{MB}{OB}$

⇒ $\dfrac{1}{2} = \dfrac{MB}{r}$

⇒ MB = $\dfrac{r}{2} = \dfrac{21}{2}$ = 10.5 cm

By C.P.C.T.

MA = MB = 10.5 cm

AB = MA + MB = 21 cm.

Since, OA = OB = AB.

∴ △AOB is an equilateral triangle.

We know that,

Area of equilateral triangle = $\dfrac{\sqrt{3}}{4} \times$ (Side)²

Substituting values we get :

$\text{Area of △AOB} = \dfrac{\sqrt{3}}{4} \times 21^2 \\[1em] = \dfrac{441\sqrt{3}}{4} \text{ cm}^2.$

From figure,

Area of segment APB = Area of sector AOBP - Area of triangle AOB

= $\Big(231 - \dfrac{441\sqrt{3}}{4} \text{ cm}^2\Big).$

Hence, area of segment APB = $\Big(231 - \dfrac{441\sqrt{3}}{4} \text{ cm}^2\Big).$

A chord of a circle of radius 15 cm subtends an angle of 60° at the centre. Find the areas of the corresponding minor and major segments of the circle.

(Use π = 3.14 and $\sqrt{3}$ = 1.73)

Answer

The circle of radius 15 cm with its chord AB subtending an angle of 60° at the centre is shown in the figure below:

In triangle OAB,

OM ⊥ AB

In triangle OAM and OMB,

∠OMA = ∠OMB = 90°

OA = OB (Radius of same circle)

OM = OM (Common)

∴ △OAM ≅ △OMB (By SAS axiom)

∴ ∠MOB = ∠MOA = $\dfrac{60°}{2}$ = 30°. [By C.P.C.T.]

In △MOB,

⇒ sin 30° = $\dfrac{MB}{OB}$

⇒ $\dfrac{1}{2} = \dfrac{MB}{OB}$

⇒ MB = $\dfrac{OB}{2} = \dfrac{15}{2}$ = 7.5 cm

By C.P.C.T.

MA = MB = 7.5 cm

AB = MA + MB = 15 cm.

Since, OA = OB = AB.

∴ △AOB is an equilateral triangle.

By formula,

Area of equilateral triangle = $\dfrac{\sqrt{3}}{4} \times$ (side)²

Area of sector of angle θ and radius r = $\dfrac{θ}{360°} \times πr^2$

Area of minor segment APB = Area of sector BOAP - Area of triangle AOB

Substituting values we get :

$\text{Area of minor segment APB} = \dfrac{60°}{360°} \times 3.14 \times 15^2 - \dfrac{\sqrt{3}}{4} \times 15^2 \\[1em] = \dfrac{1}{6} \times 3.14 \times 225 - \dfrac{225 \times 1.73}{4} \\[1em] = 117.75 - 97.3125 \\[1em] = 20.4375 \text{ cm}^2.$

Area of circle = πr²

= 3.14 × 15²

= 3.14 × 225

= 706.5 cm².

From figure,

Area of major segment AQB = Area of circle - Area of minor segment APB

= 706.5 - 20.4375 = 686.0625 cm².

Hence, area of minor segment = 20.4375 cm² and area of major segment = 686.0625 cm².

A chord of a circle of radius 12 cm subtends an angle of 120° at the centre. Find the area of the corresponding segment of the circle.

(Use π = 3.14 and $\sqrt{3}$ = 1.73)

Answer

Let AB be the chord subtending angle 120° at the center.

Draw OM ⊥ AB.

In triangle OAM and OMB,

∠OMA = ∠OMB = 90°

OA = OB (Radius of same circle)

OM = OM (Common)

∴ △OAM ≅ △OMB (By RHS axiom)

∴ ∠MOB = ∠MOA = $\dfrac{120°}{2}$ = 60°. [By C.P.C.T.]

In △MOB,

$\Rightarrow \text{sin 60°} = \dfrac{MB}{OB} \\[1em] \Rightarrow \dfrac{\sqrt{3}}{2} = \dfrac{MB}{OB} \\[1em] \Rightarrow MB = \dfrac{OB\sqrt{3}}{2} \\[1em] \Rightarrow MB = \dfrac{12 \times 1.73}{2} = 10.38 \text{ cm} \\[1em] \Rightarrow \text{tan 60°} = \dfrac{MB}{MO} \\[1em] \Rightarrow \sqrt{3} = \dfrac{10.38}{MO} \\[1em] \Rightarrow MO = \dfrac{10.38}{\sqrt{3}} = \dfrac{10.38}{1.73} = 6 \text{ cm}.$

By C.P.C.T.

MA = MB = 10.38 cm

AB = MA + MB = 20.76 cm.

We know that,

Area of triangle = $\dfrac{1}{2} \times \text{ base} \times \text{height}$

Substituting values we get :

Area of triangle OAB = $\dfrac{1}{2} \times AB \times OM$

= $\dfrac{1}{2} \times 20.76 \times 6$ = 62.28 cm².

We know that,

Area of sector of angle θ and radius r = $\dfrac{θ}{360°} \times πr^2$

Area of minor segment APB = Area of sector BOAP - Area of triangle AOB

Substituting values we get,

$\Rightarrow \text{Area of minor segment APB} = \dfrac{120°}{360°} \times 3.14 \times 12^2 - 62.28 \\[1em] = \dfrac{1}{3} \times 3.14 \times 144 - 62.28 \\[1em] = 150.72 - 62.28 \\[1em] = 88.44 \text{ cm}^2.$

Hence, area of corresponding segment of the circle = 88.44 cm².

A horse is tied to a peg at one corner of a square shaped grass field of side 15 m by means of a 5 m long rope. Find :

(i) the area of that part of the field in which the horse can graze.

(ii) the increase in the grazing area if the rope were 10 m long instead of 5 m.

(Use π = 3.14)

Answer

(i) From figure,

It is the quadrant of radius 5 m in which horse can graze.

We know that,

Area of quadrant = $\dfrac{1}{4} \times$ Area of circle

Substituting value we get :

$\text{Area of quadrant} = \dfrac{1}{4} \times πr^2 \\[1em] = \dfrac{1}{4} \times 3.14 \times 5^2 \\[1em] = \dfrac{1}{4} \times 3.14 \times 25 \\[1em] = 19.625 \text{ m}^2.$

Hence, area of field in which horse can graze = 19.625 m².

(ii) If rope would be 10 m long, then gazing area of horse would be a quadrant of 10 m.

$\text{Area } = \dfrac{1}{4} \times 3.14 \times 10^2 \\[1em] = \dfrac{1}{4} \times 3.14 \times 100 \\[1em] = \dfrac{314}{4} = 78.5 \text{ m}^2.$

Increase in grazing area = 78.5 - 19.625 = 58.875 m².

Hence, increase in grazing area = 58.875 m².

A brooch is made with silver wire in the form of a circle with diameter 35 mm. The wire is also used in making 5 diameters which divide the circle into 10 equal sectors as shown in the figure. Find :

(i) the total length of the silver wire required.

(ii) the area of each sector of the brooch.

Answer

(i) Given,

Diameter = 35 mm

Radius (r) = $\dfrac{\text{Diameter}}{2} = \dfrac{35}{2}$ = 17.5 mm

From figure,

Length of silver required = Circumference of circle + 5 × Diameter

= 2πr + 5 × 35

= $2 \times \dfrac{22}{7} \times 17.5$ + 175

= 2 × 22 × 2.5 + 175

= 110 + 175 = 285 mm.

Hence, total length of silver wire required = 285 mm.

(ii) There are ten sectors.

Angle subtended by each sector = $\dfrac{360°}{10}$ = 36°.

We know that,

Area of sector of angle θ and radius r = $\dfrac{θ}{360°} \times πr^2$

Substituting values we get :

$= \dfrac{36°}{360°} \times \dfrac{22}{7} \times (17.5)^2 \\[1em] = \dfrac{1}{10} \times 22 \times 2.5 \times 17.5 \\[1em] = \dfrac{962.5}{10} \\[1em] = \dfrac{9625}{100} \\[1em] = \dfrac{385}{4} \text{ mm}^2.$

Hence, area of each sector = $\dfrac{385}{4}$ mm².

An umbrella has 8 ribs which are equally spaced. Assuming umbrella to be a flat circle of radius 45 cm, find the area between the two consecutive ribs of the umbrella.

Answer

There are 8 ribs and total angle in a circle = 360°.

So, angle subtended by each rib = $\dfrac{360°}{8}$ = 45°.

Area between two consecutive ribs = Area of a rib

We know that,

Area of sector of angle θ and radius r = $\dfrac{θ}{360°} \times πr^2$

Substituting values we get :

$\text{Area subtended by a rib} = \dfrac{45}{360} \times \dfrac{22}{7} \times 45^2 \\[1em] = \dfrac{1}{8} \times \dfrac{22}{7} \times 2025 \\[1em] = \dfrac{44550}{56} \\[1em] = \dfrac{22275}{28} \text{ cm}^2.$

Hence, area between two consecutive ribs = $\dfrac{22275}{28}$ cm².

A car has two wipers which do not overlap. Each wiper has a blade of length 25 cm sweeping through an angle of 115°. Find the total area cleaned at each sweep of the blades.

Answer

Area swept by each wiper = Area of sector of radius 25 cm and angle 115°.

We know that,

Area of sector of angle θ and radius r = $\dfrac{θ}{360°} \times πr^2$

Substituting values we get :

$\text{Area swept by each wiper} = \dfrac{115°}{360°} \times \dfrac{22}{7} \times 25^2 \\[1em] = \dfrac{23}{72} \times \dfrac{22}{7} \times 625 \\[1em] = \dfrac{23 \times 11 \times 625}{36 \times 7} \\[1em] = \dfrac{158125}{252} \text{ cm}^2.$

Area swept at each sweep of blades = 2 × Area swept by each wiper

= $\dfrac{158125}{252} \times 2 = \dfrac{158125}{126}$ cm².

Hence, area cleaned at each sweep of blades = $\dfrac{158125}{126}$ cm².

To warn ships for underwater rocks, a lighthouse spreads a red coloured light over a sector of angle 80° to a distance of 16.5 km. Find the area of the sea over which the ships are warned.

(Use π = 3.14)

Answer

Given,

Lighthouse spreads a red coloured light over a sector of angle 80° to a distance of 16.5 km.

So, for sector :

Radius (r) = 16.5 km

Angle (θ) = 80°

We know that,

Area of sector of angle θ and radius r = $\dfrac{θ}{360°} \times πr^2$

Substituting values we get :

$\text{Area of sea over which ships are warned} = \dfrac{80°}{360°} \times 3.14 \times (16.5)^2 \\[1em] = \dfrac{2}{9} \times 3.14 \times 272.25 \\[1em] = \dfrac{1709.73}{9} \\[1em] = 189.97 \text{ km}^2.$

Hence, area of the sea over which the ships are warned = 189.97 km².

A round table cover has six equal designs as shown in Fig. below. If the radius of the cover is 28 cm, find the cost of making the designs at the rate of ₹ 0.35 per cm². (Use $\sqrt{3}$ = 1.7)

Answer

Since,

There are 6 equal chords so each chord will subtend $\dfrac{360°}{6}$ = 60° at the center.

Let AB be one of the chords and center be O.

In triangle OAB,

OM ⊥ AB

In triangle OAM and OMB,

∠OMA = ∠OMB = 90°

OA = OB (Radius of same circle)

OM = OM (Common)

∴ △OAM ≅ △OMB (By RHS axiom)

∴ ∠MOB = ∠MOA = $\dfrac{60°}{2}$ = 30°. [By C.P.C.T.]

In △MOB,

⇒ sin 30° = $\dfrac{MB}{OB}$

⇒ $\dfrac{1}{2} = \dfrac{MB}{r}$

⇒ MB = $\dfrac{r}{2} = \dfrac{28}{2}$ = 14 cm

By C.P.C.T.

MA = MB = 14 cm

AB = MA + MB = 28 cm.

Since, OA = OB = AB.

∴ △AOB is an equilateral triangle.

We know that,

Area of equilateral triangle = $\dfrac{\sqrt{3}}{4} \times$ (Side)²

$\text{Area of equilateral triangle AOB} = \dfrac{\sqrt{3}}{4} \times 28^2 \\[1em] = 1.7 \times 7 \times 28 \\[1em] = 333.2 \text{ cm}^2.$

We know that,

Area of sector of angle θ and radius r = $\dfrac{θ}{360°} \times πr^2$

$\text{Area of sector AOBP} = \dfrac{60°}{360°} \times \dfrac{22}{7} \times 28^2 \\[1em] = \dfrac{1}{6} \times 22 \times 4 \times 28 \\[1em] = \dfrac{1}{3} \times 11 \times 4 \times 28 \\[1em] = \dfrac{1232}{3} \\[1em] = 410.67 \text{ cm}^2.$

From figure,

Area of segment ABP = Area of sector AOBP - Area of triangle OAB

= 410.67 - 333.2 = 77.47 cm² ≈ 77.5 cm².

There are 6 such segment.

So, total area of segments = 77.5 × 6 = 464.82 cm².

Given,

Cost of making designs = ₹ 0.35 per cm²

So, total cost = Area × Cost per area

= 464.82 × 0.35 = ₹ 162.68

Hence, cost of making designs = ₹ 162.68

Tick the correct answer in the following :

Area of sector of angle p (in degrees) of a circle with radius R is

1. $\dfrac{p}{180} \times 2πR$

2. $\dfrac{p}{180} \times πR^2$

3. $\dfrac{p}{360} \times 2πR$

4. $\dfrac{p}{720} \times 2πR^2$

Answer

We know that,

Area of sector of angle θ and radius r = $\dfrac{θ}{360°} \times πr^2$

Given,

angle = p

Radius = R

Substituting values we get :

$\text{Area } = \dfrac{p}{360} \times πR^2$

Multiplying numerator and denominator by 2, we get :

$\text{Area }= \dfrac{p}{360 \times 2} \times πR^2 \times 2 \\[1em] = \dfrac{p}{720} \times 2πR^2.$

Hence, Option 4 is the correct option.