Areas Related To Circles

We know that, circumference of a circle, C = 2πr

and diameter, D = 2r

According to the question,

C = D + 16.8

⇒ 2πr = 2r + 16.8

⇒ 2πr – 2r = 16.8

⇒ 2r(π – 1)= 16.8

⇒ 15r = 8.4 × 7

⇒ r = 3.92 cm

∴ Circumference of circle = 2πr

=24.64cm

Given: Radius of a circle, r = 42cm

Central Angle of the sector, θ = 150°

To find: Length of the arc i.e. AB

Now,

Length of an arc of a sector of angle θ

⇒ Length of an arc of a sector of angle θ

⇒ Length of an arc of a sector of angle θ

⇒ Length of an arc of a sector of angle θ = 5×22 =110 cm

Given: Length of the arc = 8.8 cm

Central Angle of the sector, θ = 60°

To find: Radius of a circle

Now,

⇒ r = 8.4cm

Given: Radius of the circle = 5.6cm

So, the circumference of the circle = 2πr

= 35.2cm

Now,

The perimeter of square = Circumference of a circle

= 35.2cm

Therefore,

Side of a square

= 8.8cm

Hence, the side of a square = 8.8cm

Given: Length of the arc = 35.2 cm

Radius of a circle, r = 42cm

To find: Central Angle of the sector

Now,

Length of an arc of a sector of angle θ

⇒ θ = 48°

Hence, the angle subtended by the arc at the centre of the circle is 48°

Diameter of a car wheel = 80cm

∴ radius =

Distance covered in 1 revolution = Circumference of wheel

= 2πr

= 2 × π × 40

= 80π cm

Now, we find total distance covered

We know that,

Here, Speed = 80 km/hr

Time = 10 minutes

Putting a value in the formula, we get

⇒ Distance = 13.33 km

= 13.33 × 1000 m

= 1333.3 m

= 1333.3 × 100 cm

= 1333333.3 cm

Now,

= 5303.02

= 5303 (approx.)

Hence, number of revolutions = 5303

Area of circular park = 88704 m²

and Area of circle = πr² = 88704

⇒ r² = 28224

⇒ r = √28224

⇒ r = 168m

[taking positive root, because radius can’t be negative]

Perimeter of circle = 2πr

= 1056 m

= 1.056 km

So, total distance after 10 rounds = 1.056 × 10 = 10.56km

Now, we know that

Here, Speed = 4.5 km/hr

Distance = 10.56 km

Putting value in formula, we get

= 2.34 hour

= 2 hours 20 minutes 24 seconds

Hence, Rajeev will take 2 hours 20 minutes and 24 seconds to walk 10 rounds.

Diameter of a bus wheel = 140cm

∴ radius =

Distance covered in 1 revolution = Circumference of wheel

= 2πr

= 2 × π × 70

= 140π cm

Now, we find total distance covered

We know that,

Here, Speed = 66 km/hr

Time = 1 minutes

Putting a value in the formula, we get

⇒ Distance = 1.1 km

= 1.1 × 1000 m

= 1100 m

= 1100 × 100 cm

= 110000 cm

Now,

Number of revolutions = 250

Given: Radius of circle = 4cm

And Central angle, θ = 30°

To find: Area of the sector

Now, Area of the sector of angle θ

⇒ Area of the sector of angle θ [∵ π = 3.14]

⇒ Area of the sector of angle θ

=4.19cm²

Now, we have to find the area of major sector (unshaded region)

= Area of circle – Area of sector OAPBO

= πr² – 4.19

= {3.14 × (4)²} – 4.19

= {3.14 × 16} – 4.19

= 50.24 – 4.19

= 46.05 cm²

Hence, the area of sector = 4.19cm² and area of the major sector = 46.05cm²

Given: Circumference of a circle = 22cm

⇒ 2πr = 22

⇒ 2r = 7

Now, we have to find the area of Quadrant

= 9.625cm²

Given: Length of minute hand = 12cm

Let us look at a clock:

So, minute hand covers a total of 60 minutes in one round of the clock.

So, 60 minutes = 360˚

1 minute

35 minutes = (35 × 6)˚ = 210˚

So in 35 minutes, minute hand subtends an angle of 210˚.

Now Area of segment

Where θ is the angle subtended.

Therefore, the area covered by minute hand

The radius of the circle = Length of the minute hand = 12cm

The area covered by minute hand

The area covered by minute hand = (22 × 12) cm²

Area = 264 cm²

Hence, The area covered by minute hand in 35 minutes is 264cm².

Let OAB be a the given sector with perimeter 27.2cm

Let arc AB = l

Perimeter of sector OAB = 27.2cm

⇒ OA + AB + OB = 27.2

⇒ 5.6 + l + 5.6 = 27.2

⇒ l = 27.2 – 11.2

⇒ l = 16cm

Now, we know that

…(i)

Area of sector OAB

[from (i)]

= 8r

= 8 × 5.6

= 44.8cm²

Given: Radius of circle = 14cm

and ∠AOB = 90°

=154cm²

and Area of major sector = Area of circle - Area of minor sector

= πr² - 154

= 22 × 2 ×14 – 154

= 462cm²

Given: Area of circle = 78.5cm²

πr² = 78.5

⇒ 3.14 r² = 78.5

⇒ r² = 25

⇒ r = ± 5

⇒ r = 5cm

[taking positive root, because radius can’t be negative]

Now, circumference of circle = 2πr

= 2 × 3.14 × 5

= 31.4cm

Given: Radius of the small circle, OB = 7cm

Radius of second circle, OA = 14cm

and ∠AOC=40°

= 17.11 cm²

= 68.4 cm²

Area of the shaded region = Area of sector OAC

– Area of sector OBD

= 68.4 – 17.1

= 51.3 cm²

The diameter of a circular park = 42m

⇒ the radius of a circular park,

Width of path = 3.5m

⇒ radius of the park with path, R = 21 + 3.5 = 24.5m

∴ Area of path = Area of outer circle – Area of the inner circle

= πR² – πr²

= π (R² – r²)

= π (R – r)(R + r)

= 22 × 0.5 × 45.5

= 500.5 m²

Now, the cost of gravelling the path of 1m² = Rs 4

the cost of gravelling the path of 500.5m² = Rs 4×500.5

= Rs 2002

The diameter of a circular park = 17.5m

⇒ the radius of a circular park,

Width of path = 3.5m

⇒ radius of the park with path,

∴ Area of path = Area of outer circle – Area of inner circle

= πR² – πr²

= π (R² – r²)

= π (R – r)(R + r)

= 231 m²

Given: Let the radius of the inner circle = r

Area enclosed between two concentric circles = 770cm²

and Radius of the outer circle, R = 21cm

∴ The area enclosed between two concentric circles

= Area of the Outer circle – Area of the inner circle

770 = πR² – πr²

770 = π(21² – r²)

245 = 441 – r²

⇒ r² = 441 – 245

⇒ r² = 196

⇒ r = √196

⇒ r = ±14

⇒ r = 14cm [taking positive square root, because radius can’t be negative]

Hence, the radius of the inner circle is 14cm.

Given: Difference between circumference and diameter of circular plot = 105m

We know that,

The circumference of a circle = 2πr

and diameter of circle = 2r

According to the question

2πr – 2r = 105

⇒ 2r(π – 1) = 105

Now, Area of circular path = πr²

= 1886.5m²

Total cost of fencing a circular field = Rs 5280

Cost of fencing per meter = Rs 24

So,

= 220m

Here, the total length of the field would be the circumference of the circular field.

∴ Total length = Circumference of circular field

⇒ 220 = 2πr

Now, Area of the field = πr²

= 3850m²

Now,

Cost of ploughing 1m² of the field = Rs 0.50

Cost of ploughing 3850m² of the field = Rs 0.50 × 3850

= Rs 1925

Hence, the cost of ploughing the field is Rs 1925

Cost of ploughing 1m² of the field = Rs1.50

The total cost of ploughing the field = Rs 5775

So,

=3850m²

Given: A field is in the form of a circle.

Let the radius of the field = r

∴ Area of the field = πr²

⇒ 3850 = πr²

⇒ r² = 1225

⇒ r = √1225

⇒ r = √(5×5×7×7)

⇒ r = 35m

Now,

Circumference = 2πr

= 2 × 22 × 5

= 220 m

Cost of fencing 1m of the field = Rs 8.50

Cost of fencing 220 of the field = Rs 8.50 × 220

= Rs 1870

Given: Radius of the circle = OA = OB = 10 cm

and θ = 90°

(i) Area of the minor sector

"Area of minor sector "

= 3.14 × 25

= 78.5cm²

(ii) Area of major sector

Area of major sector = Area of circle - Area of minor sector = πr² – 78.5

= 3.14 × (10)² – 78.5

= 314 – 78.5

= 235.5 cm²

Given: Radius of the circle = OA =OB = 21cm

and θ = 60°

(i) Length of the arc

= 22cm

(ii) Area of the sector formed by this arc

= 11 × 21

= 231 cm²

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

In ΔOAB,

∠OAB = ∠OBA (As OA = OB)

∠OAB + ∠AOB + ∠OBA = 180°

2∠OAB + 60° = 180°

∠OAB = 60°

∴ ΔOAB is an equilateral triangle.

Area of segment APB = Area of sector OAPB − Area of ΔOAB

Given: Radius of the circle = OA = OB = 12cm

and θ = 120°

To find: Area of the corresponding segment of the circle

i.e. Area of segment APB = Area of sector OAPB – Area of ΔAOB

So, firstly we find the Area of sector OAPB

= 150.72cm²

Now, we have to find the area of ΔAOB

We draw OM ⊥ AB

∴∠OMB = ∠OMA = 90°

In ΔOMA and ΔOMB

∠OMA = ∠OMB [both 90°]

OA = OB [both radius]

OM = OM [common]

∴ OMA ≅ ΔOMB [by RHS congruency]

⇒ ∠AOM = ∠BOM [CPCT]

∴In right triangle OMA, we have

⇒ AM = 6√3 cm

⇒ 2AM =12√3 cm

⇒ AB =12√3 cm

and

⇒ OM = 6cm

= 36√3

= 36 × 1.73

= 62.28 cm²

Area of segment APB = Area of sector OAPB – Area of ΔAOB

= (150.72 – 62.28)

= 88.44 cm²

Here, brooch is made up of silver wire in the form of a circle.

The diameter of the brooch = 35mm

"⇒ The radius of the brooch, r"

Since the wire is used in making 5 diameters and circle

So,

The total length of the silver wire required = length of wire in circle

+ wire used in 5 diameters

= 2πr + 5 × 2r

= 2r(π + 5)

= 5×57

= 285 mm

Now, Area of each sector of the brooch

∴Area of each sector=

(i)

Let ABCD be square field

and length of rope, r = 10m

We need to find the area of the field which horse can graze, i.e. the area of sector QBP

As we know that in a square all angles are of 90°

Hence, ∠QBP = 90°

= 78.5 m²

Hence, the area of the field which horse can graze = 78.5 m²

(ii) the decrease in the grazing area if the rope was 5 m long instead of 10m

Length of rope is decreased to 5m

Area grazed by a horse now = Area of sector HBG

= 19.625 m²

So, the decrease in the grazing area

= Area of sector QBP – Area of sector HBG

= 78.5 – 19.625

= 58.875 m²

Given: Radius of circle = 32cm

Area of design = Area of circle – Area of ΔABC

Firstly, we find the area of a circle

Area of circle = πr²

…(a)

Now, we will find the area of equilateral ΔABC

Construction:

Draw OD ⊥ BC

In ΔBOD and ΔCOD

OB = OC (radii)

OD = OD (common)

∠ODB = ∠ODC (90°)

∴ ΔBOD ≅ ΔCOD [by RHS congruency]

⇒ BD = DC [by CPCT]

or BC = 2BD …(i)

and,

Now, In ΔBOD, we have

⇒ BD = 16√3 cm

From (i), BC = 2BD ⇒ BC = 32√3 cm

Now, Area of equilateral ΔABC

= 768√3 cm² …(b)

Therefore, Area of design = Area of circle – Area of ΔABC

[from (a) and (b)]

Given: PQ = 24cm and PR = 7cm

Since QR is a diameter, it forms a semicircle

We know that angle in a semicircle is a right angle.

Hence, ∠RPQ = 90°

Hence, ΔRPQ is a right triangle

In ΔRPQ, by Pythagoras theorem

(Hypotenuse)² = (Perpendicular)² + (Base)²

(QR)² = (PQ)² + (PR)²

⇒ (QR)² = (24)² + (7)²

⇒ (QR)² = 576 + 49

⇒ (QR)² = 625

⇒ (QR)² = (25)²

⇒ QR = 25cm

∴ Diameter, QR = 25cm

So,

Now, Area of ΔPQR

= 12 × 7

= 84cm²

Area of shaded region = Area of semicircle – Area of ΔPQR

Area of the shaded region

= Area of sector AOB – Area of sector COD

Area of sector AOB

Here, radius = 21cm and θ = 30°

Area of sector COD

Here, radius = 7cm and θ = 30°

Now, shaded region = Area of sector AOB – Area of sector COD

Hence, area of shaded region is cm²

Area of shaded region = Area of square ABCD

– Area of semicircle APD

– Area of semicircle BPC

Area of square ABCD

Given: Side of square = 14cm

Area of square = Side × Side

= 14 × 14

= 196 cm²

Area of semicircle APD

Diameter = AD = 14cm

So,

= 11 × 7

= 77cm²

Similarly, Area of semicircle BPC = 77cm²

Area of shaded region = Area of square ABCD

– Area of semicircle APD

– Area of semicircle BPC

= 196 – 77 -77

= 42cm²

Hence, area of shaded region is 42cm²

Area of shaded region = Area of quadrant OBPQ

– Area of square OABC

Area of square OABC

Given: Side of square = 20cm

Area of square = Side × Side

= 20 × 20

= 400 cm²

Area of quadrant

We need to find the radius

Joining OB

Also, all angles of a square are 90°

∴∠BAO = 90°

Hence, ΔOBA is a right triangle

In ΔOBA, by Pythagoras Theorem

(Hypotenuse)² = (Perpendicular)² + (Base)²

(OB)² = (AB)² + (OA)²

⇒ (OB)² = (20)² + (20)²

⇒ (OB)² = 400 + 400

⇒ (OB)² = 800

⇒ OB = √(10×10×2×2×2)

⇒ OB = 20√2cm

= 628 cm²

Area of shaded region = Area of quadrant OBPQ

– Area of square OABC

= 628 – 400

= 228cm²

ABCD is a square lawn of side 58m. AED and BFC are two circular ends.

Now, diagonal of the lawn = √(58)² + (58)² = 58√2m

It is given that diagonal of square = Diameter of circle

∴The radius of a circle having a centre at the point of intersection of diagonal

It is given that square ABCD is inscribed by the circle with centre O.

∴Area of 4 segments = Area of circle – Area of square

= πr² – (side)²

m²

= 961.14m²

Area of whole lawn = Area of circle – Area of two segments

=5286.28 – 961.14

= 4325.14 m²

The side of hexagonal plot = 72m

= 1296√3cm²

∴ Area of hexagonal plot = 6 × Area of triangle OAB

= 6 × 1296√3

= 7776(1.732)

=13468.032m²

In ΔOCA, by Pythagoras theorem

(Hypotenuse)² = (Perpendicular)² + (Base)²

(OA)² = (OC)² + (AC)²

⇒ (72)² = (OC)² + (36)²

⇒ (OC)² = 5184 – 1296

⇒ (OC)² = 3888

⇒ r² = 3888

∴Area of inscribed circular swimming tank = πr²

= 12219.429m²

∴ Required difference = 13468.032 – 12219.429

= 1248.603m²

Hence, the difference between the area of a regular hexagonal plot and the area of the circular swimming tank inscribed in it is 1248.60.3m²

Area of shaded region = Area of segment with chord BC

Now, AC = BA = 14cm

∴ BC = √(14)² + (14)² = 14√2 cm

= 154cm²

= 98

So, Area of shaded region = 154 – 98

= 56cm²

Area of the shaded region

= Area of an equilateral triangle – Area of 3 sectors

Given: Area of equilateral ΔABC = 100√3cm²

⇒ a² = 400

⇒ a = 20cm

It is given that radius is equal to half the length of the side

i.e.

Now,

= 157.14cm²

Hence, the area of the shaded region

= Area of ΔABC – Area of 3 sectors

= 100√3 – 157.14

= 100×1.732 – 157.14

=173.2 – 157.14

= 16.06 cm²

Area of shaded region = Area of ΔABC – Area of circle

Given side of triangle = 12cm

= 36√3cm²

Now, we have to find the area of a circle. For that we need a radius.

Draw AD ⊥ BC

So, In BDO

⇒ r = 2√3cm

Now, Area of circle = πr²

= 3.14 × (2√3)²

= 37.68cm²

Area of shaded region = Area of ΔABC – Area of circle

= 36√3 – 37.68

= 36(1.73) – 37.68

= 24.6cm²

Hence, the area of the portion of the triangle not included in the circle is 24.6cm²

Given: Radius of circle = 16cm

Area of shaded region = Area of circle – Area of ΔABC

Firstly, we find the area of a circle

Area of circle = πr²

…(a)

Now, we will find the area of equilateral ΔABC

Construction:

Draw OD ⊥ BC

In ΔBOD and ΔCOD

OB = OC (radii)

OD = OD (common)

∠ODB = ∠ODC (90°)

∴ ΔBOD ≅ ΔCOD [by RHS congruency]

⇒ BD = DC [by CPCT]

or BC = 2BD …(i)

and,

Now, In ΔBOD, we have

⇒ BD = 8√3 cm

From (i), BC = 2BD ⇒ BC = 16√3 cm

Now, Area of equilateral ΔABC

= 192√3 cm² …(b)

Therefore, Area of design = Area of circle – Area of ΔABC

[from (a) and (b)]

= 804.57 – 332.544

= 472.03cm²