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Science > Mathematics > Trigonometry > You are Here |

#### Example – 01:

- Find area of sector of circle which subtends an angle of 120° at the centre, if the radius of circle is 6 cm.
**Given:**Angle subtended at centre = θ = 120° = 120 x π/180 = (2π/3)^{c}, Radius of circle = r = 6 cm.**To find:**Area of sector = A = ?**Solution:**

Area of sector = ½ r^{2}θ= ½ x 6^{2} x (2π/3) =12π sq. cm

**Ans:** Area of sector is 12π sq. cm

#### Example – 02:

- The area of circle is 81π sq. cm. Find the length of its arc subtending an angle of 150° at the centre. Also find area of the corresponding sector.
**Given:**Area of circle = 81π sq. cm, Angle subtended at centre = θ = 150° = 150 x π/180 = (5π/6)^{c},**To find:**Length of arc = S = ?, Area of sector = A = ?**Solution:**

Area of circle = πr^{2} = 81π

∴ r^{2} = 81

∴ r = 9 cm

Length of arc = S = r θ = 9 x (5π/6) = 7.5 π cm

Area of sector = ½ r^{2}θ= ½ x 9^{2} x (5π/6) = 33.75 π sq. cm

**Ans:** length of arc is 7.5 π cm and area of sector is 33.75 π sq. cm

#### Example – 03:

- The area of circle is 25π sq. cm. Find the length of its arc subtending an angle of 144° at the centre. Also find area of the corresponding sector.
**Given:**Area of circle = 25π sq. cm, Angle subtended at centre = θ = 144° = 144 x π/180 = (4π/5)^{c},**To find:**Length of arc = S = ?, Area of sector = A = ?

Area of circle = πr^{2} = 25π

∴ r^{2} = 25

∴ r = 5 cm

Length of arc = S = r θ = 5 x (4π/5) = 4π cm

Area of sector = ½ r^{2}θ= ½ x 5^{2} x (4π/5) = 10π sq. cm

**Ans:** length of arc is 4π cm and area of sector is 10π sq. cm

#### Example – 04:

- The area of circle is 81π sq. cm. Find the length of its arc subtending an angle of 300° at the centre. Also find area of the corresponding sector.
**Given:**Area of circle = 81π sq. cm, Angle subtended at centre = θ = 300° = 300 x π/180 = (5π/3)^{c},**To find:**Length of arc = S = ?, Area of sector = A = ?

Area of circle = πr^{2} = 81π

∴ r^{2} = 81

∴ r = 9 cm

Length of arc = S = r θ = 9 x (5π/3) = 15π cm

Area of sector = ½ r^{2}θ= ½ x 9^{2} x (5π/3) = 67.5 π sq. cm

**Ans:** length of arc is 15π cm and area of sector is 67.5π sq. cm

#### Example – 05:

- The perimeter of a sector of a circle of area 25π sq. cm is 20 cm. Find the area of sector.
**Given:**Area of circle = 25π sq. cm, Perimeter = 20 cm**To find:**Area of sector = A = ?

Area of circle = πr^{2} = 25π

∴ r^{2} = 25

∴ r = 5 cm

Perimeter of sector = r + r + s = 20

∴ 2r + r θ = 20

∴ r (2 + θ) = 20

∴ 5 (2 + θ) = 20

∴ 2 + θ = 4

∴ θ = 2^{c}

Area of sector = ½ r^{2}θ= ½ x 5^{2} x 2 = 25 sq. cm

#### Example – 06:

- The perimeter of a sector of a circle of area 64π sq. cm is 56 cm. Find the area of sector.
**Given:**Area of circle = 64π sq. cm, Perimeter = 56 cm**To find:**Area of sector = A = ?

Area of circle = πr^{2} = 64π

∴ r^{2} = 64

∴ r = 8 cm

Perimeter of sector = r + r + s = 56

∴ 2r + r θ = 56

∴ r (2 + θ) = 56

∴ 8 (2 + θ) = 56

∴ 2 + θ = 7

∴ θ = 5^{c}

Area of sector = ½ r^{2}θ= ½ x 8^{2} x 5 = 160 sq. cm

#### Example – 07:

- Find area of sector whose arc length is 30π cm and the angle of sector is 40°.
**Given:**Length of arc = 30π cm, angle of sector = θ = 40° = 40 x π/180 = (2π/9)^{c},**To find:**Area of sector = A = ?

Length of arc = S = r θ

∴ 30π = r x (2π/9)

∴ r = 135 cm

Area of sector = ½ r^{2}θ= ½ x 135^{2} x (2π/9) = 2025π sq. cm

#### Example – 08:

- In a circle of radius 12 cm, an arc PQ subtends the angle of 30° at the centre. Find the area between arc PQ and chord PQ.
- Given radius of circle = r = 12 cm, angle subtended at the centre = θ = 30° = 30 x π/180 = (π/6)
^{c}, - To find: area between arc PQ and chord PQ.
**Solution:**

Area of sector = ½ r^{2}θ= ½ x 12^{2} x (π/6) = 12π sq. cm

In Δ OOR, sin 30° = QR/OQ

∴ OR = OQ sin 30° = 12 x 1/2 = 6 cm

Area of Δ POQ = ½ x base x height = ½ x OP x QR = ½ x 12 x 6 = 36 sq.cm.

Area of shaded region = Area of sector – Area of Δ POQ

∴ Area of shaded region = 12π – 36 = 12(π – 3) sq. cm

**Ans:** The area between arc PQ and chord PQ is 12(π – 3) sq. cm

#### Example – 09:

- OPQ is sector of a circle with centre O and radius 12 cm. if m ∠ POQ= 60°, find the difference between the areas of sector POQ and Δ POQ.
- Given radius of circle = r = 12 cm, angle subtended at the centre = θ = 60° = 60 x π/180 = (π/3)
^{c}, - To find: the difference between the areas of sector POQ and Δ POQ.
**Solution:**

Area of sector = ½ r^{2}θ= ½ x 12^{2} x (π/3) = 24π sq. cm

In Δ OQR, sin 60° = QR/OQ

∴ OR = OQ sin 60° = 12 x √3 /2 = 6√3 cm

Area of Δ POQ = ½ x base x height = ½ x OP x QR = ½ x 12 x 6√3 = 36√3 sq.cm.

Area of shaded region = Area of sector – Area of Δ POQ

∴ Area of shaded region = 24π – 36√3 = 12(2π – 3√3) sq. cm

**Ans:** The difference between the areas of sector POQ and Δ POQ. is 12(2π – 3√3) sq. cm

#### Example – 10:

- OPQ is a sector of a circle with centre O and radius 12 cm. if m∠OPQ = 30°, Find the area between arc PQ and chord PQ.
- Given radius of circle = r = 12 cm,
- To find: the area between arc PQ and chord PQ.
**Solution:**

Δ OPQ is isosceles triangle

m∠ OPQ = m∠ OQP = 30°

m∠ POQ = θ = 120° = 120 x π/180 = (2π/3)^{c}

Area of sector = ½ r^{2}θ= ½ x 12^{2} x (2π/3) = 48π sq. cm

Δ OQR is 30°-60°-90° triangle

OR = ½OQ = ½ x 12 = 6 cm

QR = √3 /2 OQ = √3 /2 x 12 = 6√3

PQ = 2 QR = 2 x 6√3 = 12√3

Area of Δ POQ = ½ x base x height = ½ x PQ x OR = ½ x 12√3 x 6 = 36√3 sq.cm.

Area of shaded region = Area of sector – Area of Δ POQ

∴ Area of shaded region = 48π – 36√3 = 12(4π – 3√3) sq. cm

**Ans:** The area between arc PQ and chord PQ. is 12(4π – 3√3) sq. cm

#### Example – 11:

- Two circles each of radius 7 cm intersect each other such that the distance between their centres is 7√2 cm. Find area common to both the circles.
- Given: radius of circle = r = 7 cm, Distance between centres = 7√2 cm
- To find: the area of common portion = ?
**Solution:**

In quadrilateral ADBC

AD = DB = BC = CA = 7cm

diagonal AB = 7√2 cm

Hence quadrilateral ADBC is a square with each angle 90°

This is central angle subtended for sectors of both the circles = θ = 90° = 90 x π/180 = (π/2)^{c}

Area of common region = area of sector (A-CED) + Area of sector (B-CFD) – area of square ADBC

∴ Area of common region = ½ r^{2}θ + ½ r^{2}θ – r^{2}

∴ Area of common region = r^{2 }θ – r^{2}

∴ Area of common region = r^{2 }( θ – 1)

∴ Area of common region = 7^{2 }( π/2 – 1)

∴ Area of common region = 49(π/2 – 1) sq. cm

Ans: The area common to both the circle is 49(π/2 – 1) sq. cm

Science > Mathematics > Trigonometry > You are Here |

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