Chapter 15: Areas Related To Circles Exercise – 15.2
Question: 1
Find in terms of π, the length of the arc that subtends an angle of 30 degrees, at the center of O of the circle with a radius of 4 cm.
Solution:
Given Data:
Radius = 4 cm
Angle subtended at the centre 'O' = 30°
Formula to be used: Length of arc = θ/360 × 2πr cm
Length of arc = 30/360 × 2π∗4 cm = 2π/3 cm
Therefore, the Length of arc the length of the arc that subtends an angle of 60 degrees is 2π/3 cm
Question: 2
Find the angle subtended at the centre of circle of radius 5 cm by an arc of length 5π/3 cm.
Solution:
Given data: Radius = 5 cm
Length of arc = 5π/3 cm
Formula to be used: Length of arc = θ/360 ∗ 2πr cm
5π/3 cm = θ/360 ∗ 2πr cm
Solving the above equation, we have:
θ = 60°
Therefore, angle subtended at the centre of circle is 60°
Question: 3
An arc of length cm subtends an angle of 144° at the center of the circle.
Solution:
Given Data:
Length of arc = 25 cm
θ = Angle subtended at the centre of circle = 144°
Formula to be used:
Length of arc = θ/360 ∗ 2πr cm
θ/360 ∗ 2πr cm = 144/360 east 2πr cm = 4π/5 ∗ r cm
As given in the question, length of arc = cm,
Therefore, cm = 4π/5 ∗ r cm
Solving the above equation, we have r = 25 cm.
Therefore the radius of the circle is found to be 25 cm.
Question: 4
An arc of length 25 cm subtends an angle of 55° at the center of a circle. Find in terms of radius of the circle.
Solution:
Given Data:
Length of arc = 25 cm
θ = Angle subtended at the centre of circle = 55°
Formula to be used:
Length of arc = θ/360 ∗ 2πr cm
= 55/36 ∗ 2πr cm
As given in the question length of arc = 25 cm, hence,
25 cm = 55/360 ∗ 2π ∗ r cm
25 = 11πr/36
radius = 25∗36/11∗π = 900/11π
Therefore, the radius of the circle is 900/11π
Question: 5
Find the angle subtended at the center of the circle of radius 'a' cm by an arc of π/4 length cm.
Solution:
Given data:
Radius = a cm
Length of arc = aπ/4 cm
θ = angle subtended at the centre of circle
Formula to be used:
Length of arc = θ/360 ∗ 2πr cm
Length of arc = θ/360 ∗ 2πa cm
θ/360 ∗ 2πa cm = aπ/4 cm
Solving the above equation, we have
θ = 45°
Therefore, the angle subtended at the centre of circle is 45°
Question: 6
A sector of the circle of radius 4 cm subtends an angle of 30°. Find the area of the sector.
Solution:
Given Data: Radius = 4 cm
Angle subtended at the centre 'O' = 30°
Formula to be used:
Area of the sector = θ/360 ∗ πr2
Area of the sector = 30/360 ∗ π42
Solving the above equation, we have: Area of the sector = 4.9 cm2
Therefore, Area of the sector is found to be 4.9 cm2
Question: 7
A sector of a circle of radius 8 cm subtends an angle of 135. Find the area of sector.
Solution:
Given Data: Radius = 8 cm
Angle subtended at the centre 'O' = 135°
Formula to be used:
Area of the sector = θ/360 ∗ πr2
Area of the sector = 135/360 ∗ π82
= 528/7 cm2
Therefore, Area of the sector calculated is 528/7 cm2
Question: 8
The area of sector of circle of radius 2 cm is cm2. Find the angle subtended by the sector.
Solution:
Given Data: Radius = 2 cm
Angle subtended at the centre 'O'
Area of sector of circle = cm2
Formula to be used:
Area of the sector = θ/360 ∗πr2
Area of the sector = θ/360 ∗ π32
= πθ/90
As given in the question area of sector of circle = cm2
cm2 = π θ/90
Solving the above equation, we have
θ = 90°
Therefore, the angle subtended at the centre of circle is 90°
Question: 9
PQ is a chord of circle with centre 'O' and radius 4 cm. PQ is of the length 4 cm. Find the area of sector of the circle formed by chord PQ.
Solution:
Given Data:
PQ is chord of length 4 cm. Also, PO = QO = 4 cm OPQ is an equilateral triangle.
Angle POQ = 60°
Area of sector (formed by the chord (Shaded region)) = (Area of sector)
Formula to be used:
Area of the sector = θ/360 ∗ πr2
Area of the sector = 60/360 ∗ π42 = 32π/3
Therefore, Area of the sector is 32π/3 cm2
Question: 10
In a circle of radius 35 cm, an arc subtends an angle of 72° at the centre. Find the length of arc and area of sector.
Solution:
Given Data: Radius = 35 cm
Angle subtended at the centre 'O' = 72°
Area of sector of circle =?
Formula to be used: Length of arc = θ/360 ∗ 2πr cm
Length of arc = 108/360 ∗ 2πx = 42 cm
Solving the above equation we have,
Length of arc = 44 cm
We know that,
Area of the sector = θ/360 ∗ πr2
Area of the sector = 72/360 ∗ π352
Solving the above equation, we have,
Area of the sector = (35 × 22) cm2
Therefore, Area of the sector is 770 cm2
Question: 11
The perimeter of a sector of a circle of radius 5.7 m is 27.2 m. find the area of the sector.
Solution:
Given Data:
Radius = 5.7 cm = OA = OB [from the figure shown above]
Perimeter = 27.2 m
Let the angle subtended at the centre be θ
Perimeter = θ/360 ∗ 2πr cm + OA + OB
= θ/360 ∗ 2πx 5.7 cm + 5.7 + 5.7
Solving the above equation we have,
θ = 158.8°
We know that, Area of the sector = θ/360 ∗ πr2
Area of the sector = 158.8/360 ∗ π 5. 72
Solving the above equation we have,
Area of the sector = 45.048 cm2
Therefore, Area of the sector is 45.048 cm2
Question: 12
The perimeter of a certain sector of a circle of radius is 5.6 m and 27.2 m. find the area of a sector.
Solution:
Given data:
Radius of the circle = 5.6 m = OA = OB
(AB arc length) + OA + OB = 27.2
Let the angle subtended at the centre be θ
We know that,
Length of arc = θ/360 ∗ 2πr cm
θ/360 ∗ 2πr cm + OA + OB = 27.2 m
θ/360 ∗ 2πr cm + 5.6 + 5.6 = 27.2 m
Solving the above equation, we have,
θ = 163.64°
We know that, Area of the sector = θ/360 ∗ πr2
Area of the sector = 163.64/360 ∗ π 5.62
On solving the above equation, we have,
Area of the sector = 44.8 cm2
Therefore,
Area of the sector is 44.8 cm2
Question: 13
A sector was cut from a circle of radius 21 cm. The angle of sector is 120°. Find the length of its arc and its area.
Solution:
Given data: Radius of circle (r) = 21 cm
θ = angle subtended at the centre of circle = 120°
Formula to be used:
Length of arc = θ/360 ∗ 2πr cm
Length of arc = 120/360 ∗ 2πx 21cm
On solving the above equation, we get, Length of arc = 44 cm
We know that, Area of the sector = θ/360 ∗ πr2
Area of the sector = 120/360 ∗ π212
Area of the sector = (22 × 21) cm2
Therefore, Area of the sector is 462 cm2
Question: 14
The minute hand of a circle is √21 cm long. Find the area described by the minute hand on the face of clock between 7:00 a.m to 7:05 a.m.
Solution:
Given data: Radius of the minute hand (r) = √21cm
Time between 7: 00 a. m to 7: 05 a. m = 5 min
We know that, 1 hr = 60 min, minute hand completes
One revolution = 360° 60 min = 360°
θ = Angle subtended at the centre of circle = 5 × 6° = 30°
Area of the sector = θ/360 ∗ πr2
Area of the sector = 30/360 ∗ pi352
Area of the sector = 5.5 cm2
Therefore, Area of the sector is 5.5 cm2
Question: 15
The minute hand of clock is 10 cm long. Find the area of the face of the clock described by the minute hand between 8 a. m to 8:25 a.m.
Solution:
Given data: Radius of the circle = radius of the clock = length of the minute hand = 10 cm
We know that, 1 hr = 60 min 60 min = 360° 1 min = 6°
Time between 8:00 a. m to 8:25 a. m = 25 min
Therefore, the subtended = 6° × 25 = 150°
Formula to be used:
Area of the sector = θ/360 ∗ πr2
Area of the sector = 150/360 ∗ π102
Area of the sector = 916.6 cm2 = 917 cm2
Therefore, Area of the sector is 917 cm2
Question: 16
A sector of 56° cut out from a circle subtends area of 4.4 cm2. Find the radius of the circle.
Solution:
Given data: Angle subtended by the sector at the centre of the circle,
θ = 56°
Let the radius of the circle be = r cm
Formula to be used:
Area of the sector = 56/360 ∗ πr2
On solving the above equation, we get, r2 = √(9/1) cm
r = 3 cm
Therefore, radius of the circle is r = 3 cm
Question: 17
In circle of radius 6 cm. Chord of length 10 cm makes an angle of 110° at the centre of circle. Find:
(i) Circumference of the circle
(ii) Area of the circle
(iii) Length of arc
(iv) The area of sector
Solution:
Given data: Radius of the circle = 6 cm
Chord of length = 10 cm
Angle subtended by chord with the centre of the circle = 110°
Formulae to be used: Circumference of a circle = 2
Area of a Circle = Area of the sector = θ/360 ∗ πr2
Length of arc = 90/360 ∗ 2πx 28cm
Circumference of a circle = 2
= 2 × 3.14 × 8 = 37.7 cm
Area of a Circle = 3.14 × 6 × 6 = 113.14 cm2
Area of the sector = θ/360 ∗ πr/2
Area of the sector = 110/360 ∗ π62
On solving the above equation we get,
Area of the sector = 33.1 cm2
Length of arc = θ/360 ∗ 2πr cm [latex]
Length of arc = 110/360 2π6cm
On solving the above equation we get, Length of arc = 22.34 cm.
Therefore, Circumference = 37.7 cm
Area of a Circle = 113.14 cm2
Area of the sector = 33.1 cm2
Question: 18
The given figure shows a sector of a circle with centre 'O' subtending an angle θ°. Prove that:
1. Perimeter of shaded region is
2. Area of the shaded region is
Solution:
Given Data: Angle subtended at the centre of the circle = θ°
Angle OAB = 90° [at point of contact, tangent is perpendicular to radius]
OAB is a right angle triangle
cos θ = adjside/hypotenuse = r/OB = OB = r sec θ
sec θ = opposite/adjside = AB/r = AB = r tan θ
Perimeter of the shaded region = AB + BC + CA (arc)
= r tan? + (OB - OC) + θ/360 ∗ 2πr cm
= r(tan θ + sec θ + π θ/180 − 1)
Area of the shaded region = (Area of triangle AOB) - (Area of sector)
(1/2 ∗ OA ∗ AB) − θ/360 ∗ πr2
On solving the above equation we get, r2/2[tan θ − πθ/180]
Question: 20
The diagram shows a sector of circle of radius 'r' cm subtends an angle θ. The area of sector is A cm2 and perimeter of sector is 50 cm. Prove that θ = 360/π (25/r - 1) and A = 25 r - r2
Solution:
Given Data: Radius of circle = 'r' cm
Angle subtended at centre of the circle = θ
Perimeter = OA + OB + (AB arc)
As given in the question, perimeter = 50
Area of the sector = θ/360 ∗ πr2
On solving the above equation, we have A = 25r - r2
Hence, proved.