AP 10th Class Maths Chapter 11 Important Questions Areas Related to Circles
1th Lesson Areas Related to Circles Class 10 Important Questions with Solutions
10th Class Maths Areas Related to Circles 1 Mark Important Questions
Question 1.
Find the circumference and area of a circle of radius 8.4 cm.
Solution:
Given radius of the circle = 8.4 cm
Circumference C = 2πr
= 2 × 22/7 × 8.4
= 52.8 cm
Area of the circle A = πr2
= 22/7 × 8.4 × 8.4
= 221.76 cm2
Question 2.
Find area of a circle whose circumference is 44 cm.
Solution:
Give circumference of the circle C = 44 cm
Let radius = 4
Circumference = 2πr = 44
= 2 × 22/7 r = 44
r = 44×7/2×22
r = 7 cm
Area of circle = πr2
= 22/7 × 7 × 7 = 154 cm2
Question 3.
Define minor sector.
Solution:
A sector of a circle is called a minor sector if the minor arc of the circle is a part of the boundary.
Shaded part of AOB is the minor sector.
Question 4.
Define major sector.
Solution:
A sector of a circle is called a major if the major arc of the circle is a part of its boundary.
Shaded part of POQ is the major sector.
Question 5.
Write the formula to find the length of arc.
Solution:
Length of arc l = x°/360° 2πr (or) x/180° πr.
Question 6.
Write the formula to find the area of a sector.
Solution:
Area of sector A = x°/360° πr2.
Question 7.
A pendulum swings through an angle of 30° and describe an arc 8.8 cm in length. Find the length of the pendulum.
Solution:
Given angle θ = 30°, l = 8.8 cm, r = ?
Length of an are
r = 0.8 × 7 × 3 = 16.8 cm
Question 8.
The radius of a circle is same as the side of a square. Find the ratio of their perimetrers.
Solution:
r = s
Perimeter of circle: Perimeter of square
2πr : 4r
πr : 2r
π : 2
Question 9.
Find the length of the arc of a circle of radius 14 cm which subtends an angle of 60° at the centre of the circle.
Solution:
θ = 60°, r = 14 cm
l = θ/360 × 2πr
= 60/360 × 2 × 22/7 × 14
1/6 × 2 × 22 × 2
= 22×2/3 = 44/3
Question 10.
If the radius of a semi-circular protractor is 7 cm, then find perimeter.
Solution:
r = 7 cm
p = 36/7 r = 36/7 × 7 = 36 cm
Question 11.
Find the area of the sector of a circle of radius 6 cm whose centred angle is 30°. (Take π = 3.14)
Solution:
r = 6 cm
θ = 30°
Area of sector = θ/360 × πr2
= 30/360 22/7 × 6 × 6
= 1/12 × 22/7 × 36
= 11×36/6×7
= 11×6/7
= 66/7 cm2
Question 12.
Find the area of a quadrant of a circle where the circumference of circle is 176 m.
Solution:
θ = 90°
Circumference = 176 m
2πr = 176
2 × 227 × r = 176
r = 176 × 7/22 × 12
r = 8×7/2
r = 28 cm
A = θ/360 × πr2
= 90/360 × 22/7 × 28 × 28
= 1/4 × 22 × 4 × 28
= 11 × 2 × 28
A = 616 cm2
Question 13.
The minute hand of clock is 84 cm long. Find the distance covered by the tip of minute hand from 10 : 10 am to 10 : 25 am.
Solution:
60 min → 360
1 min → 6°
10 : 10 to 10 : 25 AM mean 15 min
15 min = 15 × 6° = 90°
θ = 90°
l = θ/360 × 2πr
= 90/360 × 2 × 22/7 × 84
1/4 × 2 × 22/7 × 84
= 11 × 12
l = 132 cm
Question 14.
The diameter of a car wheel is 42 cm. Find the number of complete revolutions it will make in moving 132 km.
Solution:
d = 42 cm
r = 21 cm
Circumference = 2πr
= 2 × 22/7 × 21
= 44 × 3
= 132 cm
No. of revolutions
= 132×104/132 = 104 (∵ 132 km = 1320000 = 132 × 104)
Question 15.
Find the area of a square that can be inscribed in a circle of area 14087 cm2.
Solution:
πr2 = 1408/7
22/7 × r2 = 1408/7
r2 = 1408/22
r2 = 704/11
r2 = 64
r = 8 cm
Question 16.
A circular arc of length 22 cm subtends an angle θ at the centre of the circle 21 cm. Find the value of θ.
Solution:
l = 22 cm
r = 21 cm
l = θ/360 × 2πr
22 × θ/360 × 2 × 22/7 × 21
1 = θ/180 × 3
θ/60 = 1
θ = 60°
Question 17.
Find the perimeter of the sector of a circle of a radius 14 cm and central angle 45°.
Solution:
r = 14 cm
θ = 45°
Perimeter = l + 2r
= θ/360 × 2πr + 2r
= 45/360 × 2 × 22/7 × 14 + 2 × 14
= 1/8 × 2 × 22 × 2 + 28
= 11 + 28
= 39 cm
Question 18.
Two concentric circles are at O. Find the area of shaded region, if outer and inner radii are 14 cm and 7 cm respectively.
Solution:
R = 14 m
r = 7 m
Area of shaded region = π(R2 - r2)
= 22/7(142 - 72)
= 22/7(196 - 49)
= 22/7 × 147
= 22 × 21
= 462 cm2
Question 19.
The circumference of two circles are in the ratio 4 : 5. What is the ratio of their radii?
Solution:
2πr1 : 2πr2= 4 : 5
r1 : r2 = 4 : 5
Question 20.
If the perimeter of a circle is equal to that of a square, then find the ratio of their areas.
Solution:
4s = 2πr
⇒ s = πr/2 ⇒ s2 = πr2
⇒ (πr/2)2 : πr2 ⇒ π2r2/4 : πr2
⇒ π/4 : 1 [π = 22/7] ⇒ 14 : 11
Question 21.
If the radii of two circles are in the ratio of 4:3, then find the ratio of their areas.
Solution:
r2 : r2 = 4 : 3
⇒ πr12 : πr22
⇒ π 42 : π 32
⇒ 16 : 9
Question 22.
Find the area of a quadrant of a circle of radius 7 cm.
Solution:
r = 7
Area of quadrant = πr2/4
= 22×7×7/7×4
= 11×7/2
= 77/2 cm
Question 23.
Find the area of a sector of a circle with radius 6 cm if angle of the sector is 60°. (Take π = 227)
Solution:
r = 6 cm
θ = 60°
Area of sector = θ/360 × πr2
= 60/360 × 22/7 × 6 × 6 cm2
= 132/7 cm2
Question 24.
A horse tied to a pole with 28m long rope. Find the perimeter where the horse can graze. (Take π = 22/7)
Solution:
r = 28 cm
θ = 90°
Perimeter = 2πr
= 2 × 22/7 × 28 = 44 × 4 = 176 cm
Question 25.
In a circle of diameter 42 cm, if an arc subtends an angle of 60° at the centre where π = 22/7 then find the length of the arc.
Solution:
θ = 60°
d = 42 cm
r = 21 cm
l = θ/360 × 2πr
= 60/360 × 2 × 22/7 × 21
= 16 × 2 × 22 × 3 = 22 cm.
Question 26.
Find the perimeter of a semicircular protractor whose radius is r.
Solution:
P = πr + 2r
Question 27.
If the circumference of a circle increases from 2π it to 4π then its area ....... the original area.
Solution:
C = 2πr
If r = 1
C = 2π
A = π(r)2 = π(1)2 = π
2πr = 4π
r = 2
New area = πr2 = π(2)2
= 4π = 4 × old area
Question 28.
If the difference between the circumference and the radius of a circle is 37cm, π = 22/7, find the circumference (in cm) of the circle.
Solution:
2πr - r = 37 cm
(2 × 22/7 - 1) r = 37
(44-7/7)r = 37
37/7 r = 37
r = 7
C = 2πr
= 2 × 22/7 × 7
= 44 cm
Question 29.
Area of a sector of a circle is 1/6 to the area of circle. Find the degree measure of its minor arc.
Solution:
θ/360 × πr2 = 1/6 × πr2
θ/360 = 1/6
θ= 60°
Question 30.
Figure is a sector of circle of radius 10.5 cm, find the perimeter of the sector. (Take π = 22/7)
Solution:
r = 10.5 cm
θ = 60°
P = l + 2r
= θ/360 × 2πr + 2r
= 60/360 × 2 × 22/7 × 10.5 + 2(10.5)
= 1/6 × 2 × 22 × 1.5 + 21
= 22 × 0.5 + 21
= 11 + 21
= 32 cm
Question 31.
Assertion (A) : Area of sector if l = 10 cm, r = 9 cm is 10.5 cm
Reason (F) : Area of sector = θ/360 × πr2
A) Assertion (A) is true, Reason (R) is true and R is the correct explanation of A.
B) A is true, R is true, but R is not the correct explanation of A.
C) A is true, R is false.
D) A is false, R is true.
Solution:
D) A is false, R is true.
Question 32.
Assertion (A) : Area of circle with 1 cm radius is ? cm2.
Reason (R) : Length of the arc of the sector (l = θ/180 × 2πr
A) Assertion (A) is true, Reason (R) is true and R is the correct explanation of A.
B) A is true, R is true, but R is not the correct explanation of A.
C) A is true, R is false.
D) A is false, R is true.
Solution:
C) A is true, R is false.
Question 33.
Assertion (A): Rddius of a circle is = 7/√π cm then area is 49 sq. units.
Reason (R) : Area of circle = πr2.
A) Assertion (A) is true, Reason (R) is true and R is the correct explanation of A.
B) A is true, R is true, but R is not the correct explanation of A.
C) A is true, R is false.
D) A is false, R is true.
Solution:
A) Assertion (A) is true, Reason (R) is true and R is the correct explanation of A.
Question 34.
Assertion (A): If a chord of length equal to the radius in a triangle then the triangle formed with that chord is an equilateral triangle.
Reason (R) : Area of ring π(R2 - r2).
A) Assertion (A) is true, Reason (R) is true and R is the correct explanation of A.
B) A is true, R is true, but R is not the correct explanation of A.
C) A is true, R is false.
D) A is false, R is true.
Solution:
B) A is true, R is true, but R is not the correct explanation of A.
Question 35.
Observe the figure, match the column.
A) Area of segment AYB
B) Area of sector OAYB
i) 441/4√3
ii) 21/4(88 - 21√3)
iii) 462
iv) 21/2
Solution:
A - i, B - iii
Question 36.
Choose correct matching.
Answer:
A - (i), B - (ii).
Question 37.
Draw a rough diagram of minor segment of a circle and shade it.
Answer:
Question 38.
What do we call the part a and b in the below circle ?
Solution:
‘a’ is minor segment and ‘b’ is major segment.
10th Class Maths Areas Related to Circles 2 Marks Important Questions
Question 1.
Find the area of quadrant of a circle whose circumference is 22 cm.
Solution:
Let radius = 4 cm
Circumference C = 2πr = 22 cm
= 2 × 22/7 × r = 22
r = 22×7/2×22 = 7/2 cm
Area of a quadrant = 14 πr2
= 1/4 × 22/7 × 7/2 × 7/2
Therefore, area of quadrant = 77/8 = 9.625 cm2
Question 2.
The circumference of a circle exceeds the diameter by 16.8 cm. Find the radius of the circle.
Solution:
Let the radius = r cm
Circumference = diameter + 16.8
2 × 22/7 r = 2r + 16.8
44r = 7 (2r + 16.8) = 14r + 16.8 × 7
44r - 14r = 117.6
30r = 117.6
r = 117.6/30 = 3.92 cm
Therefore, radius r = 3.92 cm.
Question 3.
A bicycle wheel makes 5000 revolutions in moving 11 km. Find the diameter of the wheel.
Solution:
Let the radius of the circle r cm
Distance covered by wheel in 1 revolution = Total distance moved/Number of revolutions
= 11/5000km = 11×1000×100/5000
That is circumference of wheel = 220 cm
So, 2πr = 220
2 × 22/7 r = 220
r = 220 × 7/2×22 = 35 cm
Diameter d = 2 × r = 2 × 35 = 70 cm
Question 4.
A wheel has diameter 84 cm. Find how many complete revolutions must it take to cover 792 m.
Solution:
Let radius of wheel
r = Diameter/2 = 84/2 = 42 cm
Circumference of the wheel
2πr = 2 × 22/7 × 42 = 264 cm = 2.64 m
No. of revolutions made to 792 m = 792/2.64 = 300
Question 5.
Find the perimeter of a quadrant of a circle of radius 14 cm.
Solution:
θ = 90°
r = 14 m
Perimeter of quadrant of cricle
= l + 2r =θ/360 × 21r + 2r
= 90/360 × 2 × 22/7 × 14 + 2 × 14
= 1/4 × 2 × 22 × 2 + 28
= 22 + 28 = 50 cm
Question 6.
Find the diameter of a circle whose area is equal to the sum of the areas of the two circles of radii 24 cm and 7 cm.
Solution:
Given r1 = 24 cm, r2 = 7 cm
πR2 = πr12 + πr22
R2 = 242 + 72
R2 = 576 + 49
R2 = 625
R = 25 cm
Diameter, d = 2R = 50 cm
Question 7.
Find the area of circle whose circumference is 22.
Solution:
Circumference of circle = 22 m
2πr = 22
2 × 22/7 × r = 22
2r/7 = 1
r = 72 m
Area of circle =πr2
= 22/7 × 7/2 × 7/2 cm2
= 77/2 cm2
Question 8.
Find the diameter of a circle whose area is equal to the sum of the areas of two circles of radii 40 cm and 9 cm.
Solution:
Given r1 = 40 cm, r2 = 9 cm
πR2 = πr12 + πr22
R2 = r12 + r22
R2 = 402+ 92
R2 = 1600 + 81
R2 = 1681
R = √1681
R = 41 cm
Diameter = 2 × 41 cm = 82 cm
Question 9.
A chord of a circle of radius 10 cm subtends a right angle at the centre. Find the area of minor segment. (Use π = 3.14)
Solution:
Radius, r = 10 cm
θ= 90°
Area of right-angled triangle
= 1/2 × 10 × 10 = 50 cm2
Area of minor segment = Area of sector - Area of right triangle
= 78.5 - 50 = 28.5 cm2
Question 10.
A piece of wire 22 cm long is bent into the form an arc of a circle subtending an angle of 60° at its centre. Find the radius of the circle.
Solution:
Given l = 22 cm
θ = 60°
l = ?/360 × 2πr
22 = 60/360 × 2 × 227 × r
1 = 1/6 × 2 × 1/7 × r
1 = r/21
r = 21 m
radius = 21 cm
Question 11.
An iron wire when bent in the form of a square encloses an arrea of 121 cm2. If the same wire is bent in the form of a circle, then find the circumference of circle.
Solution:
Area of square = 121 m2
S2 = 121
⇒ S = √121
S = 11 cm
⇒ 4S = 4 × 11 = 44 cm
2πr = 44
⇒ 2 × 22/7 × r = 44
r = 7 cm
Area of circle = πr2
= 22/7 × 7 × 7 cm2
= 154 cm2
Question 12.
If the radius of the circle is 6 cm and the length of an arc is 12 cm. Find the area of sector.
Solution:
Radius, r = 6 cm
Length of arc, l = 12 cm
Area of sector = ?r/2
= 12×6/2
= 6 × 6
= 36 cm2
Question 13.
PA is the tangent drawn to a circle of whose centre is ‘O’, OA is the radius and P is the external point of the circle. If PA = 24 cm, OP = 25 cm., then find its radius.
Solution:
Given PA = 24 cm, OP = 25 cm
We have OP2 = OA2 + PA2
⇒ (25)2 = OA2 + (24)2
⇒ 625 = OA2 + 576
⇒ OA2 = 49
⇒ OA = 7 cm
Question 14.
OABC is a Rhombus whose three vertices A, B and C lie on a circle with centre ‘O’. If the radius of the circle is 10 cm. Find the area of the Rhombus.
Solution:
Join OB
Now OA = OB (radii)
OA = OB (Sides of Rhombus)
∴ ∆OAB is an equilateral triangle.
Area of equilateral ∆OAB
= √3/4 a2 = √3/4 × (10)2 = 25√3 cm2
∴ Area of Rhombus = 2 × area of ∆OAB
= 2 × 25√3 cm2
= 50√3 cm2
Question 15.
The minute hand of a clock is √21 cm long. Find the area of described by the minute hand on the face of the clock between 7 : 00 am to 7 : 05 am (π = √22/7)
Solution:
60 min → 360°
1 min → 360°/60° = 6°
Time between 7 am to 1 : 05 am = 5 min
Angle = 5 × 6°= 30°
Radius, r = √21 cm
Area swept = θ/360 × πr2
= 30°/360° × 227 × √21 × √21 = 1/12 2/27 × 21
= 11/2 m2 = 5.5 m2
Question 16.
Find the area of quadrant of a circle, where the circumference of a circle is 44 cm. (π =√22/7)
Solution:
2πr = 44
2 × 22/7 × r = 44
r = 7 cm,
θ = 90°
Area of quadrant = θ/360 × πr2
= 90°/360° × 22/7 × 7 × 7 cm2
= 1/4 × 22 × 7cm2 = 77/2 = 38.5 cm2
Question 17.
Find the area of a sector of a circle whose radius is 7 cm and angle at the centre is 60°.
Solution:
Radius = 7 cm, Angle at centre = 60°
Question 18.
Find the area of a sector whose radius is 10 cm. and which make right angle at the center. (Take π = 3.14).
Solution:
Radius of the sector = 10 cm
Angle of sector = x° = 90°
Question 19.
Find the area of a sector whose radius is 10 cm. and which make right angle at the center. (Take π = 3.14).
Solution:
Radius of the sector =10 cm
Angle of sector = x° = 90°
Area of sector = A = x°/360° × πr2
= 90/360 × 3.14 × (10)2
= 1/4 × 3.14 × 100
= 1/4 × 314 = 78.5 cm2
Question 20.
Find the area of the shaded part in the given figure.
Solution:
Area of shaded part = Area of semi-circle - Area of triangle
= πr2/2 - 1/2 bh
= 22/7×7×7/2 - 1/2 × 14 × 6
= 11 × 7 - 7 × 6
= 77 - 42 = 35 sq. cm.