Ncert – Class 10 – Area Related to Circles– Hots Question
Question 1
A chord PQ of the circle of radius 10 cm subtends an angle of 60 degrees at the centre of the circle. Find the area of minor and major segment of the circle.
Solution
25/21(220 + 21√3)𝑐𝑚2
Question 2
Rithika bought a pendulum clock for her living room. That clock contains a small pendulum of
length 15cm. the minute hand and hour hand are 9cm and 6cm long respectively. Based on the
above information answer the following questions.
i) Find the area swept by the hour hand in 1 hour?
ii) Find the area swept by the hour hand between 11am and 5pm?
iii) Find the area swept by the minute hand in 10minutes?
Solution
(i)Length of hour hand = 6cm
In 12 hours central angle = 360°
1 hour “ “ = 𝟑𝟔𝟎°/𝟏𝟐= 30°
ii) Arc hour hand 21 am to 5 pm
(11a.m to 5 p.m) = 6 places
Hour hand moves central angle = 6 x 30° = 180 °
Area = (𝟏𝟖𝟎/𝟑𝟔𝟎)x(𝟐𝟐/𝟕)x 6 x 6 = 𝟑𝟗𝟔/𝟕= 56.571cm2
iii) length of minute hand = 9cm
In 10 mins. = 2 places = 30° x 2 = 60°
Area (formed in 10 mins.) = 𝟔𝟎/𝟑𝟔𝟎°x𝟐𝟐/𝟕x 9 x 9 = 𝟐𝟗𝟕/𝟕
= 42.42cm2
Question 3
Pendulum Clock : It is a clock that uses a pendulum, a swinging weight, as its timekeeping element. From its invention in 1656 by Christiaan Huygens, the pendulum clock was the world’s most precise timekeeper, accounting for its widespread use. Their greater accuracy allowed for the faster pace of life which was necessary for the Industrial Revolution. The home pendulum clock was replaced by less-expensive, synchronous, electric clocks in the 1930s and 40s. Pendulum clocks are now kept mostly for their decorative and antique value.
Dhriti bought a pendulum clock for her living room. the clock contains a small pendulam of lenght 45 cm. the minute hand and hour hand of the clock are 9 cm and 6 cm long respectively.
(i) Find the area swept by the minute hand in 14 minutes.
(ii) Find the angle described by hour hand in 10 minutes.
(iii) Find the distance covered by the tip of hour hand in 3.5
hours.
(iv) If the tip of pendulum covers a distance of 66 cm in
complete oscillation, then find the angle described by
pendulum at the centre.
Solution
i)In 14minutes, minutehand will rotate=360° /6014=84° Therefore the area swept by the minute hand in 14 minutes will be the area of a sector of 84° in a circle of 9cm radius =Area of sector of angle
=84° /360° π r2 =84° /360° 22/7 99=59.4cm2
ii)Hour hand completes one full revolution in 12 hours
Angle described in 1 hour=360°/12=30°
Angle described in 60 minutes=30°
Angle described in 10 minutes=(30/60)10=5°
iii) Arc length=Angleradius
1 hour=60° =360°
3.5 hour=3.5360π /180
7π
length=7π 6 = 42π
iv) 2d=66 cm d =33cm p= angle/3602π r
33=angle/360°222/7*45
42°
Question 4
A chord subtends an angle of 90°at the centre of a circle whose radius is 20 cm. Compute the area of the corresponding major segment of the circle
Solution
Area of the sector = θ/360 × π × r
2
Base and height of the triangle formed will be = radius of the circle
Area of the minor segment = area of the sector – area of the triangle formed
Area of the major segment = area of the circle – area of the minor segment
Now,
Radius of circle = r = 20 cm and
Angle subtended = θ = 90°
Area of the sector = θ/360 × π × r
2 = 90/360 × 22/7 × 202
Or, area of the sector = 314.2 cm2
Area of the triangle = ½ × base × height = ½ × 20 × 20 = 200 cm2
Area of the minor segment = 314.2 – 200 = 114.2 cm2
Area of the circle = π × r
2 = (22/7) × 202 = 1257.14
Area of the major segment = 1257.14 – 114.2 = 1142 .94 cm2
So, the area of the corresponding major segment of the circle = 1142 .94 cm2
Question 5
Calculate the perimeter of an equilateral triangle if it inscribes a circle whose area is 154 cm2
Solution
Here, as the equilateral triangle is inscribed in a circle, the circle is an incircle.
Now, the radius of the incircle is given by,
r = Area of triangle/semi-perimeter
In the question, it is given that area of the incircle = 154 cm2
So, π × r2 = 154
Or, r = 7 cm
Now, assume the length of each arm of the equilateral triangle to be “x” cm
So, the semi-perimeter of the equilateral triangle = (3x/2) cm
And, the area of the equilateral triangle = (√3/4) × x2
We know, r = Area of triangle/semi-perimeter
So, r = [x2
(√3/4)/ (3x/2)]
=> 7 = √3x/6
Or, x = 42/√3
Multiply both numerator and denominator by √3
So, x = 42√3/3 = 14√3 cm
Now, the perimeter of an equilateral triangle will be = 3x = 3 × 14√3 = 72.7 cm.
Question 6
A chord PQ of length 12 cm subtends an angle 120 degree at the centre of a circle. Find the area of the minor segment cut off by the chord PQ.
Solution
Now,
PD = PQ × (0.5) = 12 × 0.5 = 6 cm
Since, ∠QOP = 120o
Also,
∠DOP = ∠QOD = 60o
We have,
In triangle OPD, we have
Sin θ = PD/ OP
Sin 60o = 6/ OP
Substituting the values, we get,
√3/2 = 6/ OP
OP = 12/√3 = 4√3 = r
Now, calculating the area of the minor segment
Area of the segment = area of sector OPUQO – area of △OPQ
= (θ/360) x πr2 – ½ x PQ x OD
= (120/360) x π(4√3)2 – ½ x 12 x 2√3
{OD=√[(4√3)X(4√3) − (6𝑋6)] = 2√3}
= 16π – 12√3 = 4(4π – 3√3)
Hence, Area of the minor segment = 4(4π – 3√3) cm2
Question 7
A fountain is enclosed by a circular fence of circumference11m and is surrounded by a circular path .The circumference of the outer boundary of the path is16m.A gardener increase the area of the path to decrease the area enclosed by the fence such that the length of the fence decrease by 3m.The path is covered by bricks which cost Rs125 per m2. What will be the total cost to the nearest whole number required to cover the area by the bricks (use 𝜋 =22/7) .
Solution
We have
Circumference of the outer boundary of the path=16m
2πr1=16
r1=8/πm
And length of decreased fence=11-8=3m
2πr2=8
r2=4/πm
Area of path=π[r1^2-r2^2]
=[64-16]/πm^2
=48/πm^2
Total cost to cover the path by bricks= Rs
48/π×125=48/22×7×125=1909.09=1910.
Question 8
Two circular flower beds are made on a pair of opposite sides of a square lawn ABCD of side
56 m as shown in the figure. If centre of each circular flower bed is the point of intersection
O of the diagonals AC and BD, find the cost of preparing the flower beds at the rate of 25 perm²
Solution
Side of square ABCD=56 m
AC=BD(diagonals of a square are equal in lengths)
Diagonal of square (AC) =√ 2× side of square
= √2×56=56√2 m
OA=OB= 1/2AC=1/2 (56
√2 )=28√2 m
Let OA=OB=r m[radius of the sector]
Area of sector OAB=[ 90 °/360°]πr²
=( 1/4)πr ²
= 1/4 ×22/7×( 28/√2 )²m²
=[ 1/4 × 22/7×28×28×2] m²
=1232 m²
Area of flower bed AB=area of sector OAB− area of ΔOAB
⇒1232− 1/2 ×OB×OA
⇒1232− 1/2×28 √2 ×28 √2
⇒1232−784=448m²
Similarly, Area of other flower bed CD=448m²
Total area = Area of square ABCD+area of flower bed AB+Area of flower bed CD
=(56×56)+448+448=4032m²
Hence, sum of the area of lawns and the flower beds are 4032m²
Question 9
Find the area of the sector of a circle with a radius of 4cm and of angle 30°. Also, find the area of the corresponding major sector.
Solution
Radius = r = 4 cm, θ=30°
Area of sector = [𝜃/360]×𝜋𝑟2
= 30/360×3.14×(4)2
= 1/12×3.14×4×4
= 1/3×3.14×4
= 12.56/3 cm2
= 4.19 cm2
Area of major sector = ((360 − θ)/360)×𝜋𝑟2
= ((360 − 30))/360×3.14×(4)2
= 330/360×3.14×4×4
= 11/12×3.14×4×4
= 46.05 cm2
Question 10
A square is inscribed in a circle. Calculate the ratio of the area of the circle and
the square.
Solution
As the square is inscribed in a circle, a diagonal of the square will be = the
diameter of the circle.
Let “r” be the radius of the circle and “d” be the length of each diagonal of the
square.
We know,Length of the diagonal of a square = side (s) × √2
So,
d = 2r
And, s × √2 = 2r
Or, s = √2r
We know, the area of the square = s2
Thus, the area of the square = (√2r)2 = 2r2
Now, the area of the circle = π × r2
∴ Area of the circle : area of the square = π × r2 : 2r2 = π : 2
So, the ratio of the area of the circle and the square is π : 2.
