Important Questions: Areas related to Circles- 2

# Class 10 Maths Chapter 11 Important Question Answers - Areas Related to Circles

## Short Answer Type Questions

Q1: Find the area of the shaded region if PQ = 24cm, PR = 7cm and O is the centre of the circle.
Ans:
Given PQ = 24 cm and PR = 7 cm
Since QR is a diameter passing through the centre O of the circle
Angle of semi-circle, ∠RPQ = 90°

Diameter of the circle = 25cm
Thus radius, r = 25/2 cm
Area of the semi-circle = (1/2)πr2

From the given figure,
Area of the shaded region = Area of the semi-circle with O as centre and OQ as radius – area of ΔPQR
Hence, area of the shaded region can be calculated as,
Area of the shaded region =

Area of the shaded region = 161.54cm2.

Q2: The perimeter of a sector of a circle of radius 5.7 m is 27.2 m.Calculate:
(i) The length of arc of the sector in cm.
Ans:
Given radius, r = 5.7 m and perimeter = 27.2 m.
Then OA = OB = 5.7m
Now,

The length of the arc of the sector is 15.8 m.
(ii) The area of the sector in cm2 correct to the nearest cm2.
Ans:
Area of sector OACB=

Area of sector correct to nearest cm= 45cm2

Q3: Find the area of the shaded region in the given figure where ABCD is a square of side 10cm and semi-circles are drawn with each side of the square as diameter. [π=3.14]
Ans:
Given, side of square ABCD = 10 cm
Area of square ABCD = (side)2 = (10)2 = 100 cm2
Given that semicircle is drawn with side of square as diameter.
Diameter of semicircle = 10 cm.
Radius of semicircle = 5 cm
Area of semi-circle =
Area =
Area of 4 semi-circles – Area of shaded region = Area of square ABCD
∴Area of shaded region = Area of 4 semi-circles - Area of square ABCD
Area of shaded region = 4×39.25−100 = 57cm2

Q4: In the given figure ΔABC  is an equilateral triangle inscribed in a circle of radius 4 cm and centre O. Show that the area of the shaded region is 4/3
Ans:
Given, ΔABC as an equilateral triangle inscribed in circle with radius, r = 4 cm.
In  ΔOBD, Let BD = a, OB = 4cm

Area of shaded region = Area of sector OBPC – Area of ΔOBC

Q5: The radii of two circles are 8cm and 6cm respectively. Find the radius of the circle having its area equal to the sum of the areas of the two circles.
Ans:
Given, radii of two circles, r1=8 cm and r2=6 cm.
Area of a circle is given by πr2
Let A1 = Area of the first circle

Hence, required radius = 10cm

Q6: A chord of a circle of radius 10cm subtends a right angle at the centre. Find the area of the corresponding: (Use π=3.14)
(i) minor sector
Ans:
Given, radius of circle, r = 10cm subtending an angle 90∘ at the centre.
∴∠AOB = 90°.
Area of minor sector =

Area of minor sector is 78.5cm2.
(ii) major sector
Ans:
Area of circle = πr= π(10)2
Area of major sector = Area of circle – Area of minor sector

Area of major sector is 235.5 cm2.
(iii) minor segment
Ans:
Area of minor segment = Area of minor sector OAB – Area of ΔOAB

Area of minor segment is 28.5 cm2.
(iv) major segment
Ans:
Area of major segment = Area of the circle – Area of minor segment

Area of major segment is 285.5 cm2.

Q7: In Akshita’s house, there is a flower pot. The sum of radii of circular top and bottom of a flowerpot is 140 cm and the difference of their circumference is 88cm, find the diameter of the circular top and bottom.
Ans:
Given, sum of radii of circular top and bottom = 140cm
Let radius of top = r cm
∴ Radius of bottom = (140−r)cm
Circumference of top = 2πr cm
Circumference of bottom = 2π(140−r)cm
Difference of circumference = [2πr−2π(140−r)]cm
By the given conditions in the problem,

∴ Radius of top is obtained as 77cm
Diameter of top can be calculated as,
⇒ 2×77 = 154cm
Radius of bottom can be obtained as,
⇒ 140  r = 140  77
= 63cm
∴ Diameter of bottom 2×63=126cm .

Q8: A chord of a circle of radius 15cm subtends an angle of 60° at the centre. Find the area of the corresponding minor and major segments of the circle. (use π=3.14 and √3 =1.73 ).
Ans:
Given, radius of circle, r = 15 cm subtending an angle 60° at centre.
∴∠AOB = 60°.
Area of a sector of a circle is given by,
Area of minor segment = Area of minor sector OAB − area of ΔOAB

Area of major segment = Area of the circle – area of minor segment

Area of minor segment is 20.4375 cm2 and area of major segment is 686.0625 cm2.

Q9: A round table cover has six equal designs as shown in the figure. If the radius of the cover is 28cm, find the cost of making the design at the rate of Rs.0.35 per cm2. (Use √3 =1.7)
Ans:
Given, radius of cover, r = 28 cm. Table has six equal designs and the cost of making the design is Rs. 0.35 per cm2.
Area of equilateral triangle is given by,
From the figure area of one design =

Total cost can be calculated as follows,

The total cost of making the design is Rs.16.268.

Q10: ABCD is a flower bed. If OA = 21m and DC = 14m. Find the area of the bed.
Ans:
Given,
OA = R = 21m
OC = r = 14m
Area of the flower bed (i.e., shaded portion) = area of quadrant of a circle of radius R of the quadrant of a circle of radius r

Hence, area of the flower bed is 192.5 m2.

Q11: In a circle of radius 21 cm, an arc subtends an angle of 60° at the centre. Find:
(i) the length of the arc.
Ans:
Given radius, r = 21 cm and angle θ = 60°.
Length of arc

(ii) area of the sector formed by the arc.
Ans:
Area of the sector =

(iii) area of the segment formed by the corresponding chord
Ans:
Area of segment formed by corresponding chord =
⇒ Area of segment =231−  Area of ΔOAB   …(1)
In right triangles OMA and OMB,

In right angled ΔOMA,

Using eq. (1),
Area of segment formed by corresponding chord =

Q12: A chord of a circle of radius 15 cmsubtends an angle of 60° at the centre. Find the area of the corresponding segment of the circle.  (Use π = 3.14 and √3 = 1.73)
Ans:
Given radius, r = 15 cm and angle θ = 60°
Area of minor sector can be obtained from

Q13: A chord of a circle of radius 12 cm subtends an angle of 120° at the centre. Find the area of the corresponding segment of the circle. (Use π = 3.14 and √3 = 1.73)
Ans:
Given radius, r  = 12cm and angle θ = 120°.
Area of corresponding sector can be obtained by using the formula

Area of corresponding segment = Area of corresponding sector – Area of ΔAOB
⇒150.72  62.28 = 88.44 cm2.

Q14: Find the area of the shaded region  in  figure,  if PQ = 24 cm, PR = 7 cm and O is  the centre of the circle.
Ans:
Given radius, r = 12cm and angle θ = 120°.
Area of corresponding sector can be obtained by using the formula

Area of corresponding segment = Area of corresponding sector – Area of ΔAOB
⇒150.72  62.28 = 88.44 cm2

## Long Answer Type Questions

Q1: Three horses are tethered with 7m long ropes at the three corners of a triangular field having sides 20m, 34m and  42 m. Find the area of the plot which can be grazed by the horses. Also, find the area of the plot which remains ungrazed.
Ans:
Given, length of ropes at the three corners, l = 33 m and the sides of the triangle as 20 m, 34 m and 42 m.
From the figure let, ∠A = θ1°, ∠B = θ2° and ∠C = θ3°
Area which can be grazed by three horses = Area of sector with central angle θ1° and radius 7 m+ Area of sector with central angle θ2° and radius 7 cm+ Area of sector with central angle θ3° and radius 7 cm

Area grazed by the horses = 77m2
∴ Ungrazed area can be calculated by subtracting area grazed by the horses from the area of the triangular plot.
⇒ Ungrazed area =(33677) = 259 m2

Q2: In the given figure, two circular flower beds have been shown on two sides of a square lawn ABCD of side 56 m.  If the corner of each circular flower bed is the point of enter section O of the diagonals of the square lawn. Find the sum of the areas of the lawn and the flower beds.
Ans:
Given, side of square lawn = 56 m.
Area of the square lawn ABCD, A = (56)2
Let OA=OB=x
∴By Pythagoras theorem,
⇒ x2+x2 = 562
⇒ 2x= 56×56
⇒ x= 28×56
Again, area of sector
Substituting for x, we get
⇒  Area of sector

(Here, ∠AOB = 90° since square ABCD is divided into 4 right triangles)
∴ Area of flowerbed AB =

Area of flower bed  AB = 448m2
Similarly, area of the other flowerbed = 448m2
Total area of the lawn and the flowerbeds = 56×56+448+448 = 4032m2

Q3: An elastic belt is placed round the rein of a pulley of radius 5 cm. One point on the belt is pulled directly away from the centre O of the pulley until it is at P, 10 cm from O. Find the length of the best that is in contact with the rim of the pulley. Also find the shaded area.
Ans:
Given, radius of rein of pulley, OA = 5 cm.
OP = 10cm
Thus, cosθ= OA/OP

Length of the belt that is in contact with the rim of the pulley can be calculated by subtracting length of arc AB from Circumference of the rim.
Length of the belt
Length of the belt = 20/3 πcm
Now, area of sector OAQB=

Area of quadrilateral OAPB is given by, 2(Area of ΔOAP) = 25√3cm2
Hence, shaded area can be obtained as,
Area of shaded region=
Hence, area of shaded region is 17.12 cm2.

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