Q1. Area of a sector of angle P (in degree) of a circle with radius R is : [SEBA 2015]
(a) (b) (c) (d)
Solution: (d)
Q2. If the sum of the areas of two circles with radii and is equal to the area of a circle of radius,then
(a) (b)
(c) (d)
Solution: (b)
Q3. If the perimeter of a circle is equal to that of a square , then the ratio of their areas is :
(a) 22 : 7 (b) 14 : 11 (c) 7 : 22 (d) 11 : 14
Solution : (b) 14 : 11 .
[ let be the radius of the circle and be side of the square .
A/Q ,
]
Q4. If the circumference of a circle is 44 cm , then its area is : [SEBA 2016]
(a) 276 (a) 44
(c) 176 (d) 154
Solution: (d) 154 .
[ A/Q,
cm
The area of the circle ]
Q5. The area of the circle that can be inscribe in a square of side 6 cm is : [ SEBA 2017]
(a) (b)
(c) (d)
Solution: (d) .
[ The square of the side cm and the radius cm
The area of the circle
]
Q6. If the circumference of a circle is 22 cm , then the area of a quadrant of the circle is : [SEBA 2018]
(a) (b) 77
(c) (d)
Solution : (a)
[ A/Q ,
cm
The area of a quadrant of the circle
]
Q7. The degree measure of the angle at the centre of a circle is 1 . The area of the sector is : [SEBA 2019]
(a) (b)
(c) (d)
Solution: (d)
Q8. The degree measure of the angle at the centre of a circle is . The length of a arc of the sector is :
(a) (b) (c) (d)
Where is the radius of the circle . [ SEBA 2020]
Solution: (b)
Q9. If the sum of the circumference of two circles with radii and is equal to the circumference of a circle of radius , then
(a) (b)
(c) (d)
Solution: (a)
Q10. If the radius of a circle is doubled , then the ratio of the areas of the new circle to the area of the given circle is :
(a) 2 : 1 (b) 1 : 2
(c) 1 : 4 (d) 4 :1
Solution: (d) 4 : 1
[ Here , and
]
Q11. In the given figure , if the angle 30° , then the area of the shaded region (in cm square) is : [CBSE 2011]
(a) 3.36 (b) 2.26 (c) 1.36 (d) 2.36
Solution: [ Here , cm and cm
The area of shaded region
]
Q12. If the area of a circle is equal to the sum of the areas of two circles of diameters 10 cm and 24 cm, then diameter of the larger circle (in cm) is : [CBSE 2012]
(a) 34 (b) 26 (c) 17 (d) 14
Solution: (b) 26
[ Here , cm and cm
A/Q ,
cm
Therefore , the diameter cm cm ]
Q13: If is taken as , the distance (in metres) covered by a wheel of diameter 35 cm , in one revolution is : [CBSE 2013]
(a) 2.2 (b) 1.54 (c) 9.625 (d) 96.25
Solution: (b) 1.54
[ Here , Diameter cm and Radius cm
Therefore, the distance covered by a wheel in one revolution
m m m ]
Q1. 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 is .
Solution: 50 cm .
[ Here , cm and cm
A/Q ,
cm
Therefore , the diameter cm cm ]
Q2. Area of a sector to circle of radius 36 cm is , then the length arc of the corresponding arc of the circle is .
Solution: .
[ Here , cm
A/Q,
Therefore, the length arc of the corresponding arc of the circle
]
Q3. If the diameter of a semi-circular protractor is 14 cm ,then its perimeter is . [CBSE 2009]
Solution : 22 cm .
[ Here , cm and cm
The perimeter of a semi-circular cm ]
Q1. In given figure , find the area of the sector of a circle with radius 4 cm .
Solution: Here , cm and
Therefore, the area of the sector of a circle
Q2. Find the length of the arc of a circle of diameter 42 cm which subtends angle of 60° at the circle ? [CBSE 2012]
Solution: Here , radius and
Therefore, the length of the arc of a circle
cm
cm
Q3. Find the area of a quadrant of a circle whose circumference is . [ CBSE 2014]
Solution: We have,
cm
Therefore , the area of quadrant of the circle
.
Q4. The difference between the circumference and radius of a circle is 37 cm . Find the area of the circle . [ CBSE 2013]
Solution: A/Q ,
cm
Therefore, the area of circle .
Q5. Find the area of a sector of a circle of radius 28 cm and central angle 45° .
Solution: Here , cm and
Therefore, the area of a sector of a circle
Q6. The radii of radius circles are 19 cm and 9 cm respectively . Find the radius of the circle which has circumference equal to the sum of the circumference of the two circles.
Solution: let be the radius of the new circle .
Here , cm and cm
A/Q ,
cm
Q1. Find the area of a sector of a circle with radius 6 cm if angle of the sector is 60° .
Solution: Here , cm and
Therefore, the area of a sector of a circle
Q2. The cost of fencing a circular field at the rate of Rs. 24 per metre is Rs. 5280 , then find the radius of field .
Solution: let be the radius of the circular field .
The length of the fence field m
A/Q ,
m
m
Q3. An umbrella has 8 ribs which are equally spaced and the umbrella to be a flat circle of radius 45 cm , find the area between the two consecutive ribs of the umbrella .
Solution: Here , cm
The area between the two consecutive ribs of the umbrella
Q4. In figure, O be the centre of two concentric circles , cm and cm .Find the area of the shaded region .
Solution: Here , , cm and cm
Therefore, the area of the shaded region
Q5. In figure, OABC is a quadrant of a circle with centre O and radius 3.5 cm . If cm ,find the area of shaded region .
Solution: Here , cm and cm
Therefore, the area of shaded region
Q6. If the area of a sector of a circle of radius 14 cm is . Find the length of the corresponding arc of the sector . [CBSE 2011]
Solution: Here , cm
A/Q ,
Therefore, the length of the corresponding arc of the sector
cm
cm
Q1. A chord of a circle of radius 10 cm subtends a right angle at the centre .Find the area of the corresponding :
(i) minor segment (ii) major sector . [ Use ]
Solution: Here , cm and
(i) The area of the corresponding minor segment Area of sector – area of triangle
(ii) Area of major sector
Q2. A chord of a circle of radius 15 cm subtends an angles of 60° at the centre . Find the areas of the corresponding minor and major segments of the circle . [SEBA 2017]
Solution: Here , radius cm ,
Area of the minor segment
Area of the sector – Area of the triangle formed by radius and chord
Area of major segment Area of the circle – Area of the minor segment
Q1. In figure, a square OPQR is inscribe in a quadrant OAQB of a circle. If the radius of the circle is cm , find the area of the shaded region . [Use π = 3.14] [CBSE 2020 standard]
Solution: Given, be a square .
So , cm
Area of square
In , we have
cm
Radius cm
Area of the quadrant
Therefore, the area of shaded region
Q3. In figure , ABCD is a square with side cm and inscribed in a circle .Find the area of the shaded region . [Use ] [CBSE 2019]
Solution: Given cm and We join AC .
InABC , we have
Radius
Therefore, the area of the shaded region Area of circle – Area of ABCD
Q4. Find the area of the shaded design in figure, where ABCD is a square of side 10 cm and semicircles are drawn with each side of the square as diameter . [ Use ]
Solution: Here , cm , Radius cm
Area of unshaded region part and part
Area of ABCD Area of two semicircle
Area of unshaded region part and part .
Therefore, the area of shaded design
Area of square ABCD Area of
Q5. Calculate the area of the designed region in figure ,common between the two quadrants of circles of radius 8 cm each . [CBSE 2012]
Solution: The radius of the quadrant of a circle cm
And the side of the square cm
Therefore, the area of the shaded region
Area of two quadrant of the circle – Area of the square
Q6. In figure , a square OABC is inscribed in a quadrant OPBQ .If OA = 15 cm , find the area of the shaded region . [ Use ] [CBSE 2019]
Solution: Since OABC is a square .
So , cm and we join OB .
In OAB we have ,
cm
Radius cm
Therefore , the area of shaded region
Area of quadrant of the circle – Area of square OABC
Q1. In figure, AB and CD are two diameter of a circle (with centre) perpendicular to each other and OD is the diameter of the smallest circle . If , find the area of the shaded region . [CBSE 2013,2020 Basic]
Solution: For big circle : Radius cm ,
Diameter cm
So, cm
For small circle : Diameter cm
and Radius cm
Area of small circle
Area of
Area of semicircle
The area of the shaded region
Area of small circle Area of semicircle –
Q2. In figure, ABC is a quadrant of a circle of radius 14 cm and a semicircle is drawn with BC as diameter . Find the area of the shaded region . [CBSE 2014]
Solution: For ABCD quadrant :
Here , Radius cm
In∆ABC , we have
cm
area of
And Area of quadrant ABCD
For semi-circle BEC :
Diameter cm ; Radius cm
Area of semi-circle BEC
Area of shaded region ar() ar(semicircle BEC) – ar(quadrant ABCD)
Q3. In figure, ABC is a right angled triangle in which , cm and cm , then the semi-circles are described on AB , BC and AC as diameters . Find the area of the shaded region .
Solution: Here , cm and cm
In ABC , we have
cm
The radius of three semicircle are : cm ; cm
And cm
Therefore, the area of the shaded region
Area of semicircle with diameter AB Area of triangle ABC
Area of semicircle with diameter AC Area of semicircle with diameter BC