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Q1. In the figure, a circle of radius 7 cm in inscribed in a square. Find the area of the shaded region.

Solution

The shaded triangle is an isosceles triangle with side Area of shaded triangle = Side of the square = 2 radius of circle = 14 cm Area of rest shaded part = (14)² - = 196 - 154 = 42 cm² Total area of shaded region in figure = 49 + 42 = 91 cm²

Q2. A field is in the form a circle. A fence is erected around the field. The cost of fencing would be Rs. 2640 at the rate of Rs. 12 per meter. Then, the field is to be ploughed at the cost of Re. 0.50 per m² what is the amount required to plough the field?

Solution

Q3. The diameter of a circle whose area is equal to the sum of the areas of the two circles of radii 40 cm and 9 cm is

• 1) 41 cm
• 2) 62 cm
• 3) 49 cm
• 4) 82 cm

Solution

Q4. Find the area of a semicircle of diameter 14 cm. 

• 1) 77 cm²
• 2) 616 cm²
• 3) 154 cm²
• 4) 38.5 cm²

Solution

Area of a circle =r² Radius = r == 7 cm Area of a circle == 77 cm²

Q5.

Solution

 

Q6. The outer and inner diameters of a circular ring are 34 cm and 32 cm respectively. The area of the ring is:

• 1) 66
• 2) 60 cm²
• 3) 33 cm²
• 4) 29 cm²

Solution

Outer radius = 17 cm Inner radius = 16 cm Area of the ring = (R + r) (R-r) = (17 + 16) (17 - 16) = (33) (1) = 33 cm²

Q7. If the radius of a circle is r and the side of a square is x, and the circle is inscribed in the square, then find the remaining area left in the square.

• 1) x²-𝜋r²
• 2) 𝜋x²- r²
• 3) 𝜋r²- x²
• 4) 𝜋r²-2x²

Solution

Area of a circle is 𝜋r². Area of the square is x². Remaining area = x²-𝜋r²

Q8. Find the circumference of the circle, whose area is 144 cm².

• 1)
• 2)
• 3)
• 4)

Solution

Area of the circle = =>Radius of the circle, r = 12 cm Hence, circumference of the circle =  

Q9. A chord of a circle of radius 12 cm subtends an angle of 120^o at the centre. Find the area of the corresponding minor segment of the circle. (Use = 3.14)

Solution

Draw OM AB in figure (RHS congruence) AOM=BOM BOM = (cpct) In OMA cos 60^o =    OM = 6 cm and sin 60^o = AM = 6 cm ar AOB = AB OM = AM OM = 36 cm² area of segment AXB = area of sector OAXB - ar AOB = 3.14 12 12 - 36 = (150.72 - 36) cm²  = 88.36 cm²

Q10. The perimeter of a quadrant of a circle of radius  cm is

• 1) 3.5 cm
• 2) 5.5 cm
• 3) 7.5 cm
• 4) 12.5 cm

Solution

Q11. Find the area of a shaded region in the figure if BC = BD = 8 cm, AC = AD = 15 cm and O is the centre of the circle.(take π = 3.14)

Solution

Q12. If a square is inscribed in a circle, find the area of the remaining region inside the circle given the circumference of circle = 14 and the perimeter of a square = 24. Find the ratio of the area of the circle to the area of the square. 

• 1) 11:18
• 2) 77:18
• 3) 7:18
• 4) None

Solution

Circumference of a circle = 14 2r = 14 r = 7 cm Perimeter of a square = 24 4x = 24 x = 6 cm

Q13. If the circumference of a circle of radius 'r' and the perimeter of square of side 'a' are equal, then the ratio of area of the circle to that of the square is:

• 1) ² : 4
• 2) ² : 16
• 3) : 4
• 4) 4 :

Solution

Let 'r' be the radius of the circle and 'a' be the side of the square. Given, Circumference of the circle = Perimeter of the square Area of the circle = Area of the square = a² = Required ratio =

Q14. In the given figure, AC = BD = 7 cm and AB = CD = 1.75 cm. Semicircles are drawn as shown in the figure. Find the area of the shaded region.

Solution

Q15. The diameter of a wheel is 35 cm. How many revolutions will it make to travel 539 m? 

• 1) 470
• 2) 500
• 3) 490
• 4) 480

Solution

Diameter of the wheel = 35 cm Circumference of the wheel = × 35 cm = Total distance = 539 m = 53900 m Number of revolutions =

Q16. In the given figure, find the area of the shaded region.

Solution

Q17. The sum of circumferences of two circles is 132 cm. If the radius of one circle is 14 cm, find the radius of the second circle.

Solution

 

Q18. A chord of a circle of radius 12 cm subtends an angle of 60^o at the centre. Find the area of the corresponding segment of the circle. [use = 3.14 and = 1.73]

Solution

r = 12 cm = 60^o triangle is equilateral Area of segment =

Q19. A horse is tied to a peg at one corner of a square field of side 5 m by means of a rope. Find the area of that part of the field in which the horse can graze.

Solution

The part of the field that the horse will be able to graze is in the form of a sector whose sector angle is 90^o and the radius of the circle will be 5 m.

Q20.

Solution

Q21. In fig., AB is a diameter of the circle with centre O and OA = 7 cm. Find the area of the shaded region. (Use = )

Solution

Area of shaded region = (Area of semi-circle CADC - Area of ACD) + Area of circle of diameter OB Area of ACD = cm² = 49 cm² Area of semi-circle CADC = Area of circle of diameter OB = = 38.5 cm² Area of shaded region = [(77 - 49) + 38.5] cm² = 66.5 cm²

Q22.

Solution

Q23. In fig., find the area of the shaded region, where a circular arc of radius 6 cm is drawn with a vertex O of an equilateral triangle OAB of side 12 cm as centre.

Solution

Required Area = Area of sector = Area of triangle = Required Area =

Q24. In a circle of radius 21 cm, an arc subtends an angle of 60^o at the centre. Find (i) Area of sector formed by the arc (ii) Area of the segment formed by the corresponding chord.

Solution

(i) Area of sector = = (ii) Area of segment = area of sector - ar ()

Q25. If the circumference of a circle increases from 2π to 4π then its area is

• 1) Halved
• 2) Doubled
• 3) Tripled
• 4) Four times

Solution

We know that the circumference of a circle is given by 2πr. Original circumference = 2π units So, Original radius = 1 units Hence, Original area = πr² = π sq. units  Increased circumference (new) = 4π units = 2π(2) units So, new radius = 2 units  Hence, new area of the circle = πr² = 4π sq. units Thus, the area has become four times. 

Q26. Area of a quadrant of circle whose circumference is 22 cm is : ( = )

• 1) 9.625 cm²
• 2) 3.5 cm²
• 3) 17.25 cm²
• 4) 3.5 cm²

Solution

circumference of the circle:

Q27.

Solution

Q28. From each corner of a square of side 4 cm a quadrant of a circle of radius 1 cm is cut and also a circle of diameter 2 cm is cut. Find the area of the remaining portion of the square.

Solution

Q29. In the given figure, PS is the diameter of a circle of length 6 cm. Q and R are the points on the diameter such that PQ, QR and RS are equal. Semicircles are drawn with PQ and QS as diameters. Find the perimeter of the shaded portion.

Solution

       

Q30. In the following figure, area of shaded region is

• 1)
• 2)
• 3)
• 4)

Solution

Q31. The diameter of a wheel is 42 cm. How many revolutions will it take to travel 462 m? 

• 1) 351
• 2) 350
• 3) 352
• 4) 353

Solution

Diameter of the wheel = 42 cm Circumference of the wheel =× 42 cm = Total distance = 462 m = 46200 m Number of revolutions =

Q32. Find the semi-perimeter of a circle of radius 21 cm. 

• 1) 132 cm
• 2) 66 cm
• 3) 108 cm
• 4) 27 cm

Solution

Radius of a circle = 21 cm Semi-perimeter of a circle = r + 2r=  = 108 cm

Q33. In fig., ABCD is square of side 8 cm CBED and ADFB are quadrants of circle. Find the area of the shaded region. (Use = 3.14).

Solution

Area of quadrant ABFD = cm² = 50.24 cm² _{Area of ABD =} Shaded area = 2(Area of quadrant ABFD - Area of _ABD) = 2(50.24 - 32) cm² = 2(18.24) cm² = 36.48 cm²

Q34. In the given figure, sectors of two concentric circles of radii 7 cm and 3.5 cm are shown. Find the area of the shaded region.

Solution

Q35. A wire is in the shape of a circle of radius 21 cm. It is bent to form a square. The side of the square is : .

• 1) 22 cm
• 2) 44 cm
• 3) 66 cm
• 4) 33 cm

Solution

Perimeter of square = Perimeter of circle Perimeter of square = 4(Side) Side of the square = = 33 cm

Q36. A child draws the figure of an aeroplane as shown. Here, the wings ABCD and FGHI are parallelograms, the tail DEF is an isosceles triangle, the cockpit CKI is a semi-circle and CDFI is a square. BP is perpendicular to CD, HQ is perpendicular to FI and EL is perpendicular to DF. If CD = 8 cm, BP = HQ = 4 cm and DE = EF = 5 cm, find the area of the whole figure.

Solution

Q37. If the area and circumference of a circle are numerically equal, then find the radius of the circle.

Solution

Let r be the radius of circle

Q38. In the given figure, diameter AB is 12 cm long. AB is trisected at points P and Q. Find the area of the shaded region.

Solution

AP = PQ = QB = = 4 cm Area of shaded region = 2 [Area of semicircle with radius AW - area of semicircle with diameter AP] = 2 = (36 - 4) = 32 cm²

Q39. The minute hand of a clock is 12 cm long. Find the area of the face of the clock described by the minute hand in 35 minutes.

Solution

Q40. In the figure, the chord AB of a circle of radius 10 cm subtends an angle of 90^o at the centre O. Find the area of the segment ACBA. (take = 3.14)

Solution

Area of the segment = Area of the sector - area of triangle          _{(Since the triangle is a right angled triangle , Area = )} = 78.5 - 50 = 28.5 sq. cm

Q41. In triangle ABC, angle A = 90^o, AB = 12 cm and BC = 20 cm. Three semi-circles are drawn with AB, AC and BC as diameters. Find the area of the shaded portion.

Solution

 

Q42. In figure, find the area of the shaded design, where ABCD is a square of side 10 cm and semi circles are drawn with each side of square as diameter. (use = 3.14)

Solution

Let us mark the four unshaded areas as I, II, III and IV. Area of I + Area III = Area of square - Area of two semi-circles = (100 - 3.14 25) cm² = (100 - 78.5) cm² = 21.5 cm² Similarly, Area of II + Area of IV = 21.5 cm² So, area of shaded region = ar of ABCD - ar (I + II + III + IV) = (100 - 2 21.5) cm² = 100 - 43 = 57 cm²

Q43. Find the area of shaded region in fig., if PQ = 16 cm, PR = 12 cm and O is the centre of the circle. ( = 3.14)

Solution

RPQ = 90^o (angle in a semi circle) RP² + PQ² = RQ² (Pythagoras theorem) RQ² = 12² + 16² = 144 + 256 = 400 = 20² Radius of the circle is 10 cm. Area of shaded region is = Area of semicircle - area of RPQ Required are = ½r² - ½ b h = ½ 10 10 - ½ 12 16 cm² = 3.14 50 - 96 cm² = 157- 96 cm² = 61 cm²

Q44.

Solution

Q45. The shape of the top of a table in a restaurant is that of a sector of a circle with centre O and angle BOD = 90^o, as shown in the figure. If OB = OD = 60 cm, find the perimeter of the table top.

Solution

Angle made by the sector at the centre = 360^o - 90^o = 270^o r = 60 cm Perimeter of the table top (sector) = = 2r + = 2 60 + = 120 + 282.6 = 402.6 cm

Q46. In fig., OPQR is a rhombus whose three vertices P,Q,R lie on a circle of radius 8 cm. Find the area of the shaded region.

Solution

Let PR and OQ intersect at S. Clearly, OP = OQ = OR = 8 cm Since, the diagonals of a rhombus bisect each other at right angles, OS = 4 cm and OSP = 90^o RS = = cm Now, PR = 2RS PR = 2RS PR = cm Area of rhombus OPQR =

Q47. The length of a rope by which a cow is tethered in increased from 16 m to 23 m. How much additional area can the cow graze now?

Solution

Additional area covered by cow = 858 m²

Q48. In the figure, arcs are drawn by taking vertices A, B and C of an equilateral triangle of side 10 cm, to intersect the sides BC, CA and AB at their respective mid-points D, E and F. Find the area of the shaded region (Use = 3.14)

Solution

Since ABC is equilateral A = B = C = 60^o Area of sector Areas of all three sectors are equal. Total area of shaded region = cm² cm² = 39.25 cm²

Q49. What will be the increase in area of circle if its radius is increased by 40%?

Solution

Q50. The diameter of a cycle wheel is 21 cm. How many revolutions will it make to travel 1.98 km?

Solution

Q51. The area of an equilateral triangle ABC is 17320.5 cm². With each vertex of the triangle as centre, a circle is drawn with radius equal to half the length of the sides of the triangle. Find the area of the shaded region. (Use = 3.14 and = 1.73205)

Solution

Q52. In the given figure, find the area of the shaded region, where ABCD is a square of side is 14 cm. APD and BPC are semi-circles.

Solution

Q53. The inside perimeter of a running track shown in the figure is 400 m. The length of each of the straight portions is 90 m, and the ends are semi-circles. If the track is 14 m wide every where, find the area of the track. Also, find the length of the outer boundary of the track.

Solution

Q54. A car has two wipers which do not overlap. Each wiper has a blade of length 25 cm sweeping through an angle of 115^o. Find the total area cleaned at each sweep of the blades.

Solution

Q55.

Solution

Q56. In the figure, a square OABC is inscribed in a quadrant OPBQ. If OA = 20 cm, find the area of the shaded region.

Solution

 

Q57. In the figure, two circular flower beds have been shown on two sides of a square lawn ABCD of side 56 m. If the centre of each circular flower bed is the point of intersection O of the diagonals of the square lawn, find the sum of the areas of the lawn and flower beds.

Solution

Total Area = Area of sector OAB + Area of sector OCD + Area of OAD + Area of OBC = = 4032 m²