∴ Radius, r = d/2
Area of semi circle
=
Q2: Case Study : Governing council of a local public development authority of Dehradun decided to build an adventurous playground on the top of a bill, which will have adequate space for parking.
After survey, it was decided to build rectangular playground, with a semicircular area allotted for parking at one end of the playground. The length and breadth of the rectangular playground are 14 units and 7 units, respectively. There are two quadrants of radius 2 units on one side for special seats.Based on the above information, answer the following questions:(i) What is the total perimeter of the parking area ?
(ii) (a) What is the total area of parking and the two quadrants?
(b) What is the ratio of area of playground to the area of parking area ?
(iii) Find the cost of fencing the playground and parking area at the rate of ? 2 per unit. ^{ }[2023, 4,5,6 Mark]
Ans: (i) Length of play ground . AB = 14 units, Breadth of play ground. AD = 7 units
Radius of semi  circular part is 7/2 units
Total perimeter of parking area = πr + 2r
=
= 11 + 7 = 18 Units
(ii) (a): Area of parking = πr^{2} / 2
=
= 19.25 sq. units
Area of two quadrants (I) a n d [II) =1/2 x 1/4 x πr^{2}
=
= 6.29 sq. units
Total area of parking and two quadrant
= 19.25 + 6.29
= 25.54 sq. units
Q3: A chord of a circle of radius 14 cm subtends an angle of 60° at the centre. Find the area of the corresponding minor segment of the circle. Also find the area of the major segment of the circle. ^{ }[2023, 4,5,6 Mark]
Ans: Here, radius t(r) = 14 cm and Sector angle (θ) = 60°
∴ Area of the sector
= 102.67 cm^{2}
Since ∠O = 60° and OA = OB = 14 cm
∴ AOB is an equilateral triangle.
⇒ AS = 14 cm and ∠A = 60°
Draw OM ⊥ AB.
In ΔAMO
Now,
Now, area of the minor segment= (Area of minor sector)  (ar ΔAOB)
= 102.67  84.87 cm^{2}
= 17.8 cm^{2}
Area of the major segment
= Area of the circle  Area of the minor segment
= (616  17.8) cm^{2 }= 598.2 cm^{2}
Q2:The number of revolutions made by a circular wheel of radius 0.25m in rolling a distance of 11 km is
(a) 2800
(b) 4000
(c) 5500
(d) 7000 [2022, 1 Mark]
Ans: (d)
In one revolution wheel covers distance of 2πr.
So. in n revolution it will cover 2πrn distance.
∴ S = 2πrn
According to question, S = 11 km. r = 0.25 m so,
⇒ n = 7000
Q3: The length of the minute hand of a dock is6 cm. Find the area swept by it when it moves from 7 :05 p.m. to 7:40 pm. [2022, 2 Mark]
Ans: We know that, fn 60 minutes, the tip of minute hand moves 360°.
In 1 minute, it will move = 360° / 60 = 6°
From 7 : 05 pm to 7: 40 pm i.e. 35 min, it wilt move through = 35 x 6° =210°
∴ Area swept by the minute hand in 35 min = Area of sector with sectorial angle θ of 210° and radius of 6 cm
=
= 66 cm^{2}
Q4: In the given figure, arcs have been drawn of radius 7 cm each with vertices A. B, C and D of quadrilateral ABCD as centres. Find the area of the shaded region. [2022, 2 Mark]
Ans: Let the measure of ∠A, ∠B, ∠C and ∠D be θ_{1}, θ_{2}, θ_{3} & θ_{4} respectively.
Required area = Area of sector with centre A + Area of sector with centre C + Area of sector with centre C + Area of sector with centre D
(By angle sum property of a quadrilateral)
= 154 cm^{2}
Q2: Given below is the picture of the Olympic rings made by taking five congruent circles of radius 1 cm each, intersecting in such a way that the chord formed by joining the point of intersection of two circles is also of length 1 cm. Total area of all the dotted regions assuming the thickness of the rings to be negligible is(a)
(b)
(c)
(d) [2021, 1 Mark]Ans: (d)
Let O be the centre of the circle. So. OA = OB = AB = 1 cm
So ΔOAB is an equilateral triangle.
∴ ∠AOB = 60°
∴ Required area = 8 x area of one segment with r = 1 cm,θ = 60°
= 8 x (area of sector  area of ΔAOB)
Q2: A piece of wire 22 cm long is bent into the form of an arc of a circle subtending an angle of 60° at its centre. Find the radius of the circle. [Use π = 22/7] [2020, 2 Marks]
Ans: Let AB be the wire of length 22 cm in the form of an art of a circle so blending an ∠AOB  60° at centre O.∵ Length of arc =
⇒
= 21 cm
Hence, radius of the circle is 21cm.
Q2: Find the area of the segment shown in the given figure, if radius of the circle is 21 cm and ∠AOB = 120°. (Take π = 22/7) [2020, 3 Marks]
Ans: Given. O is the centre of the circle of radius 21cm and AB is the chord that subtends an angle of 120° at the centre.Draw OM ⊥ AB,Area of the minor segment AMBP = Area of sector OAPB  Area of ΔAOB
Now, area of sector OAPB
=
= 462 cm^{2}
Since, OM ⊥ AB.
[∵ Perpendicular from the centre to the chord bisects the angle subtended by the chord at the centre.]
In ΔAOM, sin60° = AM/AO, cos60° = OM/OA
⇒
⇒
∵
Area of ΔAOB =
Hence, Required Area =
= 462  381.92= 80.08 cm^{2 }
Q3: In the given figure, three sectors of a circle of radius 7 cm, making angles of 60°, 80° and 40° at the centre are shaded. Find the area of the shaded region. [2020, 23 Marks]Ans: Radius (r) of circle = 7 cm
Area of shaded region =
=
=
= 77 cm^{2}
120 videos463 docs105 tests

1. What is the formula to find the area of a circle? 
2. How do you calculate the circumference of a circle? 
3. Can you find the area of a sector of a circle? 
4. What is the relationship between the circumference and the diameter of a circle? 
5. How can I find the length of an arc in a circle? 
120 videos463 docs105 tests


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