Chapter 11 Area Related to Circles NCERT Exemplar Solutions Exercise 11.4 Class 10 Maths
Chapter Name  NCERT Maths Exemplar Solutions for Chapter 11 Area Related to Circles Exercise 11.4 
Book Name  NCERT Exemplar for Class 10 Maths 
Other Exercises 

Related Study  NCERT Solutions for Class 10 Maths 
Exercise 11.4 Solutions
1. The area of a circular playground is 22176 m^{2} . Find the cost of fencing this ground at the rate of Rs 50 per metre.
Solution
Area of the circular playground = 22176 m^{2}
Let r be the radius of the circle.
Ï€r^{2} = 22176
⇒ (22/7)r^{2} = 22176
⇒ r^{2} = 22176 × (22/7)
⇒ r^{2} = 7056
⇒ r = 84
Radius of the circular playground = 84 m
Circumference of the circle = 2Ï€r
= 2 × (22/7) × 84
= 528 m
Cost of fencing 1 meter of ground = Rs. 50
Cost of fencing the total ground = Rs. 528 × 50
= Rs. 26,400
2. The diameters of front and rear wheels of a tractor are 80 cm and 2 m respectively. Find the number of revolutions that rear wheel will make in covering a distance in which the front wheel makes 1400 revolutions.
Solution
Diameter of front wheels = d_{1} = 80 cm
Diameter of rear wheels = d_{2} = 2 m = 200 cm
Let r_{1} be the radius of the front wheels = 80/2 = 40 cm
Let r_{2} be the radius of the rear wheels = 200/2 = 100 cm
Circumference of the front wheels = 2Ï€r
= 2 × (22/7) × 40
= 1760/7 cm
Circumference of the rear wheels = 2Ï€r
= 2 × (22/7) × 100
= 4400/7 cm
No. of revolutions made by the front wheel = 1400
∴ Distance covered by the front wheel = 1400 × (1760/7)
= 352000 cm
Number of revolutions made by rear wheel in covering a distance in which the front wheel makes 1400 revolutions,
= (Distance covered by front wheel)/(Circumference of rear wheel)
= 352000/4400/7
= (352000 × 7)/4400
= 560 revolutions.
3. Sides of a triangular field are 15m, 16m and 17m. With the three corners of the field a cow, a buffalo and a horse are tied separately with ropes of length 7 m each to graze in the field. Find the area of the field which cannot be grazed by the three animals.
Solution
Sides of the triangle are 15m, 16m, and 17m.
Now,
Perimeter of the triangle = (15 + 16 + 17)m
= 48 m
Semi  perimeter of the triangle = s
= 48/2
= 24 m
By Heron's formula,
Area of the triangle =
Let B, C and H be the corners of the triangle on which buffalo, cow and horse are tied respectively with ropes of 7 m each.
So, the area grazed by each animal will be in the form of a sector.
Radius of each sector = r
= 7 m
Let x, y, z be the angles at corners B, C, H respectively.
Area of field not grazed by the animals = Area of triangle – (area of the three sectors)
= 109.892 – 77
= 32.982 cm^{2}
4. Find the area of the segment of a circle of radius 12 cm whose Corresponding sector has a central angle of 60° (Use Ï€ = 3.14).
Solution
Radius of the circle = r
= 12 cm
OA = OB = 12 cm
∠ AOB = 60° (given)
As triangle OAB is an isosceles triangle,
∠ OAB = ∠ OBA = Î¸ (let)
And, Sum of interior angles of a triangle is 180°,
Î¸ + Î¸ + 60° = 180°
⇒ 2Î¸ = 120°
⇒ Î¸ = 60°
Therefore, the triangle AOB is an equilateral triangle.
AB = OA = OB = 12 cm
Area of the triangle AOB = (√3/4) × a^{2},
where a is the side of the triangle.
= (√3/4) × (12)^{2}
= (√3/4) × 144
= 36√3 cm^{2}
= 62.354 cm^{2}
Also, Central angle of the sector AOBCA = ∅
= 60°
= 60Ï€/180
= Ï€/3 radians
So, area of the sector AOBCA = (1/2)r^{2} ∅
= (1/2) × 12^{2} × Ï€/3
= 75.36 cm^{2}
Area of the segment ABCA = Area of the sector AOBCA  Area of the triangle AOB
= (75.36  62.354)cm^{2}
= 13.006 cm^{2}
5. A circular pond is 17.5 m is of diameter. It is surrounded by a 2 m wide path. Find the cost of constructing the path at the rate of Rs. 25 per m^{2} .
Solution
Diameter of the circular pond = 17.5 m
Let r be the radius of the park = r
r = 8.75 m
The circular pond is surrounded by a path of width 2 m.
Radius of the outer circle = R
= (8.75+2) m
= 10.75 m
Area of the road = Area of the outer circular path – Area of the circular pond
Ï€r^{2}  Ï€R^{2}
= 3.14 × (10.75)^{2}  3.14 × (8.75)^{2}
= 3.14 × (10.75)^{2}  (8.75)^{2}
= 3.14 × (10.75 + 8.75) × 10.75  8.75)
= 3.14 × 19.5 × 2
= 122. 46 m^{2}
So, the area of the path is 122.46 m^{2} .
Now, Cost of constructing the path per m^{2} = Rs. 25
Cost of constructing 122.46 m^{2} of the path = Rs. 25 × 122.46
= Rs. 3061.50
6. In Fig., ABCD is a trapezium with AB  DC, AB = 18 cm, DC = 32 cm and distance between AB and DC = 14 cm. If arcs of equal radii 7 cm with centers A, B, C and D have been drawn, then find the area of the shaded region of the figure.
Solution
AB = 18 cm,
DC = 32 cm
Distance between AB and DC = Height = 14 cm
Now,
Area of the trapezium = (1/2) × (Sum of parallel sides) × Height
= (1⁄2) × (18+32) × 14
= 350 cm^{2}
We have,
AB ∥ DC,
∠ A +∠ D = 180°
∠ B +∠ C = 180°
Radius of each arc = 7 cm
Area of sector with ∠A = (∠A/180) × Ï€r^{2}
Area of sector with ∠B = (∠B/360) × Ï€r^{2}
Area of sector with ∠C = (∠C/360) × Ï€r^{2}
Area of sector with ∠D = (∠D/360) × Ï€r^{2}
Therefore,
Sum of total areas of sectors =
Area of shaded region = Area of trapezium  (Total area of sectors)
= 350  154
= 196 cm^{2}
Therefore, the required area of shaded region is 196 cm^{2}.
7. Three circles each of radius 3.5 cm are draw in such a way that each of them touches the other two. Find the area enclosed between these circles.
Solution
The three circles are drawn such that each of them touches the other two.
By joining the centers of the three circles,
AB = BC = CA = 2(radius)
= 7 cm
So, triangle ABC is an equilateral triangle with each side 7 cm.
Area of the triangle = (√3/4) × a^{2} , Where a is the side of the triangle.
= (√3/4) × 7^{2}
= (49√3/4) cm^{2}
= 21.2176 cm^{2}
Central angle of each sector = 60°
= (60Ï€/180)
= Ï€/3 radians
Area of each sector = (1/2)r^{2} Î¸
= (1/2) × (3.5)^{2} × (Ï€/3)
= 12.25 × [22/(7×6)]
= 6.4167 cm^{2}
Total area of three sectors = 3 × 6.4167
= 19.25 cm^{2}
Area enclosed between three circles = Area of triangle ABC  Area of the three sectors
= 21.2176  19.25
= 1.9676 cm^{2}
Therefore, the required are enclosed between these circles is 1.967 cm^{2} (approx).
8. Find the area of the sector of a circle of radius 5 cm, if the corresponding arc length is 3.5 cm.
Solution
Radius of the circle = r = 5 cm
Arc length of the sector = l = 3.5 cm
Let the central angle (in radians ) be Î¸.
We have,
Arc length = Radius × Central angle (in radians)
Central angle (Î¸) = Arc length / Radius
= l/r
= 3.5/5
= 0.7 radians
Area of the sector = (1⁄2) × r^{2}Î¸
= (1⁄2) × 25 × 0.7
= 8.75 cm^{2}
Therefore, required area of the sector of a circle is 8.75 cm^{2}.
9. Four circular cardboard pieces of radii 7 cm are placed on a paper in such a way that each piece touches other two pieces. Find the area of the portion enclosed between these pieces.
Solution
The four circles are placed such that each piece touches the other two pieces.
By joining the centers of the circles by a line segment, we get a square ABDC with sides,
AB = BD = DC = CA = 2(Radius)
= 2(7) cm
= 14 cm
Area of the square = (Side)^{2}
= (14)^{2}
= 196 cm^{2}
ABDC is a square,
So, each angle has a measure of 90°.
∠ A = ∠ B = ∠ D = ∠ C = 90° = Ï€/2 radians
= Î¸ (let)
Radius of each sector = 7 cm
Area of the sector with central angle A = 1⁄2 r^{2}Î¸
= 1⁄2 × 49 × Ï€/2
= 1⁄2 × 49 × (22/(2×7))
= (77/2) cm^{2}
As the central angles and the radius of each sector are same, area of each sector is 77/2 cm^{2}
Area of the shaded portion = Area of square – Area of the four sectors
= 196 – (4 × (77/2))
= 196 – 154
= 42 cm^{2}
Therefore, required area of the portion enclosed between these pieces is 42 cm^{2}.
10. On a square cardboard sheet of area 784 cm^{2} , four congruent circular plates of maximum size are placed such that each circular plate touches the other two plates and each side of the square sheet is tangent to two circular plates. Find the area of the square sheet not covered by the circular plates.
Solution
Area of the square = 784 cm^{2}
Side of the square = √Area
= √784
= 28 cm
The four circular plates are congruent,
Then diameter of each circular plate = 28/2
= 14 cm
Radius of each circular plate = 7 cm
We know that,
Area of the sheet not covered by plates = Area of the square – Area of the four circular plates
As all four circular plates are congruent,
We have,
Area of all four plates equal.
Area of one circular plate = Ï€r^{2} .
= (22/7)× 7^{2}
= 154 cm^{2}
Area of four plates = 4 × 154
= 616 cm^{2}
Area of the sheet not covered by plates = 784  616
= 168 cm^{2}
11. Floor of a room is of dimensions 5 m × 4 m and it is covered with circular tiles of diameters 50 cm each as shown in Fig. Find the area of floor that remains uncovered with tiles. (Use Ï€ = 3.14).
Solution
Length = 5 m
Breadth = 4m
Area = l × b
= 20 m^{2}
Diameter of each circular tile = 50 cm
Radius = 25 cm
= 1/4 m
Now,
Area of circular tile ;
= Ï€r^{2}
= 3.14 × (1/4)^{2}
= 3.14/6
Area of 80 circular tiles = 80 × (3.14/16) = 1.57 m^{2}
Area which remain uncovered Area of floor  Area of 80 circular tiles
= 20  15.7
= 4.3 m^{2}
12. All the vertices of a rhombus lie on a circle. Find the area of the rhombus, if area of the circle is 1256 cm^{2} . (Use Ï€ = 3.14).
Solution
Let the radius of the circle be r.
Area of circle = 1256 cm^{2}
Ï€r^{2} = 1256
⇒ r = 20 cm
So, Diameter = 40 cm
As, all the vertices of a rhombus lie on a circle that means each diagonal of a rhombus must pass through the center of a circle that is why both diagonals are equal and same as the diameter of the given circle.
Let d_{1} and d_{2} be the diagonals of rhombus,
And, Diameter of circle = 40 cm
Area of rhombus = (1/2) × d_{1} × d_{2}
= (1/2) × 40 × 40
= 800 cm^{2}
13. An archery target has three regions formed by three concentric circles as shown in Fig.. If the diameters of the concentric circles are in the ratio 1: 2:3, then find the ratio of the areas of three regions.
Required ratio = A_{1} : A_{2} : A_{3}
= 1 : 3 : 5
In 1 min = 360/60 In (6 : 05 am to 6 :40 am ) = 35 minutes,
= 210
Length of minute hand (r) = 5 cm
Area of sector AOBA with angle O = (Ï€r^{2} /360) × ∠O
= 275/6
= 45(5/6) cm^{2}
⇒ 770 = (Ï€r^{2} /360) × 200
⇒ r = 21 cm
Length of corresponding arc of this sector = Central angle × Radius
= 200 × 21 × Ï€/180
= 220/3
= 77(1/3) cm
Hence, we observe that arc lengths of two sectors of two different circles may be equal but their area need not be equal.
There are four equally semi  circles and LMJK formed a square.
FH = 14  (3 + 3)
= 8 cm
Area of semicircle HJM = Ï€r^{2}/2
= Ï€2^{2}/2 = 2Ï€
= 6.28 cm^{2}
Area of four semicircle = 4 × 6.28
= 25.12 cm^{2}
Now,
Area of square ABCD = 14^{2}
= 196 cm^{2}
Area of shaded region = area of square ABCD  (Area of 4 semicircle + area of square JKLM)
= 196  (8Ï€ + 16)
= 180  8Ï€ cm^{2}
⇒ (22/7)r^{2} = 1.54
⇒ r = 0.7 m
Distance travelled by circular wheel in one revolution = Circumference of circular wheel
= 2Ï€r
= 2 × (22/7)× 0.7
= 4.4 m
Number of revolutions = (Total distance)/(Distance of revolution)
= 176/4.4
= 40
Central angle of AOBA = 90°
In Î” AOB, AB^{2} = OA^{2} + OB^{2}
5^{2} = r^{2} + r^{2}
⇒ r = 5/√2 cm
OA^{2} = OD^{2} + AD^{2}
= [5/√2)^{2}  [5/2]^{2}
= 5/2 cm
= 25/4 cm^{2}
Area of minor segment = Area of sector AOBA  Area of isosceles Î”AOB
= [25Ï€/8  25/4]
Now, Area o circle = Ï€r^{2}
= Ï€(5/√2)^{2}
= 25Ï€/2
Area of major segment = Area of circle  Area of minor segment
= (75Ï€/8 + 25/4)
Now, Difference = Area of major segment  Area of minor segment
= (75Ï€/8 + 25/4)  (25Ï€/8  254)
= 25Ï€/4 + 25/4 cm^{2}
Radius of the circle (r) = 21 cm and
= (22/7) × (21)^{2}
= 1386 cm^{2}
= (22/7) × [(21)^{2}/360] × 120°
= 462 cm^{2}
= 924 cm^{2}
= 924  462
= 462 cm^{2}