Chapter 11 Area Related to Circles NCERT Exemplar Solutions Exercise 11.3 Class 10 Maths
Chapter Name  NCERT Maths Exemplar Solutions for Chapter 11 Area Related to Circles Exercise 11.3 
Book Name  NCERT Exemplar for Class 10 Maths 
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Related Study  NCERT Solutions for Class 10 Maths 
Exercise 11.3 Solutions
1. Find the radius of a circle whose circumference is equal to the sum of the circumferences of two circles of radii 15 cm and 18 cm.
Solution
Radius of first circle = r_{1} = 15 cm
Radius of second circle = r_{2} = 18 cm
Circumference of 1st circle = 2Ï€r_{1} = 30Ï€ cm
Circumference of 1st circle = 2Ï€r_{2} = 36Ï€ cm
Let the radius of the circle = R
We have,
Circumference of circle = Circumference of first circle + Circumference of second circle
2Ï€R = 2Ï€r_{1} + 2Ï€r_{2}
⇒ 2Ï€R = 30Ï€ + 36Ï€ = 66Ï€
⇒ R = 33
⇒ Radius = 33 cm
Therefore, required radius of a circle is 33 cm.
2. In Fig., a square of diagonal 8 cm is inscribed in a circle. Find the area of the shaded region.
Solution
Let us take a be the side of square.
Diameter of a circle = Diagonal of the square = 8 cm
In right angled triangle ABC,
Using Pythagoras theorem,
(AC)^{2} = (AB)^{2} + (BC)^{2}
⇒ (8)^{2} = a^{2} + a^{2}
⇒ 64 = 2a^{2}
⇒ a^{2} = 32
Area of square = a^{2}
= 32 cm^{2}
Radius of the circle = Diameter/2
= 4 cm
Area of the circle = Ï€r^{2}
= Ï€(4)^{2}
= 16 cm^{2}
So, the area of the shaded region = Area of circle  Area of square
The area of the shaded region = 16Ï€ – 32
= 16 × (22/7) – 32
= 128/7
= 18.286 cm^{2}
3. Find the area of a sector of a circle of radius 28 cm and central angle 45°.
Solution
Area of a sector of a circle = (1/2)r^{2} Î¸
(In which r = radius and Î¸ = angle in radians subtended by the arc at the center of the circle)
We have,
Radius of circle = 28 cm
Angle subtended at the center = 45°
Angle subtended at the center (in radians) = Î¸
= 45Ï€/180
= Ï€/4
Area of a sector of a circle = (1/2) × r^{2}Î¸
= (1/2) × (28)^{2} × (Ï€/4)
= 308 cm^{2}
Therefore, the required area of a sector of a circle is 308 cm^{2}.
4. The wheel of a motor cycle is of radius 35 cm. How many revolutions per minute must the wheel make so as to keep a speed of 66 km/h?
Solution
Radius of wheel = r = 35 cm
1 revolution of the wheel = Circumference of the wheel
= 2Ï€r
= 2 × (22/7) × 35
= 220 cm
Speed of the wheel = 66 km/hr
= (66×1000×100) × 1/60 cm/min
= 110000 cm/min
∴ Number of revolutions in 1 min = 110000/220 = 500
Thus, required number of revolutions per minute is 500.
5. A cow is tied with a rope of length 14 m at the corner of a rectangular field of dimensions 20m × 16m. Find the area of the field in which the cow can graze.
Solution
ABCD is a rectangular field.
Length of field = 20 m
Breadth of the field = 16 m
A cow is tied at a point A.
Let length of rope be AE = 14 m
Angle subtended at the center of the sector = 90°
Angle subtended at the center (in radians) Î¸ = 90Ï€/180
= Ï€/2
Area of a sector of a circle = (1/2) × r^{2}Î¸
= (1/2) × (14)^{2} × Ï€/2
= 154 m^{2}
So, the required area of a sector of a circle is 154 m^{2}.
6. Find the area of the flower bed (with semicircular ends) shown in Fig.
Solution
Length and breadth of the rectangular portion AFDC of the flower bed are 38 cm and 10 cm Area of the flower bed = Area of the rectangular portion + Area of the two semicircles.
Area of rectangle AFDC = Length × Breadth
= 38 × 10
= 380 cm^{2}
Both ends of flower bed are semicircle in shape.
Diameter of the semicircle = Breadth of the rectangle AFDC
= 10 cm
Radius of the semi  circle = 10/2 = 5 cm
Area of the semi  circle = Ï€r^{2}/2
= 25(Ï€/2) cm^{2}
As, there are two semi  circles in the flower bed,
Area of two semi  circles = 2 × ( Ï€r^{2}/2)
= 25Ï€ cm^{2}
Therefore, total area of flower bed = (380 + 25Ï€) cm^{2}.
7. In Fig., AB is a diameter of the circle, AC = 6 cm and BC = 8 cm. Find the area of the shaded region (Use Ï€= 3.14).
Solution
AC = 6 cm and BC = 8 cm
A triangle in a semi  circle with hypotenuse as diameter is right angled triangle.
By Pythagoras theorem in right angled triangle ACB,
(AB)^{2} = (AC)^{2} + (CB)^{2}
⇒ AB^{2} = 6^{2} + 8^{2}
⇒ AB^{2} = 36 + 64
⇒ AB^{2} = 100
⇒ AB = 10
Diameter of the circle = 10 cm
Radius of the circle = 5 cm
Area of circle = Ï€r^{2}
= Ï€(5)^{2}
= 25Ï€ cm^{2}
= 25 × 3.14 cm^{2}
= 78.5 cm^{2}
Area of the right angled triangle = ( 1⁄2 ) × Base × Height
= (1⁄2) × AC × CB
= (1⁄2) × 6 × 8
= 24 cm^{2}
Area of the shaded region = Area of the circle  Area of the triangle
= (78.5  24) cm^{2}
= 54.5 cm^{2}
8. Find the area of the shaded field shown in Fig.
Solution
Construction : Join ED
Radius of semi  circle DFE = 6  2 = 4M
Area of rectangle ABCD = BC × AB
= 8 × 4 = 32m^{2}
Area of semicircle DFE = Ï€r^{2}/2
= Ï€(2)^{2}/2
= 2Ï€ m^{2}
Area of shaded region = Area of rectangle ABCD + Area of semicircle DFE
= (32 + 2Ï€) m^{2}.
9. Find the area of the shaded region in Fig.
Solution
Therefore, area of two semicircles = 2 × Ï€r^{2}/2
= Ï€(2)^{2}/2
= 4Ï€ m
= 16 × 4
= 64 m^{2}
Area of outer rectangle (ABCD) = 26 × 13
= 312 m^{2}
Area of shaded region = Area of outer rectangle  (area of two semicircles + area of inner rectangles)
= 312  (64 + 4Ï€)
= 248  4Ï€) m^{2}
Therefore,
In Î”OAB,
AOB + OAB + OBA = 180°
⇒ 60° + Î¸ + Î¸ = 180°
⇒ Î¸ = 60°
Area of Î”OAB = (√3/4) (side)^{2}
= (√3/4) (14)^{2}
= 49√3 cm^{2}
Now, Area of square (ABCD) = (12)^{2}
= 144 cm^{2}
Area of shaded region = Area of square  Area of four quadrants
= 30.93 cm^{2}
⇒ ∠A = ∠B = ∠C = 60°
Area of sector CDE = Î¸/360 × Ï€r^{2}
= 60/360 × Ï€(5)^{2}
= 3 × 13.0833
= 39.25 cm^{2}
= Ï€(126^{2}  105^{2})
= (22/7) (126  105)(126 + 105)
= 15246 m^{2}
= (∠A/360) × Ï€(21)^{2}
Area of sector with ∠B = (∠B/360) × Ï€r^{2}
= (∠B/360) × Ï€(21)^{2}
= (∠C/360) × Ï€(21)^{2}
Area of sector with ∠D = (∠D/360) × Ï€r^{2}
= (∠D/360) × Ï€(21)^{2}
Therefore,
Now,
20 = (60/360) × 2Ï€r
⇒ r = 60/Ï€ cm