NCERT Solutions for Class 10 Maths Chapter 12 - Areas Related to Circles
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Chapter 12 - Areas Related to Circles Exercise Ex. 12.1
Let the radius of the circle be r
Circumference of circle = 2r
Area of circle = r2
Given that circumference and area of the circle are equal.
So, 2r = r2
2 = r
Hence, the radius of the circle will be 2 units
Chapter 12 - Areas Related to Circles Exercise Ex. 12.2
Radius (r) of the circle = 15
Area of sector OPRQ =
= 117.75 cm2
.... (Since OP = OQ)
Area of ∆OPQ =
Area of segment PRQ = Area of sector OPRQ –Area of ∆OPQ
= 117.75 – 97.3125
= 20.4375 cm2
Area of major segment PSQ
= Area of circle – Area of segment PRQ
= 152p – 20.4375
= 3.14 × 225 – 20.4375= 686.0625 cm2
Draw a perpendicular OV on chord ST. It will bisect the chord ST.
SV = VT
OV = 6
ST = 2SV =
Area of ∆OST =
= 36 × 1.73
Area of sector OSUT =
Area of segment SUT = Area of sector OSUT
= 150.72 – 62.28
= 88.44 cm2
The horse can graze a sector of 90° in a circle of 5 m radius.
i. So area that can be grazed by horse = area of sector OACB
= 19.63 m2
ii. Area that can be grazed by the horse when the length of rope is 10 m long =
Change in grazing area = 78.5 – 19.63 = 58.87 cm2
The figure shows that each blade of the wiper will sweep an area of a sector of 115° in a circle of 25 cm radius.
Chapter 12 - Areas Related to Circles Exercise Ex. 12.3
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Other Chapters for CBSE Class 10 MathematicsChapter 1- Real Numbers Chapter 2- Polynomials Chapter 3- Pairs of Linear Equations in Two Variables Chapter 4- Quadratic Equations Chapter 5- Arithmetic Progressions Chapter 6- Triangles Chapter 7- Coordinate Geometry Chapter 8- Introduction to Trigonometry Chapter 9- Some Applications of Trigonometry Chapter 10- Circles Chapter 11- Constructions Chapter 13- Surface Areas and Volumes Chapter 14- Statistics Chapter 15- Probability
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