plz answer soon tomorrow is exam plzzzzzzzzzzz
Please post queries individually.
Solution to Q 6 of Exercise 12.3 of NCERT.
R = 32 cm
The side of an equilateral triangle circumscribed in a circle of radius R, is given by,
2Rsin(60)
Area of the equilateral triangle = (3/4)side2
= (3/4)(4x32x32x3/4)
= 1330.17 cm2
Area of circle = 3.14x32x32 = 3215.36 cm2
Area of design = Area of circle - Area of triangle = 3215.36 - 1330.17 = 1885.2 cm2
Regards,
Team,
TopperLearning.
- 1
- 2
- 3
- 4
- 5
- 6
- 7
- 8
- 9
- 10
You have rated this answer /10
Browse free questions and answers by Chapters
- 1 Polynomials
- 2 Coordinate Geometry
- 3 Surface Areas and Volumes
- 4 Statistics
- 5 Probability
- 6 Triangles
- 7 Circles
- 8 Pair of Linear Equations in 2 Variables
- 9 Quadratic Equations
- 10 Arithmetic Progression
- 11 Some Applications of Trigonometry
- 12 Introduction to Trigonometry
- 13 Constructions
- 14 Areas Related to Circles
- 15 Real Numbers
- 16 Pair of Linear Equations in Two Variables