Last updated at May 29, 2018 by Teachoo

Transcript

Example 2 The decorative block shown in figure is made of two solids — a cube and a hemisphere. The base of the block is a cube with edge 5 cm, and the hemisphere fixed on the top has a diameter of 4.2 cm. Find the total surface area of the block.(Take π = 22/7) Total surface area of block = Total Surface area of cube + Curved Surface area of hemisphere – Base area of hemisphere Total Surface area of cube Given that side of cube is 5 cm Total Surface area of the cube = 6 (side)2 = 6 (5)2 = 6×5×5 = 150 cm2 Curved surface area of hemisphere Diameter of hemisphere = 4.2 cm So, radius = r = Diameter/2 = 4.2/2 = 2.1 cm Curved surface area of hemisphere = 2𝜋𝑟2 = 2×22/7×(2.1)2 = 2×22/7×2.1×2.1 = 27.72 cm2 Base area of hemisphere Base of hemisphere is a circle with radius 2.1 cm Base area of hemisphere = Area of circle = 𝜋𝑟2 = 22/7×(2.1)2 = 22/7×2.1×2.1 = 22×0.3×2.1 = 13.86 cm2 Therefore, Total surface area of block = Total Surface area of cube + Curved Surface area of hemisphere – Base area of hemisphere = (150) + (27.72) – (13.86) = 177.72 – 13.86 = 163.86 cm2

Example 2
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Example 3 Important

Ex 13.1, 4 Important

Ex 13.1, 9 Important

Example 6 Important

Example 7 Important

Example 9 Important

Example 11 Important

Ex 13.3, 8 Important

Ex 13.3, 9 Important

Example 12 Important

Ex 13.4, 4 Important

Ex 13.4, 5 Important

Surface Area and Volume Formulas Important

Class 10

Important Questions for Exam - Class 10

- Chapter 1 Class 10 Real Numbers
- Chapter 2 Class 10 Polynomials
- Chapter 3 Class 10 Pair of Linear Equations in Two Variables
- Chapter 4 Class 10 Quadratic Equations
- Chapter 5 Class 10 Arithmetic Progressions
- Chapter 6 Class 10 Triangles
- Chapter 7 Class 10 Coordinate Geometry
- Chapter 8 Class 10 Introduction to Trignometry
- Chapter 9 Class 10 Some Applications of Trignometry
- Chapter 10 Class 10 Circles
- Chapter 11 Class 10 Constructions
- Chapter 12 Class 10 Areas related to Circles
- Chapter 13 Class 10 Surface Areas and Volumes
- Chapter 14 Class 10 Statistics
- Chapter 15 Class 10 Probability

About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.