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Chapter 13 Class 10 Surface Areas and Volumes
Example 3 Important
Ex 12.1, 4 Important
Ex 12.1, 9 Important
Example 6 Important
Example 7 Important
Question 2 Important Deleted for CBSE Board 2024 Exams
Question 4 Important Deleted for CBSE Board 2024 Exams
Question 8 Important Deleted for CBSE Board 2024 Exams
Question 9 Important Deleted for CBSE Board 2024 Exams
Question 5 Important Deleted for CBSE Board 2024 Exams
Question 4 Important Deleted for CBSE Board 2024 Exams
Question 5 Important Deleted for CBSE Board 2024 Exams
Surface Area and Volume Formulas Important
Chapter 13 Class 10 Surface Areas and Volumes
Last updated at Aug. 17, 2023 by Teachoo
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) Now, 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 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