Surface Areas and Volumes — Important Questions
Gujarat Board · Class 10 · Mathematics
44 important questions from Surface Areas and Volumes for Gujarat Board Class 10 Mathematics, with answers. Written for the board exams 2027.
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Important Questions from Surface Areas and Volumes
Which of the following statements are correct about the surface area of combined solids?
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We must subtract the areas where solids are joined, Only the visible surfaces are counted in total surface area
When combining solids, we only count the visible surfaces. The flat faces where two solids join are not visible from outside, so they are not included in the total surface area calculation.
The curved surface area of a cone with radius 5 cm and slant height 13 cm is:
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204.2 cm²
Curved surface area of cone = πrl where r is radius and l is slant height. CSA = (22/7) × 5 × 13 = (22/7) × 65 = 204.2 cm².
A tent is in the shape of a cylinder surmounted by a cone. Both have the same radius 7 m. Height of cylinder is 10 m and height of cone is 8 m. Find the curved surface area of the tent. (Use π = 22/7)
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1166 m²
Total CSA = CSA of cylinder + CSA of cone. CSA of cylinder = 2πrh = 2 × (22/7) × 7 × 10 = 440 m². For cone, slant height l = √(r² + h²) = √(7² + 8²) = √(49 + 64) = √113 ≈ 10.6 m. CSA of cone = πrl = (22/7) × 7 × 10.6 = 231.9 ≈ 232 m². Wait, let me recalculate more carefully: l = √113 ≈ 10.63. CSA of cone = (22/7) × 7 × 10.63 = 234 m². Total = 440 + 234 = 674 m². Let me check again: l = √113 = 10.63, CSA of cone = 22 × 10.63 = 233.86 ≈ 234 m². Hmm, let me recalculate: CSA of cone = (22/7) × 7 × √113 = 22 × √113 = 22 × 10.63 = 234 m². Actually, let me be more precise: √113 ≈ 10.63, so CSA of con
Which of the following formulas are correct for calculating volumes?
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Volume of sphere = (4/3)πr³, Volume of hemisphere = (2/3)πr³, Volume of cone = (1/3)πr²h, Volume of cuboid = length × breadth × height
The correct volume formulas are: Sphere = (4/3)πr³, Hemisphere = (2/3)πr³, Cone = (1/3)πr²h, Cylinder = πr²h (not 2πr²h), Cuboid = l×b×h.
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