Surface Areas and Volumes
Gujarat Board · Class 9 · Mathematics
Most important questions from Surface Areas and Volumes for Gujarat Board Class 9 Mathematics board exam 2026. MCQs, short answer, and long answer questions with marks.
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Sample Questions
Which of the following statements about cones are correct?
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The slant height can be found using l² = r² + h², Total surface area = πr(l + r), Curved surface area = πrl
Correct statements: 1) l² = r² + h² (Pythagoras theorem), 2) Total surface area = πrl + πr² = πr(l + r), 3) Curved surface area = πrl. Incorrect: Volume of cone = (1/3)πr²h (not πr²h), and cone has only one circular base.
If the volume of a sphere is 4851 cm³, find its radius. (Use π = 22/7)
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10.5 cm
Volume of sphere = (4/3)πr³ = 4851. So (4/3) × (22/7) × r³ = 4851. Solving: r³ = 4851 × 3 × 7 ÷ (4 × 22) = 4851 × 21 ÷ 88 = 1157.625. Therefore r³ = 1157.625, so r = 10.5 cm. Step 1: Set up equation. Step 2: Solve for r³. Step 3: Find cube root.
A cone has height 12 cm and slant height 13 cm. What is its base radius?
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5 cm
Using Pythagoras theorem: l² = r² + h². So 13² = r² + 12². Therefore 169 = r² + 144. So r² = 169 - 144 = 25. Therefore r = 5 cm. Step 1: Apply Pythagoras theorem. Step 2: Substitute known values. Step 3: Solve for r.
Which of the following are correct formulas for surface areas and volumes?
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Surface area of sphere = 4πr², Volume of hemisphere = (2/3)πr³, Curved surface area of hemisphere = 2πr²
Correct formulas: Surface area of sphere = 4πr², Volume of hemisphere = (2/3)πr³, Curved surface area of hemisphere = 2πr². Incorrect: Volume of cone = (1/3)πr²h (not 1/2), Total surface area of hemisphere = 3πr² (not 2πr²).
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