Surface Areas And Volumes Of Solid Figures — Important Questions
NIOS · Class 10 · Maths
34 important questions from Surface Areas And Volumes Of Solid Figures for NIOS Class 10 Maths, with answers. Written for the board exams 2027.
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Important Questions from Surface Areas And Volumes Of Solid Figures
Find the slant height of a cone with base radius 5 cm and height 12 cm.
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13 cm
Step 1: Use Pythagoras theorem: slant height l = √(r² + h²) where r=5 cm, h=12 cm. Step 2: Calculate r² = 5² = 25 and h² = 12² = 144. Step 3: Add: r² + h² = 25 + 144 = 169. Step 4: Take square root: l = √169 = 13 cm. This forms a 5-12-13 right triangle, which is a common Pythagorean triplet.
What is the total surface area of a hemisphere with radius 14 cm? (Use π = 22/7)
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1848 cm²
Step 1: Total surface area of hemisphere = 3πr² where r=14 cm. Step 2: Calculate r² = 14² = 196. Step 3: Substitute: TSA = 3 × (22/7) × 196 = 3 × 22 × 28 = 1848 cm². Step 4: This includes curved surface (2πr²) plus flat circular base (πr²). Common mistake: Using only 2πr² which is just curved surface.
A cubical box has a volume of 216 cm³. What is the length of its edge?
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6 cm
Step 1: Volume of cube = a³ = 216 cm³, need to find edge length a. Step 2: Take cube root of both sides: a = ∛216. Step 3: Find factors: 216 = 6 × 6 × 6 = 6³. Step 4: Therefore a = 6 cm. To verify: 6³ = 6 × 6 × 6 = 216 ✓. Common mistake: Taking square root instead of cube root.
Find the volume of a cylinder with diameter 14 cm and height 8 cm. (Use π = 22/7)
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1232 cm³
Step 1: Volume of cylinder = πr²h. Given diameter = 14 cm, so radius r = 7 cm, height h = 8 cm. Step 2: Calculate r² = 7² = 49. Step 3: Substitute: V = (22/7) × 49 × 8 = 22 × 7 × 8 = 1232 cm³. Step 4: The volume is 1232 cm³. Common mistake: Using diameter instead of radius in the formula.
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