3D Solids (Volume & Surface Area)
SSC GD Exam Preparation - Quantitative Aptitude
3D solids are objects with length, breadth, and height. We mainly calculate Volume (space occupied) and Surface Area (total area covering the solid).
🔹 1. Cube
Formulas:
Volume = a³
Surface Area = 6a²
Example:
Side = 5 cm
Volume = 5³ = 125 cm³
Surface Area = 6 × 5² = 150 cm²
🔹 2. Cuboid (Rectangular Box)
Formulas:
Volume = L × B × H
Surface Area = 2(LB + BH + HL)
Example:
L=10 cm, B=5 cm, H=4 cm
Volume = 10 × 5 × 4 = 200 cm³
Surface Area = 2(50 + 20 + 40) = 220 cm²
🔹 3. Sphere
Formulas:
Volume = 4/3 × π × r³
Surface Area = 4 × π × r²
🔹 4. Hemisphere
Radius = r
Formulas:
Volume = 2/3 × π × r³
Surface Area = 3 × π × r²
(curved + base)
🔹 5. Cylinder
Radius = r, Height = h
Formulas:
Volume = π × r² × h
Surface Area = 2πr(h + r)
🔹 6. Cone
Radius = r, Height = h, Slant Height = l
Formulas:
Volume = 1/3 × π × r² × h
Surface Area = πr(l + r)
🔹 7. Frustum of a Cone
Top radius = r₁, Bottom radius = r₂, Slant height = l, Height = h
Formulas:
Volume = 1/3 × π × h × (r₁² + r₂² + r₁r₂)
Surface Area = π(r₁ + r₂)l + π(r₁² + r₂²)
🔹 8. Pyramid
Base area = B, Height = h
Formulas:
Volume = 1/3 × B × h
Surface Area = Base Area + Lateral Area
🧠 Practice Section: 15 Questions (With Answers)
Each Q followed by its Answer (SSC style). Click on "View Answer" to check your understanding.
Q1. Cube side = 6 cm. Find volume.
View Answer
Volume = 6³ = 216 cm³
Q2. Cube side = 6 cm. Find surface area.
View Answer
Surface Area = 6 × 6² = 6 × 36 = 216 cm²
Q3. Cuboid L=10, B=5, H=4. Volume?
View Answer
Volume = 10 × 5 × 4 = 200 cm³
Q4. Cuboid L=10, B=5, H=4. Surface area?
View Answer
Surface Area = 2(50 + 20 + 40) = 220 cm²
Q5. Sphere radius = 7 cm. Volume?
View Answer
Volume = 4/3 × π × 343 = 4/3 × 22/7 × 343 = 1436.57 cm³
Q6. Sphere radius = 7 cm. Surface area?
View Answer
Surface Area = 4πr² = 4 × 22/7 × 49 = 616 cm²
Q7. Hemisphere radius = 14 cm. Volume?
View Answer
Volume = 2/3 × π × 14³ = 2/3 × 22/7 × 2744 = 5724.19 cm³
Q8. Hemisphere radius = 14 cm. Surface area?
View Answer
Surface Area = 3πr² = 3 × 22/7 × 196 = 1848 cm²
Q9. Cylinder r=7, h=10. Volume?
View Answer
Volume = πr²h = 22/7 × 49 × 10 = 1540 cm³
Q10. Cylinder r=7, h=10. Surface area?
View Answer
Surface Area = 2πr(h+r) = 2 × 22/7 × 7 × (10+7) = 1588 cm²
Q11. Cone r=5, h=12. Volume?
View Answer
Volume = 1/3πr²h = 1/3 × 22/7 × 25 × 12 = 314.29 cm³
Q12. Cone r=5, h=12. Slant height l=13. Surface area?
View Answer
Surface Area = πr(l+r) = 22/7 × 5 × (13+5) = 502.86 cm²
Q13. Pyramid base area = 36, height = 9. Volume?
View Answer
Volume = 1/3 × B × h = 1/3 × 36 × 9 = 108 cm³
Q14. Frustum r₁=3, r₂=6, h=5. Volume?
View Answer
Volume = 1/3 × π × 5 × (9+36+18) = 1/3 × π × 5 × 63 = 105π = 329.87 cm³
Q15. Cylinder radius = 7, height = 10. Curved surface area?
View Answer
Curved Surface Area = 2πrh = 2 × 22/7 × 7 × 10 = 440 cm²
✅ SSC GD Exam Strategy
🟢 Memorize key formulas for different 3D shapes
🟢 Practice unit conversions (cm³ to m³, etc.)
🟢 Understand the difference between volume and surface area
🟢 Master sphere, cylinder, and cone formulas as they frequently appear
🟢 Time yourself - aim for 45-60 seconds per 3D mensuration question
You've completed 3D Solids!
SSC GD Tip: Mastering 3D mensuration is crucial for solving geometry problems efficiently. Remember that volume measures the space occupied by a solid, while surface area measures the total area covering it. Practice identifying which formula to use for different 3D shapes. For complex solids, visualize them as combinations of simpler shapes. Regular practice with previous year SSC GD questions will build your speed and accuracy in solving these problems. Pay special attention to cylinder, cone, and sphere problems as they frequently appear in exams.
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