Class 10 Maths Surface Areas & Volumes – Formulas & NCERT Solutions

Class 10 Maths Chapter 12 Surface Area and Volume: Your Go-To Guide for Nailing It

Hey there! If you're diving into Class 10 Maths and Chapter 12 on Surface Area and Volume feels like a puzzle, I’ve got you covered. This is like chatting with a friend who’s breaking it down step by step—no fluff, just the good stuff. We’re talking about finding how much paint covers a box or how much water a tank holds. We’ll start with the basics, go over formulas, work through examples, tackle 25 NCERT questions with solutions you can toggle, and answer 15 FAQs from simple to tricky. Fuzzy Math Academy’s LMS and 24-hour AI Math Solver are here to help you anytime. Let’s get rolling!

Introduction: What’s This Chapter About?

Surface area and volume are all about measuring 3D shapes—cubes, cylinders, cones, spheres, and more. Surface area tells you how much space is on the outside (think wrapping a gift). Volume tells you how much fits inside (like water in a bottle). In Class 10, you’ll deal with solids like cubes, cuboids, cylinders, cones, spheres, and combinations of these (like a bucket or a capsule). It’s practical stuff—think designing a water tank or calculating material for a tent. Ready? Let’s lay the groundwork.

Basic Rules: The Foundations You Need

Here’s what you need to know before jumping into calculations:

• Surface Area: Total surface area (TSA) is the entire outer area. Curved surface area (CSA) or lateral surface area is just the sides, not the top or bottom.
• Volume: Measures capacity—how much a shape can hold (in cubic units).
• Units: Area in square units (cm², m²). Volume in cubic units (cm³, m³). Stick to consistent units!
• Combination of Solids: For objects like a hemisphere on a cylinder, add the surface areas or volumes, but watch for shared surfaces (they don’t count in TSA).
• Frustum: A cone with its top sliced off—special formulas apply.

Pro tip: Sketch the shape and label dimensions before calculating. It keeps you from mixing up radius and height.

Key Formulas: Your Toolkit for Surface Area and Volume

These are the formulas you’ll use. I’ll list them like a cheat sheet for each solid.

Cuboid (l = length, b = breadth, h = height)

• TSA = 2(lb + bh + hl)
• CSA = 2(l + b)h
• Volume = l × b × h

Cube (a = side)

• TSA = 6a²
• CSA = 4a²
• Volume = a³

Cylinder (r = radius, h = height)

• CSA = 2πrh
• TSA = 2πr(r + h)
• Volume = πr²h

Cone (r = radius, h = height, l = slant height, l = √(r² + h²))

• CSA = πrl
• TSA = πr(r + l)
• Volume = (1/3)πr²h

Sphere (r = radius)

• Surface Area = 4πr²
• Volume = (4/3)πr³

Hemisphere

• CSA = 2πr²
• TSA = 3πr²
• Volume = (2/3)πr³

Frustum of a Cone (r₁ = lower radius, r₂ = upper radius, h = height, l = √(h² + (r₁ - r₂)²))

• CSA = π(r₁ + r₂)l
• TSA = π(r₁ + r₂)l + πr₁² + πr₂²
• Volume = (1/3)πh(r₁² + r₂² + r₁r₂)

Memorize these, and you’re halfway there. Let’s see them in action.

Some Examples: Let’s Work Through Them

Let’s solve a few problems together, like we’re sitting at a desk.

Example 1: Cuboid Surface Area

A cuboid has l=6 cm, b=4 cm, h=3 cm. Find TSA.

TSA = 2(lb + bh + hl) = 2(6×4 + 4×3 + 6×3) = 2(24 + 12 + 18) = 2×54 = 108 cm².

Example 2: Cylinder Volume

A cylinder has r=7 cm, h=10 cm. Find volume (use π=22/7).

Volume = πr²h = (22/7)×7²×10 = (22/7)×49×10 = 1540 cm³.

Example 3: Cone CSA

A cone has r=3 cm, h=4 cm. Find CSA.

Slant height l = √(3² + 4²) = √25 = 5 cm. CSA = πrl = (22/7)×3×5 = 330/7 ≈ 47.14 cm².

Example 4: Combination (Cylinder + Hemisphere)

A solid has a cylinder (r=5 cm, h=12 cm) with a hemisphere on top. Find TSA.

Cylinder CSA = 2πrh = 2×(22/7)×5×12 = 2640/7. Hemisphere CSA = 2πr² = 2×(22/7)×5² = 1100/7. Base not included. TSA = (2640/7 + 1100/7) = 3740/7 ≈ 534.29 cm².

These show you the flow. Now, let’s hit those NCERT questions.

25 NCERT Questions: From Easy to Brain-Benders (With Toggle Solutions)

These are pulled from the NCERT book, starting simple and getting tougher. Click to see solutions—try them first!

Q1 (Easy): TSA of a cube with side 4 cm.

TSA = 6a² = 6×4² = 6×16 = 96 cm².

Q2: Volume of a cuboid l=5 cm, b=3 cm, h=2 cm.

Volume = l×b×h = 5×3×2 = 30 cm³.

Q3: CSA of a cylinder r=7 cm, h=10 cm (π=22/7).

CSA = 2πrh = 2×(22/7)×7×10 = 440 cm².

Q4: Volume of a cube with side 6 cm.

Volume = a³ = 6³ = 216 cm³.

Q5: TSA of a cuboid l=8 cm, b=6 cm, h=4 cm.

TSA = 2(lb + bh + hl) = 2(8×6 + 6×4 + 8×4) = 2(48 + 24 + 32) = 2×104 = 208 cm².

Q6: Volume of a cylinder r=3.5 cm, h=12 cm (π=22/7).

Volume = πr²h = (22/7)×(3.5)²×12 = (22/7)×12.25×12 = 462 cm³.

Q7: CSA of a cone r=5 cm, h=12 cm.

l = √(5² + 12²) = √169 = 13 cm. CSA = πrl = (22/7)×5×13 = 1430/7 ≈ 204.29 cm².

Q8: Volume of a sphere r=7 cm (π=22/7).

Volume = (4/3)πr³ = (4/3)×(22/7)×7³ = (4/3)×(22/7)×343 = 4312/3 ≈ 1437.33 cm³.

Q9: TSA of a hemisphere r=10 cm.

TSA = 3πr² = 3×(22/7)×10² = 3×(22/7)×100 = 6600/7 ≈ 942.86 cm².

Q10: Volume of a cone r=6 cm, h=8 cm.

Volume = (1/3)πr²h = (1/3)×(22/7)×6²×8 = (1/3)×(22/7)×36×8 = 2112/7 ≈ 301.71 cm³.

Q11: TSA of a cylinder r=4 cm, h=14 cm.

TSA = 2πr(r + h) = 2×(22/7)×4×(4 + 14) = 2×(22/7)×4×18 = 3168/7 ≈ 452.57 cm².

Q12: Volume of a hemisphere r=3.5 cm.

Volume = (2/3)πr³ = (2/3)×(22/7)×(3.5)³ = (2/3)×(22/7)×42.875 = 1886.5/21 ≈ 89.83 cm³.

Q13: CSA of a frustum r₁=10 cm, r₂=6 cm, h=8 cm.

l = √(8² + (10-6)²) = √(64 + 16) = √80 = 4√5 cm. CSA = π(r₁ + r₂)l = (22/7)×(10+6)×4√5 ≈ 447.34 cm².

Q14: Volume of a frustum r₁=12 cm, r₂=8 cm, h=10 cm.

Volume = (1/3)πh(r₁² + r₂² + r₁r₂) = (1/3)×(22/7)×10×(12² + 8² + 12×8) = (1/3)×(22/7)×10×(144 + 64 + 96) = 31856/21 ≈ 1516.95 cm³.

Q15: TSA of a cone r=4 cm, h=3 cm.

l = √(4² + 3²) = 5 cm. TSA = πr(r + l) = (22/7)×4×(4 + 5) = (22/7)×4×9 = 792/7 ≈ 113.14 cm².

Q16: Surface area of a sphere r=5 cm.

Surface Area = 4πr² = 4×(22/7)×5² = 4×(22/7)×25 = 2200/7 ≈ 314.29 cm².

Q17: Cylinder + cone, r=5 cm, cylinder h=10 cm, cone h=6 cm. TSA.

Cone l = √(5² + 6²) = √61 cm. Cylinder CSA = 2πrh = 2×(22/7)×5×10 = 1100/7. Cone CSA = πrl = (22/7)×5×√61. Base shared. TSA = 1100/7 + (22/7)×5×√61 + π×5² ≈ 343.48 cm².

Q18: Volume of a cuboid with diagonal 13 cm, l=12 cm, b=5 cm.

Diagonal = √(l² + b² + h²) = 13. So, 12² + 5² + h² = 169. h² = 169 - 144 - 25 = 0. h = 0 (error in data, assuming cuboid). Volume = l×b×h, needs h.

Q19: Paint needed for cylinder r=7 cm, h=15 cm (1 m² = 10000 cm², 1L paints 10 m²).

TSA = 2πr(r + h) = 2×(22/7)×7×(7 + 15) = 968 cm² = 0.0968 m². Paint = 0.0968/10 = 0.00968 L.

Q20: Frustum TSA r₁=15 cm, r₂=10 cm, h=12 cm.

l = √(12² + (15-10)²) = √169 = 13 cm. TSA = π(r₁ + r₂)l + πr₁² + πr₂² = (22/7)×[(15+10)×13 + 15² + 10²] = (22/7)×(325 + 225 + 100) = 14300/7 ≈ 2042.86 cm².

Q21: Water in a conical tank r=5 m, h=4 m.

Volume = (1/3)πr²h = (1/3)×(22/7)×5²×4 = (22/7)×(100/3) ≈ 104.76 m³.

Q22: Hemisphere bowl r=7 cm, volume of liquid.

Volume = (2/3)πr³ = (2/3)×(22/7)×7³ = (2/3)×(22/7)×343 = 7182/21 ≈ 342 cm³.

Q23: Cylinder with hemispherical ends, r=3 cm, total h=20 cm. TSA.

Cylinder h = 20 - 2×3 = 14 cm. Cylinder CSA = 2πrh = 2×(22/7)×3×14 = 264 cm². Two hemispheres = sphere = 4πr² = 4×(22/7)×3² = 792/7. TSA = 264 + 792/7 ≈ 377.14 cm².

Q24: Cost of painting sphere r=10 cm at ₹5/cm².

Surface Area = 4πr² = 4×(22/7)×10² = 8800/7 ≈ 1257.14 cm². Cost = 1257.14×5 ≈ ₹6285.7.

Q25 (Higher): Frustum bucket r₁=20 cm, r₂=12 cm, h=15 cm. Volume and TSA.

l = √(15² + (20-12)²) = √289 = 17 cm. Volume = (1/3)πh(r₁² + r₂² + r₁r₂) = (1/3)×(22/7)×15×(20² + 12² + 20×12) = (22/7)×5×784 = 86240/7 ≈ 12320 cm³. TSA = π(20+12)×17 + π×20² + π×12² = (22/7)×(544 + 400 + 144) = 23936/7 ≈ 3419.43 cm².

These questions build from basic to complex, covering all solids. For visuals, picture a cone as a party hat (base radius, height to tip) or a frustum as a bucket (wider bottom, narrower top). If you need a diagram, imagine a cone with r, h, l labeled—NCERT has good sketches!

15 FAQs: From “What’s CSA?” to “Why Frustum?”

These are questions students often ask, from simple to advanced. Click to expand.

FAQ1: What’s the difference between TSA and CSA?

TSA is the total outer area, including bases. CSA is just the curved or lateral sides, like the walls of a cylinder.

FAQ2: How do I find slant height?

For a cone, use l = √(r² + h²). For a frustum, l = √(h² + (r₁ - r₂)²).

FAQ3: Why is volume of a cone 1/3 of a cylinder?

A cone tapers to a point, using only a third of the cylinder’s space with the same base and height.

FAQ4: When to use combination formulas?

When a solid is made of two shapes, like a cylinder with a cone on top. Add areas or volumes, but subtract shared surfaces for TSA.

FAQ5: What’s a frustum?

A cone with its top cut off parallel to the base, like a bucket or lamp shade.

FAQ6: How to avoid unit mistakes?

Convert all dimensions to the same unit (cm or m) before calculating. Area in square units, volume in cubic.

FAQ7: Why is hemisphere TSA 3πr²?

Curved part = 2πr², base = πr². Total = 3πr².

FAQ8: Can volume be zero?

No, unless dimensions are zero, which isn’t a real solid.

FAQ9: How to handle decimals in π?

Use π = 22/7 for simplicity unless asked for 3.14 or exact form.

FAQ10: Common errors in combination solids?

Forgetting to subtract shared surfaces or mixing up radii of different parts.

FAQ11: Real-life use of frustum?

Buckets, traffic cones, or water tanks—anything tapered.

FAQ12: Why is sphere volume (4/3)πr³?

It’s derived from calculus, but think of it as filling a sphere with tiny cubes—takes more space than a cylinder.

FAQ13: Advanced: Surface area of irregular solids?

Break into known shapes, calculate each part, adjust for overlaps.

FAQ14: How to check calculations?

Double-check units, dimensions, and shared surfaces. Use approximate π values to estimate.

FAQ15 (Higher): Link to other chapters?

Uses geometry (circles, triangles for slant height) and algebra (solving for dimensions). Real-life data can tie to statistics.

These FAQs should clear up most doubts. If something’s still tricky, our AI Math Solver is ready 24/7.

Wrapping Up: You’ve Got This in the Bag

Surface area and volume are like building blocks—once you know the shapes, it’s just plugging in numbers and watching for tricks like shared surfaces. Work through those NCERT questions, test yourself with the FAQs, and you’ll ace Chapter 12. Fuzzy Math Academy’s LMS and AI Math Solver are here for extra practice or quick fixes. Got a question or want more on frustums or real-life applications? Drop a comment. Keep measuring those solids!

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