Is it correct to say: Missing the flat circular base while computing TSA of a hemisphere?

TSA of hemisphere = 2πr² (curved) + πr² (base) = 3πr². Only 2πr² if the base is not exposed (as in a solid resting on ground).

Is it correct to say: Confusing height with slant height in cone problems?

Height h is the vertical distance; slant height l is √(r² + h²). Draw the right triangle every time.

Is it correct to say: Applying wrong π value?

Use π = 22/7 when radii/heights end in 7 or 14; otherwise π = 3.14. Follow the question.

Is it correct to say: Adding the areas of both faces at the junction of two solids?

When a cone sits on a hemisphere of the same radius, the shared circular face is INSIDE ⇒ not counted in TSA of the combined solid.

Is it correct to say: Equating surface areas on melting/recasting?

Only VOLUME is conserved. Surface area generally changes.

Is it correct to say: Mixing units — using r in cm and h in m?

Convert all lengths to the SAME unit before computing V or SA.

Surface Areas and Volumes

Surface areas and volumes of cuboid, cube, cylinder, cone, sphere, hemisphere, and combinations of solids — NCERT Class 10 Maths Ch 12

📐 9 formulas✏️ 3 examples🎯 6 practice⚖️ 5-6 marks🏫 CBSE📚 Class 10✓ 2025–26 syllabus

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Board Exam Tips

→Combination-of-solids problem (cone on top of cylinder / hemisphere on cuboid) is a guaranteed 4-mark question.
→State units in the final answer: cm² for surface area, cm³ for volume.
→For a hemisphere, TSA = 3πr² (curved + circular base). Do not miss the base.
→When one solid is melted and recast into another, VOLUME is conserved — never surface area.
→Frustum problems use l = √(h² + (R − r)²) — memorise the slant-height formula.

📐 Formulas(9)

Cuboid★ Board fav

V = l bh, TSA = 2 (l b + bh + h l)

Symbol	Meaning
$l$	Length
$b$	Breadth
$h$	Height

Cube

V = a^{3}, TSA = 6 a^{2}

Cylinder★ Board fav

V = π r^{2} h, CSA = 2 π r h, TSA = 2 π r (r + h)

Cone

V = \frac{1}{3} π r^{2} h, CSA = π r l, TSA = π r (r + l)

Sphere★ Board fav

V = \frac{4}{3} π r^{3}, Surface Area = 4 π r^{2}

Hemisphere

V = \frac{2}{3} π r^{3}, CSA = 2 π r^{2}, TSA = 3 π r^{2}

Slant Height of a Cone

l = r^{2} + h^{2}

Frustum of a Cone

V = \frac{1}{3} π h (R^{2} + R r + r^{2}), CSA = π (R + r) l

Volume Conservation on Recasting★ Board fav

V_{old} = V_{new}

✏️ Solved Examples

1Solved Exampleeasy3 steps

Find the volume of a cylinder with radius 7 cm and height 10 cm. (Take π = 22/7.)

Apply V = πr²h

V = \frac{22}{7} \times 7^{2} \times 10

2Solved Exampleboard5 steps

A toy is in the shape of a cone of radius 3.5 cm mounted on a hemisphere of the same radius. If the total height of the toy is 15.5 cm, find its total surface area.

Radius r = 3.5 cm; height of cone h = 15.5 − 3.5 = 12 cm.

3Solved ExampleHOTS5 steps

A solid metallic sphere of radius 10.5 cm is melted and recast into small right circular cones each of radius 3.5 cm and height 3 cm. Find the number of cones formed. (π = 22/7)

Volume conservation

V_{sphere} = n \cdot V_{cone}

⚠️ Traps & Common Mistakes

⚠️Common Mistakes6

1
Missing the flat circular base while computing TSA of a hemisphere
✓TSA of hemisphere = 2πr² (curved) + πr² (base) = 3πr². Only 2πr² if the base is not exposed (as in a solid resting on ground).
2
Confusing height with slant height in cone problems
✓Height h is the vertical distance; slant height l is √(r² + h²). Draw the right triangle every time.
3
Applying wrong π value
✓Use π = 22/7 when radii/heights end in 7 or 14; otherwise π = 3.14. Follow the question.
4
Adding the areas of both faces at the junction of two solids
✓When a cone sits on a hemisphere of the same radius, the shared circular face is INSIDE ⇒ not counted in TSA of the combined solid.
5
Equating surface areas on melting/recasting
✓Only VOLUME is conserved. Surface area generally changes.
6
Mixing units — using r in cm and h in m
✓Convert all lengths to the SAME unit before computing V or SA.

🎯 Practice Yourself

🎯Practice Yourself6 questions

Q1
Find the volume of a sphere of diameter 14 cm. (π = 22/7)
Q2
The radius and height of a cylinder are 7 cm and 20 cm. Find its CSA and TSA.
Q3
A conical tent has base radius 5 m and slant height 13 m. Find the area of canvas required.
Q4
Water flows through a pipe of internal diameter 2 cm at 3 m/s. How much water (in litres) flows out in 1 minute?
Q5
A frustum of a cone has R = 20 cm, r = 8 cm, h = 16 cm. Find its slant height.
Q6
A cube of side 6 cm is cut into 8 smaller cubes. Find the total surface area of all small cubes and compare with the original.

📝 Notes

Surface Areas and Volumes

3-D geometry with formulas you can plug straight into. Every shape has an area (skin) and a volume (stuff inside).

One-look master table

| Solid | Volume | Curved SA | Total SA | |---|---|---|---| | Cuboid | l·b·h | 2h(l+b) | 2(lb + bh + hl) | | Cube | a³ | 4a² | 6a² | | Cylinder | πr²h | 2πrh | 2πr(r + h) | | Cone | ⅓πr²h | πrl | πr(r + l) | | Sphere | 4/3 πr³ | 4πr² | 4πr² | | Hemisphere | 2/3 πr³ | 2πr² | 3πr² | | Frustum | ⅓πh(R² + Rr + r²) | π(R+r) l | π(R+r)l + πR² + πr² |

Two situations you must be alert to

Combination of solids (cone on cylinder, hemisphere on cuboid, etc.) — hidden internal faces are NOT counted in TSA. Just add the CSA of each part.
Melting and recasting — VOLUME is preserved; use V_old = n · V_new.

Slant heights

Cone: l = √(r² + h²)
Frustum: l = √(h² + (R − r)²)

Both come from Pythagoras applied to the cross-section.

Quick sanity checks

Sphere: SA/V ratio = 3/r ⇒ small sphere has much more surface per unit volume (why raindrops evaporate fast).
Cone volume = ⅓ of a cylinder with same base and height.
Doubling all linear dimensions multiplies SA by 4 and V by 8.