Class 9 — Mathematics (NCERT)

Surface Areas and Volumes of Cones, Spheres, and Hemispheres

Class 9

✓By the end of this lesson students will be able to recall and apply formulas for the curved surface area, total surface area, and volume of a cone.
✓By the end of this lesson students will be able to recall and apply formulas for the surface area and volume of a sphere.
✓By the end of this lesson students will be able to recall and apply formulas for the curved surface area, total surface area, and volume of a hemisphere.
✓By the end of this lesson students will be able to distinguish between curved surface area and total surface area for cones and hemispheres.
✓By the end of this lesson students will be able to solve problems involving surface areas and volumes of these three-dimensional solids.

Key concepts

Cone

A cone is a three-dimensional geometric shape that tapers smoothly from a flat base (usually circular) to a point called the apex or vertex. The height (h) is the perpendicular distance from the apex to the centre of the base. The slant height (l) is the distance from the apex to any point on the circumference of the base.

l² = r² + h² (Relationship between slant height, radius, and height)\nCurved Surface Area (CSA) = πrl\nTotal Surface Area (TSA) = πr(l + r)\nVolume (V) = (1/3)πr²h

Sphere

A sphere is a perfectly round three-dimensional object in which every point on its surface is equidistant from its centre. This equidistant measure is called the radius (r).

Surface Area (SA) = 4πr²\nVolume (V) = (4/3)πr³

Hemisphere

A hemisphere is exactly half of a sphere. It has a curved surface and a flat circular base. The radius (r) is the distance from the centre of the flat base to any point on its circumference, or from the centre of the curved surface to any point on its edge.

Curved Surface Area (CSA) = 2πr²\nTotal Surface Area (TSA) = 3πr² (CSA + Area of circular base)\nVolume (V) = (2/3)πr³

Key facts to remember

1The slant height (l) of a cone is related to its radius (r) and height (h) by the Pythagorean theorem: l² = r² + h².
2The unit of surface area is always square units (e.g., cm², m²).
3The unit of volume is always cubic units (e.g., cm³, m³).
4A hemisphere is exactly half of a sphere.
5The total surface area of a hemisphere includes its curved surface area (2πr²) and the area of its circular base (πr²), making it 3πr².
6Always use the correct value of π as specified in the problem (usually 22/7 or 3.14).

Worked examples

Example 1

A conical tent has a base radius of 7 m and a height of 24 m. Find its slant height, curved surface area, and volume. (Use π = 22/7)

IGiven: Radius (r) = 7 m, Height (h) = 24 m.

IITo find slant height (l):

IIIWe know, l² = r² + h²

IVl² = (7)² + (24)²

Vl² = 49 + 576

VIl² = 625

VIIl = √625

VIIIl = 25 m

9To find Curved Surface Area (CSA):

10CSA = πrl

11CSA = (22/7) × 7 × 25

12CSA = 22 × 25

13CSA = 550 m²

14To find Volume (V):

15V = (1/3)πr²h

16V = (1/3) × (22/7) × (7)² × 24

17V = (1/3) × (22/7) × 49 × 24

18V = 22 × 7 × (24/3)

19V = 22 × 7 × 8

20V = 154 × 8

21V = 1232 m³

Answer

Slant height = 25 m, Curved Surface Area = 550 m², Volume = 1232 m³

Example 2

Find the surface area and volume of a sphere whose diameter is 14 cm. (Use π = 22/7)

IGiven: Diameter (d) = 14 cm.

IIRadius (r) = d/2 = 14/2 = 7 cm.

IIITo find Surface Area (SA):

IVSA = 4πr²

VSA = 4 × (22/7) × (7)²

VISA = 4 × (22/7) × 49

VIISA = 4 × 22 × 7

VIIISA = 88 × 7

9SA = 616 cm²

10To find Volume (V):

11V = (4/3)πr³

12V = (4/3) × (22/7) × (7)³

13V = (4/3) × (22/7) × 343

14V = (4/3) × 22 × 49

15V = (88 × 49) / 3

16V = 4312 / 3

17V = 1437.33 cm³ (approx.)

Answer

Surface Area = 616 cm², Volume = 1437.33 cm³ (approx.)

Example 3

A hemispherical bowl has a radius of 3.5 cm. Find the capacity of the bowl and the total surface area of the bowl. (Use π = 22/7)

IGiven: Radius (r) = 3.5 cm = 7/2 cm.

IITo find Capacity (Volume, V):

IIIV = (2/3)πr³

IVV = (2/3) × (22/7) × (7/2)³

VV = (2/3) × (22/7) × (343/8)

VIV = (2 × 22 × 343) / (3 × 7 × 8)

VIIV = (2 × 22 × 49) / (3 × 8) (Cancelling 7 with 343)

VIIIV = (11 × 49) / (3 × 2) (Cancelling 2 with 8, and 22 with 2)

9V = 539 / 6

10V = 89.83 cm³ (approx.)

11To find Total Surface Area (TSA):

12TSA = 3πr²

13TSA = 3 × (22/7) × (7/2)²

14TSA = 3 × (22/7) × (49/4)

15TSA = (3 × 22 × 7) / 4 (Cancelling 7 with 49)

16TSA = (3 × 11 × 7) / 2 (Cancelling 22 with 4)

17TSA = 231 / 2

18TSA = 115.5 cm²

Answer

Capacity = 89.83 cm³ (approx.), Total Surface Area = 115.5 cm²

Common mistakes

✗Confusing the vertical height (h) with the slant height (l) in cone formulas, or vice versa.
✗Using the diameter instead of the radius in formulas, especially for spheres and hemispheres.
✗Forgetting to include the area of the circular base when calculating the Total Surface Area of a hemisphere.
✗Incorrectly applying the fractional coefficients (1/3 for cone volume, 4/3 for sphere volume, 2/3 for hemisphere volume).
✗Making calculation errors, particularly with fractions, squaring, or cubing numbers.

Exam tips

★Always draw a neat diagram for each problem to visualise the solid and label its given dimensions clearly.
★Write down all given values and the relevant formulas you intend to use before starting your calculations.
★Pay close attention to the units of measurement and ensure consistency throughout the problem. Convert units if necessary.
★Simplify calculations by cancelling common factors, especially when π = 22/7 is used, to avoid large numbers.
★Double-check your calculations and ensure the final answer is presented with the correct units.

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