Ganita Manjari · Class 9 · Part II · Chapter 14

Ganita Manjari Class 9 Maths Chapter 14 Math of Space: Surface Area and Volume Exercise 14.3 Solutions

Nine questions on cones: surface areas, volume, a joker's cap, a tent, a solid of revolution and a cup filled to half its depth.

Last updated 6 October 2026

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Ch 14Math of Space: Surface Area and Volume

How to answer every question

CSA = πrl, V = ⅓πr²h

Find the slant height l = √(h² + r²) by the Baudhāyana–Pythagoras theorem.
Choose the formula CSA = πrl, TSA = πr(l + r), V = ⅓πr²h.
Check the open end A cap or tent has no base, so use the curved surface only.
Keep the unit Areas are in square units, volumes in cubic units.

Slant height: the length along the cone's surface from the apex to the rim.

CSA of a cone: πrl.

Volume of a cone: one third of the base area times the height.

Cones

Chapter 14 Exercise 14.3 Question 1 Solution : Find the total surface area of a cone with slant height 21 m and base diameter 24 m. (π ≈ 227)

Answer: ≈ 1244.57 m²

Radius and formula

r = 12 m, l = 21 m.

TSA = πr(l + r) = 227 × 12 × 33 = 87127 ≈ 1244.57 m²

With π = 3.14 the answer is 1243.44 m².

The total surface area is about 1244.57 m² (87127 m²).

Chapter 14 Exercise 14.3 Question 2 Solution : Find the curved surface area of a cone with slant height 10 cm and base radius 7 cm.

Answer: 220 cm²

Use CSA = πrl

227 × 7 × 10 = 220 cm²

The curved surface area is 220 cm².

Chapter 14 Exercise 14.3 Question 3 Solution : The height of a cone is 16 cm and its base radius is 12 cm. Find the curved and total surface areas.

Answer: 240π and 384π

Slant height first

l = √(h² + r²) = √(256 + 144) = √400 = 20 cm

Areas

CSA = πrl = π × 12 × 20 = 240π cm²

TSA = πr(l + r) = π × 12 × 32 = 384π cm²

With π = 3.14: CSA = 753.6 cm² and TSA = 1205.76 cm². With π = 227: about 754.29 cm² and 1206.86 cm².

CSA = 240π ≈ 753.6 cm² and TSA = 384π ≈ 1205.76 cm².

Chapter 14 Exercise 14.3 Question 4 Solution : A cone has height 15 cm and volume 1570 cm³. Find the radius of the base. (π = 3.14)

Answer: 10 cm

Use V = ⅓πr²h

1570 = 13 × 3.14 × r² × 15 = 15.7 r²

r² = 157015.7 = 100 ⟹ r = 10 cm

The radius is 10 cm.

Chapter 14 Exercise 14.3 Question 5 Solution : The curved surface area of a cone is 308 cm² and its slant height is 14 cm. Find (i) the radius, (ii) the total surface area.

Answer: r = 7 cm; TSA = 462 cm²

Part (i)

πrl = 308 ⟹ 227 × r × 14 = 308 ⟹ 44r = 308 ⟹ r = 7 cm

Part (ii)

TSA = πr(l + r) = 227 × 7 × 21 = 462 cm²

Check: the base is πr² = 154, and 308 + 154 = 462 ✓.

(i) 7 cm, (ii) 462 cm².

Chapter 14 Exercise 14.3 Question 6 Solution : A joker's cap is a cone of base radius 7 cm and height 24 cm. Find the area of sheet for 10 caps.

Answer: 5500 cm²

Slant height and one cap

l = √(24² + 7²) = √625 = 25 cm

CSA = πrl = 227 × 7 × 25 = 550 cm²

A cap is open at the bottom (the head goes in), so only the curved surface needs sheet.

Ten caps

10 × 550 = 5500 cm²

The sheet needed is 5500 cm².

Chapter 14 Exercise 14.3 Question 7 Solution : What length of tarpaulin 3 m wide is needed for a conical tent of height 8 m and base radius 6 m? Allow an extra 20 cm length for stitching and wastage. (π ≈ 227)

StarredAnswer: about 63.06 m

Area of the curved surface

l = √(8² + 6²) = 10 m

CSA = πrl = 227 × 6 × 10 = 13207 ≈ 188.57 m²

Length of the strip

Area = width × length, so:

length = 13207 × 3 = 4407 ≈ 62.86 m

Add the extra 0.2 m: 62.86 + 0.2 = 63.06 m.

About 63.06 m of tarpaulin is needed.

Chapter 14 Exercise 14.3 Question 8 Solution : A right triangle with sides 6 cm, 8 cm and 10 cm is rotated through 360° about the side of 8 cm. Find the volume and the curved surface area of the solid.

Answer: V = 96π; CSA = 60π

The solid is a cone

The 8 cm side is the axis, so h = 8 cm, r = 6 cm, and the slant height is the hypotenuse l = 10 cm.

Volume and curved surface area

V = 13πr²h = 13π × 36 × 8 = 96π ≈ 301.44 cm³

CSA = πrl = π × 6 × 10 = 60π ≈ 188.4 cm²

(Taking π = 3.14. With π = 227 these are about 301.71 cm³ and 188.57 cm².)

Volume 96π ≈ 301.44 cm³ and curved surface area 60π ≈ 188.4 cm².

Chapter 14 Exercise 14.3 Question 9 Solution : A cup is a right circular cone. It is filled with water to half the depth. What fraction of the cup's volume does the water fill?

StarredAnswer: one eighth

The cup

A girl thinking next to a conical cup that is partly filled with water. — Q9 · a cone-shaped cup filled to half its depth (from the book)

The water is a smaller cone

The water forms a cone of the same shape, with half the height. By similar triangles, its radius is also half.

Q9 · half the depth means half the radius

water : cup = ⅓ × π × (R over 2)² × (H over 2)⅓ × π × R² × H = 14 × 12 = 18

The water fills 18 of the cup.

Answers at a glance

For a cone, find l first.

Question	What is asked	Key idea	Answer
Q1	TSA of a cone	πr(l + r)	1244.57 m²
Q2	CSA	πrl	220 cm²
Q3	From h and r	l = 20	CSA 240π, TSA 384π
Q4	Radius from volume	⅓πr²h = 1570	10 cm
Q5	From CSA	πrl = 308	r = 7 cm; TSA 462 cm²
Q6	Ten caps	10 × πrl	5500 cm²
Q7	Tarpaulin	Area over width + 0.2	63.06 m
Q8	Rotate 6-8-10 triangle	h = 8, r = 6, l = 10	96π; 60π
Q9	Cup half full	Similar cones	1/8

Source of the questions: NCERT, Ganita Manjari, Grade 9, Part II, Chapter 14, Exercise Set 14.3. The solutions, explanations and diagrams on this page are our own working.

How to answer every question

Cones

Answers at a glance

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