How many exercises are in NCERT Class 9 Maths Chapter 11?

Chapter 11 has nine exercises covering surface areas and volumes of cuboids, cubes, cylinders, cones, spheres, and hemispheres. Exercises 11.1 to 11.4 focus on surface areas, and the remaining exercises cover volumes.

What is the difference between curved surface area and total surface area?

Curved surface area (CSA) is the area of only the curved portion of a solid (excluding flat bases). Total surface area (TSA) includes the curved surface plus all flat bases. For example, for a cylinder: CSA $= 2\pi rh$, TSA $= 2\pi r(h+r)$. The flat bases contribute $2\pi r^2$ to the TSA.

How do you find the slant height of a cone?

Use the Pythagorean theorem: $l = \sqrt{r^2 + h^2}$, where $r$ is the radius of the base and $h$ is the vertical height. The slant height is the distance from the tip of the cone to any point on the circumference of the base. You need the slant height for surface area calculations (CSA $= \pi rl$).

What is the volume of a hemisphere?

The volume of a hemisphere is $\dfrac{2}{3}\pi r^3$, which is exactly half the volume of a full sphere ($\dfrac{4}{3}\pi r^3$). The TSA of a hemisphere is $3\pi r^2$ (curved surface $2\pi r^2$ plus flat base $\pi r^2$).

How do you convert between litres and cubic centimetres?

$1$ litre $= 1000$ cm$^3$ and $1$ m$^3 = 1000$ litres. These conversions are essential for word problems involving tanks, vessels, and containers. Be careful: $1$ m$^3 = 10^6$ cm$^3$ (not $1000$ cm$^3$).

How do you solve melting and recasting problems?

When a solid is melted and recast into a different shape, the total volume remains the same. Set up the equation: volume of original solid $=$ volume of new solid(s). If it is recast into multiple identical solids, divide the total volume by the number of solids to get the volume of each.

Why is the volume of a cone one-third the volume of a cylinder?

When a cone and a cylinder have the same base radius and same height, the cone's volume is exactly $\dfrac{1}{3}$ of the cylinder's volume. This can be demonstrated experimentally by filling a cone with water and pouring it into a cylinder of the same dimensions — it takes exactly $3$ cones to fill the cylinder.

How many exercises are in NCERT Class 9 Maths Chapter 11?

Chapter 11 has nine exercises covering surface areas and volumes of cuboids, cubes, cylinders, cones, spheres, and hemispheres. Exercises 11.1 to 11.4 focus on surface areas, and the remaining exercises cover volumes.

What is the difference between curved surface area and total surface area?

Curved surface area (CSA) is the area of only the curved portion of a solid (excluding flat bases). Total surface area (TSA) includes the curved surface plus all flat bases. For example, for a cylinder: CSA $= 2\pi rh$, TSA $= 2\pi r(h+r)$. The flat bases contribute $2\pi r^2$ to the TSA.

How do you find the slant height of a cone?

Use the Pythagorean theorem: $l = \sqrt{r^2 + h^2}$, where $r$ is the radius of the base and $h$ is the vertical height. The slant height is the distance from the tip of the cone to any point on the circumference of the base. You need the slant height for surface area calculations (CSA $= \pi rl$).

What is the volume of a hemisphere?

The volume of a hemisphere is $\dfrac{2}{3}\pi r^3$, which is exactly half the volume of a full sphere ($\dfrac{4}{3}\pi r^3$). The TSA of a hemisphere is $3\pi r^2$ (curved surface $2\pi r^2$ plus flat base $\pi r^2$).

How do you convert between litres and cubic centimetres?

$1$ litre $= 1000$ cm$^3$ and $1$ m$^3 = 1000$ litres. These conversions are essential for word problems involving tanks, vessels, and containers. Be careful: $1$ m$^3 = 10^6$ cm$^3$ (not $1000$ cm$^3$).

How do you solve melting and recasting problems?

When a solid is melted and recast into a different shape, the total volume remains the same. Set up the equation: volume of original solid $=$ volume of new solid(s). If it is recast into multiple identical solids, divide the total volume by the number of solids to get the volume of each.

Why is the volume of a cone one-third the volume of a cylinder?

When a cone and a cylinder have the same base radius and same height, the cone's volume is exactly $\dfrac{1}{3}$ of the cylinder's volume. This can be demonstrated experimentally by filling a cone with water and pouring it into a cylinder of the same dimensions — it takes exactly $3$ cones to fill the cylinder.

Solved Examples

NCERT Solutions for Class 9 Maths Chapter 11: Surface Areas and Volumes — Free PDF

Complete solutions for all exercises — surface areas and volumes of cubes, cuboids, cylinders, cones, and spheres.

CBSEClass 9

The SparkEd Authors (IITian & Googler)15 March 202640 min read

Chapter 11 Overview: Surface Areas and Volumes

$Chapter 11 is one of the longest and most formula-heavy chapters in Class 9 Maths. It covers the surface areas (lateral/curved and total) and volumes of five types of 3D solids: cuboids, cubes, cylinders, cones, and spheres (including hemispheres).$

$This chapter builds directly on the mensuration concepts from Class 8 (Chapter 9), where you studied cubes, cuboids, and cylinders. In Class 9, the new additions are cones, spheres, and hemispheres, each with their own set of formulas for CSA, TSA, and volume.$

$The chapter has nine exercises — Exercises 11.1 to 11.4 cover surface areas of different solids, while the remaining exercises cover volumes. Memorising the formulas is essential, but equally important is knowing when to apply CSA versus TSA, and handling unit conversions correctly. This chapter typically carries 8-12 marks in CBSE Class 9 exams and forms the foundation for Class 10 Chapter 12 (Areas Related to Circles) and Chapter 13 (Surface Areas and Volumes of combined solids).$

All Formulas at a Glance

$Solid$	$Lateral/Curved SA$	$Total SA$	$Volume$
$l, b, h$	$2h(l+b)$	$2(lb+bh+hl)$	$lbh$
$a$	$4a^2$	$6a^2$	$a^3$
$r, h$	$2\pi rh$	$2\pi r(h+r)$	$\pi r^2 h$
$r, h, l$	$\pi r l$	$\pi r(l+r)$	$\dfrac{1}{3}\pi r^2 h$
$r$	$4\pi r^2$	$4\pi r^2$	$\dfrac{4}{3}\pi r^3$
$r$	$2\pi r^2$	$3\pi r^2$	$\dfrac{2}{3}\pi r^3$

$l = \sqrt{r^2 + h^2}$

Key Concepts and Definitions

$Curved Surface Area (CSA): The area of only the curved portion of a solid, excluding any flat faces. For a cylinder, this is the side surface; for a cone, the sloping surface; for a sphere, the entire surface.$

$Total Surface Area (TSA): The area of all surfaces, including flat bases and the curved surface.$

$Volume: The space occupied by the solid, measured in cubic units.$

$l$

$+$

$1$

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Exercise 11.1 — Surface Area of Cuboids and Cubes (Solved)

$1.5$

$l = 1.5$

$Since the box is open at the top:$

\text{Area} = lb + 2(bh + hl)

= (1.5)(1.25) + 2[(1.25)(0.65) + (0.65)(1.5)]

= 1.875 + 2[0.8125 + 0.975]

= 1.875 + 2(1.7875)

= 1.875 + 3.575 = 5.45 \text{ m}^2

$---$

$10$

$Solution:$

\text{TSA} = 6a^2 = 6 \times 10^2 = 600 \text{ cm}^2

$---$

$5$

$l = 15$

\text{TSA} = 2(lb + bh + lh) = 2(75 + 25 + 75) = 2(175) = 350 \text{ cm}^2

$3 \times 150 = 450$

Exercise 11.2 — Surface Area of Cylinders (Solved)

$14$

$Solution:$

\text{CSA} = 2\pi rh

88 = 2 \times \dfrac{22}{7} \times r \times 14

88 = 88r

r = 1 \text{ cm}

$= 2r = 2$

$---$

$3.5$

$r = \dfrac{3.5}{2} = 1.75$

\text{CSA} = 2\pi rh = 2 \times \dfrac{22}{7} \times 1.75 \times 12 = 132 \text{ m}^2

$= 132 \times 12.50 = \text{Rs } 1650$

$---$

$1$

$r = 70$

\text{TSA} = 2\pi r(r + h) = 2 \times \dfrac{22}{7} \times 0.7 \times (0.7 + 1) = 2 \times \dfrac{22}{7} \times 0.7 \times 1.7

= 2 \times 22 \times 0.1 \times 1.7 = 7.48 \text{ m}^2

Exercises 11.3 & 11.4 — Cones and Spheres (Solved)

$6$

$Solution:$

\text{TSA} = \pi r(l + r) = \dfrac{22}{7} \times 6 \times (8 + 6) = \dfrac{22}{7} \times 6 \times 14 = 264 \text{ cm}^2

$---$

$10$

$Solution:$

l = \sqrt{r^2 + h^2} = \sqrt{24^2 + 10^2} = \sqrt{576 + 100} = \sqrt{676} = 26 \text{ m}

\text{CSA} = \pi rl = \dfrac{22}{7} \times 24 \times 26 = \dfrac{13728}{7} \approx 1961.14 \text{ m}^2

$= 1961.14 \times 70 \approx \text{Rs } 1{,}37{,}280$

$---$

$5.6$

$Solution:$

\text{SA} = 4\pi r^2 = 4 \times \dfrac{22}{7} \times (5.6)^2 = 4 \times \dfrac{22}{7} \times 31.36 = 394.24 \text{ cm}^2

$---$

$3.5$

$Solution:$

\text{TSA} = 3\pi r^2 = 3 \times \dfrac{22}{7} \times (3.5)^2 = 3 \times \dfrac{22}{7} \times 12.25 = 115.5 \text{ cm}^2

$---$

$7 : 4$

$l = 7k$

\text{CSA} = \pi rl = \dfrac{22}{7} \times 4k \times 7k = 88k^2

88k^2 = 792 \implies k^2 = 9 \implies k = 3

$= 4 \times 3 = 12$

Volumes — Key Solved Problems

$6$

$Solution:$

V = lbh = 6 \times 5 \times 4.5 = 135 \text{ m}^3

$1 \text{ m}^3 = 1000$

V = 1{,}35{,}000 \text{ litres}

$---$

$3.5$

$Solution:$

V = \dfrac{1}{3}\pi r^2 h = \dfrac{1}{3} \times \dfrac{22}{7} \times (3.5)^2 \times 12

= \dfrac{1}{3} \times \dfrac{22}{7} \times 12.25 \times 12 = \dfrac{1}{3} \times \dfrac{3234}{7} = \dfrac{1078}{7} = 154 \text{ cm}^3

$---$

$10.5$

$Solution:$

V = \dfrac{4}{3}\pi r^3 = \dfrac{4}{3} \times \dfrac{22}{7} \times (10.5)^3 = \dfrac{4}{3} \times \dfrac{22}{7} \times 1157.625 = 4851 \text{ cm}^3

$---$

$7$

$Solution:$

V = \dfrac{2}{3}\pi r^3 = \dfrac{2}{3} \times \dfrac{22}{7} \times 7^3 = \dfrac{2}{3} \times \dfrac{22}{7} \times 343 = \dfrac{2 \times 22 \times 49}{3} = \dfrac{2156}{3} \approx 718.67 \text{ cm}^3

$= \dfrac{718.67}{1000} \approx 0.719$

$---$

$3$

$= 2$

$=$

= 40 \times 3 \times \dfrac{100}{3} = 4000 \text{ m}^3/\text{min}

$4000 \times 1000 = 40{,}00{,}000$

Additional Worked Examples

$4.2$

$=$

\dfrac{4}{3}\pi (4.2)^3 = \pi (6)^2 h

\dfrac{4}{3} \times (4.2)^3 = 36h

\dfrac{4 \times 74.088}{3} = 36h

98.784 = 36h \implies h = 2.744 \text{ cm}

$---$

$84$

$= 42$

= 2 \times \dfrac{22}{7} \times 42 \times 120 = 31680 \text{ cm}^2

$= 500 \times 31680 = 1{,}58{,}40{,}000$

$---$

$27$

$= 27 \times \dfrac{4}{3}\pi r^3$

$R$

\dfrac{4}{3}\pi R^3 = 27 \times \dfrac{4}{3}\pi r^3

R^3 = 27r^3 \implies R = 3r

$3$

Common Mistakes to Avoid

$Mistake 1: Confusing CSA and TSA. CSA excludes the base(s). TSA includes everything. Read the question carefully. If it says "curved surface" or "lateral surface", use CSA. If it says "total surface", use TSA.$

$\pi r l$

$= 2\pi r^2$

$1$

$\pi = 22/7$

Tips for Scoring Full Marks

$1 \text{ m}^3 = 1000$

Practice Questions with Answers

$5$

$l = \sqrt{25 + 144} = \sqrt{169} = 13$

$---$

$616$

$4\pi r^2 = 616$

$---$

$7$

$\dfrac{V_{\text{cone}}}{V_{\text{cylinder}}} = \dfrac{\frac{1}{3}\pi r^2 h}{\pi r^2 h} = \dfrac{1}{3}$

$---$

$14$

$r = 7$

$---$

$7$

$= \dfrac{22}{7} \times 49 \times 10 = 1540$

Key Takeaways

$l = \sqrt{r^2 + h^2}$

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