MathsICSE Class 9 · Mensuration

Circles, Rings and Paths, Cube and Cuboid: Area, Surface Area, Volume and Cost (ICSE Class 9)

Updated 17 October 2026 · Fact-checked

A circle has circumference 2πr and area πr². A ring or path between radii R and r has area π(R² − r²). A cuboid has volume lbh, total surface area 2(lb + bh + lh) and lateral surface area 2h(l + b); a cube has volume a³ and surface area 6a². For thick-walled containers, subtract the inner volume from the outer.

A ring's area is the big circle minus the small one; a cuboid's volume is l×b×h, and its surface area adds up all three face pairs.

Illustration drawn for QuestionBuddy.

Understand Circles, Rings and Paths, and the Cube and Cuboid

Use $π = \frac{22}{7}$ unless the question says otherwise.

Circle. Circumference $= 2 π r$ , area $= π r^{2}$ . If the circumference is $44$ cm, $r = 7$ cm and area $= 154$ cm².

Rings and paths. The area between two concentric circles of radii $R > r$ is $π (R^{2} - r^{2}) = π (R - r) (R + r)$ . A path of width $w$ around a circular lawn of radius $r$ has outer radius $r + w$ . For a path *inside*, the inner radius is $r - w$ .

Cuboid (length $l$ , breadth $b$ , height $h$ ):
$ $V = l bh, TSA = 2 (l b + bh + l h), LSA = 2 h (l + b), diagonal = l^{2} + b^{2} + h^{2} .$ $

Cube (edge $a$ ): $V = a^{3}$ , TSA $= 6 a^{2}$ , LSA $= 4 a^{2}$ . If the edge doubles, the volume becomes $8$ times and the surface area $4$ times.

Thickness problems. For an *open* box of external size $l \times b \times h$ and wall thickness $t$ , the inner size is $(l - 2 t) \times (b - 2 t) \times (h - t)$ (the top is open, so only the base thickness is removed from the height). Volume of material $=$ outer volume $-$ inner volume. If a closed box, subtract $2 t$ from the height.

Cost and capacity. Surface area gives cost of painting or tiling; volume gives capacity. $1 m^{3} = 1000$ litres. A pool without a top has a base and four walls, so the area to tile is $l b + 2 h (l + b)$ .

Level. Class IX is a school-exam class. Thickness problems and multi-part cost questions are at the hardest school-exam level.

Circles, Rings and Paths, and the Cube and Cuboid: formulas to remember

Circle: $C = 2 π r, A = π r^{2}$; π = 22/7 unless told.
Ring: $A = π (R^{2} - r^{2})$; Concentric circles.
Cuboid: $V = l bh, T S A = 2 (l b + bh + l h)$; LSA = 2h(l + b).
Cube: $V = a^{3}, T S A = 6 a^{2}$; LSA = 4a².
Open box with thickness t: $(l - 2 t) (b - 2 t) (h - t)$; Inner volume.

How to solve Circles, Rings and Paths, and the Cube and Cuboid questions

Draw the figure, list dimensions in the same unit, choose the formula, then compute.

1Convert all lengths to one unit.
2For a ring or path, find both radii first.
3For a box with thickness, find the inner dimensions before the volume.
4Apply the formula and keep fractions with 22/7 exact where possible.
5For cost, multiply area by the rate; for capacity, volume in cubic units, then convert.
6State units: cm, cm², cm³ and so on.

Circles, Rings and Paths, and the Cube and Cuboid in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

A pond with a path

Suppose a circular pond of radius 7 m has a 3.5 m wide path around it, paved at ₹8 per m².

Find the area of the path and the cost.

Try it first, then show the solutionSolution

Outer radius $= 10.5$ m.
Area $= \frac{22}{7} (10. 5^{2} - 7^{2}) = \frac{22}{7} \times 61.25 = 192.5$ m².
Cost $= 192.5 \times 8 = ₹1540$ .

Answer: 192.5 m²; ₹1,540.

Memory hook: Outer minus inner circle.

Tiling a swimming pool

Imagine a pool 25 m long, 10 m wide and 2 m deep, with the floor and four walls tiled at ₹60 per m².

Find the capacity in litres and the cost of tiling.

Try it first, then show the solutionSolution

Volume $= 25 \times 10 \times 2 = 500$ m³ $= 500, 000$ litres.
Area $= 25 \times 10 + 2 \times 2 \times (25 + 10) = 250 + 140 = 390$ m².
Cost $= 390 \times 60 = ₹23400$ .

Answer: 500,000 litres; ₹23,400.

Memory hook: Open top: base plus lateral area.

Tricks to remember Circles, Rings and Paths, and the Cube and Cuboid

Short memory aids. Say each one aloud once, then use it on a question the same day.

Outer minus inner
Ring or path: subtract the smaller circle from the larger.
Use it when: Ring and path problems.
Open box loses one height
Only the base thickness comes off the height.
Use it when: Thickness questions.
Square the edge, cube the volume
Doubling the edge: surface ×4, volume ×8.
Use it when: Scaling.

Circles, Rings and Paths, and the Cube and Cuboid: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Circle from circumference · 1 mark · MCQBoard level

A circle has circumference $44$ cm. Its area is (use $π = \frac{22}{7}$ ):
(a) 44 cm²
(b) 154 cm²
(c) 308 cm²
(d) 616 cm²

Try it first, then show the solutionSolution

$2 \times \frac{22}{7} \times r = 44 \Rightarrow r = 7$ .
Area $= \frac{22}{7} \times 49 = 154$ cm².

Answer: (b) 154 cm²

Example 2: Cuboid · 3 marks · Short answerBoard level

A cuboid measures $12$ cm $\times$ $9$ cm $\times$ $8$ cm. Find its volume, total surface area and lateral surface area.

Try it first, then show the solutionSolution

$V = 12 \times 9 \times 8 = 864$ cm³.
TSA $= 2 (108 + 72 + 96) = 2 \times 276 = 552$ cm².
LSA $= 2 \times 8 \times (12 + 9) = 336$ cm².

Answer: 864 cm³; 552 cm²; 336 cm²

Example 3: A ring · 3 marks · Short answerHarder board level

Two concentric circles have radii $14$ cm and $7$ cm. Find the area of the ring between them.

Try it first, then show the solutionSolution

Area $= \frac{22}{7} (1 4^{2} - 7^{2}) = \frac{22}{7} \times 147$ .
$= 22 \times 21 = 462$ cm².

Answer: 462 cm²

Example 4: A path around a pond · 4 marks · Short answerHarder board level

A circular pond of radius $7$ m has a path $3.5$ m wide all around it. Find the area of the path and the cost of paving it at ₹8 per m².

Try it first, then show the solutionSolution

Outer radius $= 7 + 3.5 = 10.5$ m.
Area $= \frac{22}{7} (10. 5^{2} - 7^{2}) = \frac{22}{7} (110.25 - 49) = \frac{22}{7} \times 61.25 = 192.5$ m².
Cost $= 192.5 \times 8 = ₹1, 540$ .

Answer: 192.5 m²; ₹1,540

Example 5: Volume of metal · 5 marks · Long answerHOTS / competency

An open metal box has external dimensions $30$ cm $\times$ $20$ cm $\times$ $12$ cm and walls $1$ cm thick. Find the volume of metal used, and its mass if 1 cm³ of this metal has mass 8 g (assumed).

Try it first, then show the solutionSolution

Outer volume $= 30 \times 20 \times 12 = 7200$ cm³.
Inner dimensions: $28 \times 18 \times 11$ (the top is open, so only the base is deducted from the height).
Inner volume $= 28 \times 18 \times 11 = 5544$ cm³.
Metal $= 7200 - 5544 = 1656$ cm³.
Mass $= 1656 \times 8 = 13248$ g $= 13.248$ kg.

Answer: 1,656 cm³; 13.248 kg

Where students lose marks in Circles, Rings and Paths, and the Cube and Cuboid

The slips examiners see most often, and what to write instead.

Mistake: Using the full height for an open box
Only the base thickness is removed.
Wrong working
Inner height $= 12 - 2 = 10$ .
Right working
Inner height $= 12 - 1 = 11$ .
Do this: Check if the box is open or closed.
Usually costs: Whole answer.
Mistake: Squaring the difference of radii
Ring area is R² − r², not (R − r)².
Wrong working
$π (14 - 7)^{2}$ .
Right working
$π (1 4^{2} - 7^{2})$ .
Do this: Square each radius first.
Usually costs: Whole answer.
Mistake: Mixing units
Convert to one unit before multiplying.
Wrong working
Length in m and height in cm.
Right working
All in m or all in cm.
Do this: List units first.
Usually costs: Whole answer.

Exam tips

Class IX is examined in school papers; state π = 22/7 when you use it.
Draw the box or ring and label every length before substituting.
In thickness problems, write inner dimensions on a separate line.
Convert capacity to litres only at the end.

Practice questions on Circles, Rings and Paths, and the Cube and Cuboid

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
The total surface area of a cube of edge $6$ cm is:
(a) 36 cm²
(b) 144 cm²
(c) 216 cm²
(d) 324 cm²
Hint
$6 a^{2}$ .
Answer
(c) 216 cm²
Q2 · 3 marks · Short answerHarder board level
The volume of a cube is $343$ cm³. Find its edge, total surface area and lateral surface area.
Hint
Cube root of 343.
Answer
Edge $7$ cm; TSA $= 6 \times 49 = 294$ cm²; LSA $= 4 \times 49 = 196$ cm².
Q3 · 3 marks · Short answerHarder board level
The area of a circle is $616$ cm². Find its radius and circumference ( $π = \frac{22}{7}$ ).
Hint
$r^{2} = 616 \times \frac{7}{22}$ .
Answer
$r^{2} = 196$ , $r = 14$ cm; circumference $= 2 \times \frac{22}{7} \times 14 = 88$ cm.
Q4 · 1 mark · Assertion-reasonHarder board level
Assertion (A): If the edge of a cube is doubled, its volume becomes 8 times.
Reason (R): The volume of a cube is the cube of its edge.
(a) Both (A) and (R) are true and (R) explains (A).
(b) Both are true but (R) does not explain (A).
(c) (A) is true, (R) is false.
(d) (A) is false, (R) is true.
Hint
$(2 a)^{3} = 8 a^{3}$ .
Answer
(a)
Q5 · 4 marks · Case-basedHOTS / competency
Case study: A swimming pool is $25$ m long, $10$ m wide and $2$ m deep. The floor and four walls are to be tiled at ₹60 per m².
(a) Find the capacity in litres ( $1 m^{3} = 1000$ litres).
(b) Find the area to be tiled.
(c) Find the cost of tiling.
Hint
Open top: base + 4 walls.
Answer
(a) 500,000 litres (b) 250 + 140 = 390 m² (c) ₹23,400.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Circumference and area of a circle?Show answer
2πr and πr².
2.Area of a ring?Show answer
π(R² − r²).
3.TSA of a cuboid?Show answer
2(lb + bh + lh).
4.How many litres in a cubic metre?Show answer
1000.

Circles, Rings and Paths, and the Cube and Cuboid: frequently asked questions

Which value of π is used?

Questions state it; 22/7 is the usual choice when the radius is a multiple of 7.

How do I handle a path around a rectangle?

Outer area minus inner area, with each side increased by twice the width.

What is the difference between LSA and TSA?

LSA excludes the top and bottom faces.