Mensuration ICSE Class 9: Triangles, Quadrilaterals, Circles, Cubes and Cuboids

What Mensuration is about

Weightage: CISCE publishes no chapter weightage and no question-paper layout for ICSE Class 9 Mathematics. Class IX is examined by your school: one 80-mark paper of 3 hours plus 20 marks of internal assessment. Mensuration is in the 6. Mensuration part of the 2028 syllabus.Source: CISCE ICSE 2028 Mathematics syllabus, Class IX

Key concepts

Area of a triangle and Heron's formula

With a base and its height: area $=\frac{1}{2}\times \text{base}\times \text{height}$.

With three sides $a$, $b$, $c$ only, use Heron's formula:
$s=\frac{a+b+c}{2}$ (semi-perimeter), area $=\sqrt{s(s-a)(s-b)(s-c)}$.

Special cases:

• Right triangle: $\frac{1}{2}\times$ product of the legs.
• Equilateral triangle of side $a$: area $=\frac{\sqrt{3}}{4}{a}^2$, height $=\frac{\sqrt{3}}{2}a$.
• Isosceles triangle with equal sides $a$ and base $b$: height $=\sqrt{a^2-\frac{b^2}{4}}$.

Once you know the area, the height to any side is $\frac{2\times \text{area}}{\text{that side}}$.

Worked example

Find the area of a triangle with sides 13 cm, 14 cm and 15 cm, and the height to the 14 cm side.

Try it first, then show the solutionSolution

1. $s=\frac{13+14+15}{2}=21$.
2. Area $=\sqrt{21\times 8\times 7\times 6}=\sqrt{7056}=84$ cm².
3. Height $=\frac{2\times 84}{14}=12$ cm.

Answer: 84 cm²; 12 cm

Area and perimeter of quadrilaterals

• Rectangle: $A=lb$, $P=2(l+b)$, diagonal $=\sqrt{l^2+b^2}$.
• Square: $A=a^2=\frac{1}{2}{d}^2$, $P=4a$.
• Parallelogram: $A=\text{base}\times \text{height}$.
• Rhombus: $A=\frac{1}{2}{d}_1{d}_2$; side $=\sqrt{{\left(\frac{d_1}{2}\right)}^2+{\left(\frac{d_2}{2}\right)}^2}$.
• Trapezium: $A=\frac{1}{2}(\text{sum of parallel sides})\times \text{height}$.
• Any quadrilateral: split it along a diagonal into two triangles; area $=\frac{1}{2}\times \text{diagonal}\times (h_1+h_2)$, where $h_1$, $h_2$ are the perpendiculars from the other vertices, or use Heron for each triangle.

Worked example

The diagonals of a rhombus are 16 cm and 12 cm. Find its area, side and height.

Try it first, then show the solutionSolution

1. Area $=\frac{1}{2}\times 16\times 12=96$ cm².
2. Side $=\sqrt{8^2+6^2}=10$ cm.
3. Height $=\frac{96}{10}=9.6$ cm (rhombus is a parallelogram: area = side × height).

Answer: 96 cm²; 10 cm; 9.6 cm

Circle: circumference and area

For radius $r$ (diameter $d=2r$):

• circumference $C=2\pi r=\pi d$;
• area $A=\pi r^2$.

Semicircle: area $\frac{1}{2}\pi r^2$, perimeter $\pi r+2r$ (the curved part plus the diameter). Quarter-circle: area $\frac{1}{4}\pi r^2$, perimeter $\frac{1}{2}\pi r+2r$.

A wheel moves forward one circumference in one revolution: number of revolutions $=\frac{\text{distance}}{\text{circumference}}$.

Worked example

The circumference of a circle is 88 cm. Find its area.

Try it first, then show the solutionSolution

1. $2\times \frac{22}{7}\times r=88$, so $r=14$ cm.
2. Area $=\frac{22}{7}\times 14\times 14=616$ cm².

Answer: 616 cm²

Inner and outer areas: rings and paths

A ring (annulus) between circles of radii $R$ and $r$ has area $\pi (R^2-r^2)=\pi (R+r)(R-r)$.

A circular path of width $w$ outside a circular garden of radius $r$ has outer radius $R=r+w$; inside, $R$ is the garden and $r=R-w$.

For rectangular paths: area of path $=$ outer rectangle $-$ inner rectangle. A path of width $w$ outside an $l\times b$ plot gives outer dimensions $(l+2w)\times (b+2w)$; inside, $(l-2w)\times (b-2w)$. Cross paths through the middle overlap in a $w\times w$ square, which must be counted once.

Worked example

A circular garden of radius 14 m has a path 3.5 m wide around it. Find the area of the path.

Try it first, then show the solutionSolution

1. $R=14+3.5=17.5$ m.
2. Area $=\frac{22}{7}(17.5^2-14^2)=\frac{22}{7}(306.25-196)=\frac{22}{7}\times 110.25$.
3. $=346.5$ m².

Answer: 346.5 m²

Cube and cuboid: surface area and volume

Cuboid $l\times b\times h$:

• volume $V=lbh$;
• total surface area $=2(lb+bh+hl)$;
• area of four walls (lateral surface) $=2h(l+b)$;
• diagonal $=\sqrt{l^2+b^2+h^2}$.

Cube of edge $a$: $V=a^3$, total surface area $=6a^2$, four walls $=4a^2$, diagonal $=a\sqrt{3}$.

Open box (no lid): surface area $=lb+2h(l+b)$.

Units: 1 m³ $=1000$ litres; 1 litre $=1000$ cm³.

Worked example

An open box (no lid) measures 50 cm × 40 cm × 30 cm. Find the area of sheet needed and its volume in litres.

Try it first, then show the solutionSolution

1. Area $=lb+2h(l+b)=2000+2\times 30\times 90=2000+5400=7400$ cm².
2. Volume $=50\times 40\times 30=60000$ cm³ $=60$ litres.

Answer: 7400 cm²; 60 litres

Thickness, cost and cross-section

Thickness: a closed box with external dimensions $L\times B\times H$ and wall thickness $t$ has internal dimensions $(L-2t)\times (B-2t)\times (H-2t)$. For an open box (no lid), the height loses only $t$: $(H-t)$. Volume of material $=$ external volume $-$ internal volume.

Cost $=$ area (or volume) $\times$ rate. Match the units of the rate (per m², per m³).

Cross-section: for any solid with the same cross-section along its length (a prism: a cuboid, a beam, a channel), volume $=$ area of cross-section $\times$ length (or height). Water flowing through a channel: volume per unit time $=$ cross-section area $\times$ speed.

Worked example

A closed wooden box has external dimensions 24 cm × 18 cm × 12 cm and the wood is 1 cm thick. Find the volume of wood used.

Try it first, then show the solutionSolution

1. External volume $=24\times 18\times 12=5184$ cm³.
2. Internal: $22\times 16\times 10=3520$ cm³.
3. Wood $=5184-3520=1664$ cm³.

Answer: 1664 cm³

Maths in real life: Mensuration

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.

Topics in this chapter

• Circles, Rings and Paths, and the Cube and Cuboid
• Heron's Formula and Areas of Triangles, Parallelograms, Rhombuses and Trapeziums

Mensuration formula sheet

Heron's formula

$A=\sqrt{s(s-a)(s-b)(s-c)},s=\frac{a+b+c}{2}$

Three sides known.

Equilateral triangle

$A=\frac{\sqrt{3}}{4}{a}^2$

Height (√3/2)a.

Trapezium

$A=\frac{1}{2}(a+b)h$

a, b parallel sides.

Rhombus

$A=\frac{1}{2}{d}_1{d}_2$

Also side × height.

Circle

$C=2\pi r;A=\pi r^2$

π = 22/7 unless told.

Ring

$A=\pi (R^2-r^2)$

= π(R + r)(R − r).

Cuboid

$V=lbh;S=2(lb+bh+hl)$

Four walls: 2h(l + b).

Cube

$V=a^3;S=6a^2$

Diagonal a√3.

Prism

$V=\text{area of cross-section}\times \text{length}$

Channels, beams, pipes.

How the school exam asks Mensuration

The question patterns to expect, and the method that works for each.

Tricks to remember Mensuration

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

• s minus each side

  Heron: semi-perimeter times s minus each side, then square root.

  Use it when: Triangles with three sides.

• Half the sum times the gap

  Trapezium area = half the sum of the parallel sides × the gap between them.

  Use it when: Trapeziums.

• One turn, one circumference

  A wheel rolls forward its own circumference every turn.

  Use it when: Revolution problems.

• Thickness bites twice

  Wall thickness comes off both sides: subtract 2t from each external dimension (only t from the height of an open box).

  Use it when: Internal dimensions.

Mensuration: exam-level solved questions

Ordered from exam level to HOTS. Cover the solution, write your own, then compare step by step.

Case-study practice (Section B style)

Read the situation, then answer every part. In a school exam set on the board pattern this is a 10-mark question split into 3 + 3 + 4 marks.

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

Where students lose marks in Mensuration

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

• ✗Mistake: Forgetting the diameter in a semicircle's perimeter

  Perimeter of a semicircle is $\pi r+2r$, not $\pi r$.

  Wrong working

  Perimeter (r = 7) $=22$ cm

  Right working

  Perimeter $=22+14=36$ cm

  ✓Do this: Trace the boundary with your finger: curved part plus straight edge.

  Usually costs: 1 mark.

• ✗Mistake: Subtracting thickness once instead of twice

  A closed box loses the thickness on both sides of each dimension.

  Wrong working

  Internal $23\times 17\times 11$ for $t=1$

  Right working

  Internal $22\times 16\times 10$

  ✓Do this: Subtract $2t$ from each dimension, except $t$ from the height of an open box.

  Usually costs: Every later mark.

• ✗Mistake: Mixing units

  Using metres for one length and centimetres for another, or forgetting 1 m³ = 1000 litres.

  ✓Do this: Convert all lengths to one unit before multiplying.

  Usually costs: The answer.

• ✗Mistake: Using the slant side as the height

  In a parallelogram or trapezium the height is perpendicular to the base.

  ✓Do this: Find the perpendicular height (often with Pythagoras or Heron) first.

  Usually costs: The answer.

• ✗Mistake: Counting the cross-path overlap twice

  Two crossing paths share a square; adding both path areas counts it twice.

  ✓Do this: Area $=w(l+b)-w^2$ for two central cross paths.

  Usually costs: 1 mark.

Practice questions on Mensuration

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

1. Q1 · 1 mark · MCQExam level

   The volume of a cube is 512 cm³. Its total surface area is:
   (a) 64 cm²
   (b) 256 cm²
   (c) 384 cm²
   (d) 512 cm²

   Hint

   Edge = ∛512.

   Answer

   (c): edge 8 cm, $6\times 64=384$.

2. Q2 · 2 marks · Short answerExam level

   Find the circumference and area of a circle of diameter 14 cm.

   Hint

   r = 7.

   Answer

   44 cm; 154 cm².

3. Q3 · 2 marks · Short answerExam level

   Find the area of a trapezium with parallel sides 20 cm and 12 cm, 8 cm apart.

   Hint

   ½(a + b)h.

   Answer

   128 cm².

4. Q4 · 1 mark · Assertion-reasonHarder exam level

   Assertion (A): Doubling the radius of a circle doubles its area.
   Reason (R): The area of a circle is proportional to the square of its radius.
   (a) (A) is true and (R) is false.
   (b) (A) is false and (R) is true.
   (c) Both (A) and (R) are true, and (R) is the correct explanation for (A).
   (d) Both (A) and (R) are true, but (R) is not the correct explanation for (A).

   Hint

   (2r)² = 4r².

   Answer

   (b)

5. Q5 · 3 marks · Short answerHarder exam level

   The perimeter of a triangle is 50 cm. One side is 4 cm longer than the smallest side and the third side is 6 cm less than twice the smallest side. Find its area.

   Hint

   x + (x + 4) + (2x − 6) = 50.

   Answer

   $4x-2=50$, $x=13$: sides 13, 17, 20; $s=25$; area $=\sqrt{25\times 12\times 8\times 5}=\sqrt{12000}=20\sqrt{30}\approx 109.54$ cm².

6. Q6 · 3 marks · Short answerHarder exam level

   A rectangular lawn 60 m × 40 m has two paths, each 3 m wide, running through the middle, one parallel to each side. Find the area of the paths and the cost of gravelling them at ₹30 per m².

   Hint

   Subtract the overlap once.

   Answer

   $3\times 60+3\times 40-9=291$ m²; cost $=₹8,730$.

7. Q7 · 3 marks · Short answerExam level

   A semicircular plot has diameter 28 m. Find the length of fencing needed and the area.

   Hint

   Perimeter πr + 2r.

   Answer

   $r=14$: perimeter $=44+28=72$ m; area $=\frac{1}{2}\times \frac{22}{7}\times 196=308$ m².

8. Q8 · 3 marks · Short answerHarder exam level

   A square lawn has a quarter-circle flower bed of radius 7 m at each corner. The lawn is 30 m on each side. Find the area of the lawn not under flower beds.

   Hint

   Four quarter-circles make one full circle.

   Answer

   $900-\frac{22}{7}\times 49=900-154=746$ m².

9. Q9 · 4 marks · Long answerHarder exam level

   How many bricks 25 cm × 12.5 cm × 7.5 cm are needed for a wall 15 m long, 0.3 m thick and 2 m high, if mortar takes up $\frac{1}{10}$ of the wall's volume?

   Hint

   Volume available for bricks = 9/10 of the wall.

   Answer

   Wall $=1500\times 30\times 200=9000000$ cm³; bricks' share $=8100000$ cm³. Brick $=2343.75$ cm³. Number $=\frac{8100000}{2343.75}=3456$.

10. Q10 · 4 marks · Long answerHarder exam level

    Water flows at 10 km/h through a rectangular channel 1.5 m wide and 0.6 m deep into a pond. How long will it take to raise the water level of a 200 m × 150 m pond by 15 cm?

    Hint

    Flow per hour = 1.5 × 0.6 × 10000 m³.

    Answer

    Needed $=200\times 150\times 0.15=4500$ m³. Flow $=9000$ m³/h. Time $=0.5$ h $=30$ minutes.

11. Q11 · 4 marks · Long answerHarder exam level

    The inner circumference of a circular track is 440 m and the track is 14 m wide. Find the area of the track and the cost of levelling it at ₹4 per m².

    Hint

    Inner r from 2πr = 440.

    Answer

    $r=70$ m, $R=84$ m; area $=\frac{22}{7}(84^2-70^2)=\frac{22}{7}\times 154\times 14=6776$ m²; cost $=₹27,104$.

12. Q12 · 4 marks · Long answerHOTS / competency

    A closed cuboidal tank is made of iron sheet 1 cm thick. Its internal dimensions are 1.2 m × 80 cm × 50 cm. Find the volume of iron used, and decide whether a tank of the same outer size but made of 2 cm sheet can hold 400 litres.

    Hint

    External dimensions add 2 cm to each.

    Answer

    External $122\times 82\times 52=520208$ cm³; internal $120\times 80\times 50=480000$ cm³; iron $=40208$ cm³. With 2 cm sheet: internal $118\times 78\times 48=441792$ cm³ $\approx 441.8$ litres, so yes, it holds 400 litres.

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. 1.State Heron's formula.Show answer

   A = √(s(s − a)(s − b)(s − c)), s = (a + b + c)/2.

2. 2.Area of a rhombus with diagonals d₁ and d₂?Show answer

   ½d₁d₂.

3. 3.Perimeter of a semicircle of radius r?Show answer

   πr + 2r.

4. 4.Area of a ring with radii R and r?Show answer

   π(R² − r²).

5. 5.Total surface area of a cuboid?Show answer

   2(lb + bh + hl).

6. 6.Surface area of an open box?Show answer

   lb + 2h(l + b).

7. 7.1 m³ in litres?Show answer

   1000 litres.

8. 8.Volume of a prism?Show answer

   Area of cross-section × length.

Quick revision checklist

The night before the exam: can you do each of these without looking?

• ☐Can you use Heron's formula without arithmetic slips?
• ☐Can you find a height from an area?
• ☐Can you find the area of a path inside or outside a plot?
• ☐Do you include the diameter in a semicircle's perimeter?
• ☐Can you count wheel revolutions?
• ☐Do you subtract 2t for thickness (t for an open top)?
• ☐Do you convert m³ to litres correctly?

Mensuration: frequently asked questions