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ICSE • Class 7 • Mathematics • Ch 17
Estimated Time: 45 Mins
Study Progress: In Progress

Mensuration

In ICSE Class 7 Mathematics, "Mensuration" provides an authoritative, computational master study guide analyzing the measurement of perimeter and area of two-dimensional plane geometric figures. This comprehensive chapter explores Perimeter and Area of Rectilinear Figures (Square: Perimeter $= 4s$, Area $= s^2$, Diagonal $= s\sqrt{2}$; Rectangle: Perimeter $= 2(l + b)$, Area $= l \times b$, Diagonal $= \sqrt{l^2 + b^2}$; Parallelogram: Area $= \text{Base} \times \text{Height} = b \times h$; Triangle: Area $= \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2}bh$; Right-angled triangle: Area $= \frac{1}{2} \times p \times b$; Equilateral triangle: Area $= \frac{\sqrt{3}}{4}s^2$), Problems on Pathways, Borders and Verandahs (Area of a uniform path running around the outside of a rectangular field, area of a path running inside, area of cross-roads running parallel to sides through the center of a field), Circles (Definition of circle, radius $r$, diameter $d = 2r$; Archimedes' constant $\pi \approx \frac{22}{7} \approx 3.1416$; Circumference of a circle $C = 2\pi r = \pi d$; Area of a circle $A = \pi r^2$; Area of a circular ring / annular pathway $A = \pi(R^2 - r^2)$; Perimeter and area of a semi-circle: Perimeter $= \pi r + 2r$, Area $= \frac{1}{2}\pi r^2$; Quadrant of a circle), and Real-World Cost Estimation (Cost of fencing $= \text{Perimeter} \times \text{Rate}$; Cost of turfing/paving/carpeting $= \text{Area} \times \text{Rate}$) aligned with the 2026–27 CISCE curriculum.

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How Did an Ancient King Test If a Jeweler Stole Pure Gold from His Royal Crown Using the Mathematics of Area and Displacement?

In the 3rd century BCE, King Hiero II of Syracuse gave a goldsmith a lump of pure gold to hammer into a sacred crown for the temple of the gods. When the crown arrived, it weighed exactly the same as the gold! But the king suspected a cunning crime: had the goldsmith melted in cheap silver and kept pure gold for himself? King Hiero asked the legendary mathematician Archimedes to prove it without melting the crown! While stepping into a full public bath in Syracuse, Archimedes noticed the water spilling over the brim. In a flash of genius, he realized that a body displaces a volume of water exactly equal to its own space, regardless of shape! He leaped from the bath, ran naked through the streets shouting "Eureka! Eureka!" ("I have found it!"). But Archimedes also proved one of the greatest formulas in human history: that the ratio of a circle's circumference to its diameter is the immortal constant $\pi$, and that a circle's area is $\pi r^2$! What is the difference between Perimeter (fence) and Area (grass)? How do you calculate the area of a garden path? Let's master mensuration.

Why This Chapter Matters

Mensuration is the practical mathematical backbone of civil engineering, architecture, interior design flooring, real estate land valuation, agricultural crop planning, and packaging manufacturing. Knowing how to compute perimeter, cross-pathways, circle boundaries, and cost rates is essential for scoring 100% in ICSE school examinations and practical adult life.

Before You Begin (Prerequisites)

  • Basic geometric shapes: Squares, rectangles, triangles, circles.
  • Units of measurement: $\text{cm}, \text{m}, \text{km}$, and conversion factors ($1\text{ m}^2 = 10,000\text{ cm}^2, 1\text{ hectare} = 10,000\text{ m}^2$).
  • Multiplication and division of fractions and decimals.

What You Will Learn (Core Objectives)

  • Calculate perimeter and area of squares, rectangles, parallelograms, and triangles.
  • Compute the area of uniform pathways constructed inside or outside rectangular plots.
  • Calculate the area of rectangular crossroads passing through the middle of a park.
  • Apply circumference ($C = 2\pi r$) and area ($A = \pi r^2$) formulas for circles and semi-circles.
  • Calculate the area of circular rings (annulus) formed between concentric circles.
  • Estimate total commercial costs of fencing (perimeter) and turfing/flooring (area).

Chapter Roadmap & Progression

1 1. Rectilinear Figures: Squares, Re...
2 2. Area of Triangles: General & Spe...
3 3. Pathways, Borders & Cross-Roads
4 4. Circles & Semi-Circles: Circumfe...

Complete Concept Guide (100% Curriculum Coverage)

1. Rectilinear Figures: Squares, Rectangles & Parallelograms

Understand
A. Square (Side $s$):
  • $\text{Perimeter} = 4 \times s$
  • $\text{Area} = s \times s = s^2 = \frac{1}{2} \times (\text{diagonal})^2$
  • $\text{Diagonal } d = s\sqrt{2}$
B. Rectangle (Length $l$, Breadth $b$):
  • $\text{Perimeter} = 2(l + b)$
  • $\text{Area} = l \times b$
  • $\text{Diagonal } d = \sqrt{l^2 + b^2}$
  • Unit Conversion: $1\text{ m}^2 = 10,000\text{ cm}^2$; $1\text{ hectare} = 10,000\text{ m}^2$; $1\text{ km}^2 = 100\text{ hectares}$.
C. Parallelogram (Base $b$, Altitude/Height $h$):

A parallelogram can be rearranged into a rectangle of the same base and height:

$$\mathbf{\text{Area of Parallelogram} = \text{Base} \times \text{Height} = b \times h}$$

2. Area of Triangles: General & Special Cases

Triangles
A. General Triangle Formula:

Any triangle is exactly half of a parallelogram with the same base and height:

$$\mathbf{\text{Area of Triangle} = \frac{1}{2} \times \text{Base} \times \text{Height} = \frac{1}{2}bh}$$
B. Special Triangle Areas:
  • Right-Angled Triangle: $\text{Area} = \frac{1}{2} \times \text{Base} \times \text{Perpendicular} = \frac{1}{2} b p$.
  • Equilateral Triangle (Side $s$): $\text{Area} = \frac{\sqrt{3}}{4}s^2$, and $\text{Height } h = \frac{\sqrt{3}}{2}s$.
  • Isosceles Triangle (Equal sides $a$, Base $b$): $\text{Area} = \frac{b}{4}\sqrt{4a^2 - b^2}$.

3. Pathways, Borders & Cross-Roads

Pathways
A. Pathway Running OUTSIDE a Rectangular Plot:

For a plot of length $l$ and breadth $b$ with a path of uniform width $w$ running around its outside:

  • Outer Length $= l + 2w$, Outer Breadth $= b + 2w$.
  • $$\mathbf{\text{Area of Path} = \text{Outer Area} - \text{Inner Area} = (l + 2w)(b + 2w) - lb}$$
B. Pathway Running INSIDE a Rectangular Plot:
  • Inner Length $= l - 2w$, Inner Breadth $= b - 2w$.
  • $$\mathbf{\text{Area of Path} = \text{Outer Area} - \text{Inner Area} = lb - (l - 2w)(b - 2w)}$$
C. Two Central Cross-Roads Running Through a Park:

Two roads of width $w$ running parallel to sides through the center of a rectangular park ($l \times b$):

$$\mathbf{\text{Area of Crossroads} = (l \times w) + (b \times w) - w^2}$$

*(The central square intersection $w^2$ must be subtracted once to avoid double-counting!)*.

4. Circles & Semi-Circles: Circumference & Annulus

Circles
A. Circle Formulas (Radius $r$, Diameter $d = 2r$):
  • Circumference ($C$): The total boundary length of the circle: $$C = 2\pi r = \pi d \quad \left( \pi \approx \frac{22}{7} \approx 3.1416 \right)$$
  • Area ($A$): The planar region enclosed by the circle: $$A = \pi r^2 = \frac{\pi d^2}{4}$$
B. Semi-Circle:
  • Perimeter of Semi-Circle: Half circumference + straight diameter boundary: $$\mathbf{\text{Perimeter} = \pi r + 2r = r(\pi + 2)}$$ *(Common error: do NOT just say $\pi r$; you MUST add the diameter $2r$ to close the shape!)*.
  • Area of Semi-Circle: $\text{Area} = \frac{1}{2}\pi r^2$.
C. Circular Ring (Annulus):

Region enclosed between two concentric circles of outer radius $R$ and inner radius $r$:

$$\mathbf{\text{Area of Ring} = \pi R^2 - \pi r^2 = \pi(R^2 - r^2) = \pi(R + r)(R - r)}$$

Key Formulas, Identities & Theorems

Circumference & Area of Circle
$$C = 2\pi r, \quad A = \pi r^2$$
pi is approx 22/7 or 3.1416.
Area of Annular Circular Ring
$$A_{\text{ring}} = \pi(R^2 - r^2) = \pi(R + r)(R - r)$$
Difference between outer and inner concentric circle areas.

Mensuration: Shapes, Pathways & Circular Geometry

Mensuration: Areas, Perimeters, Pathways & Circles RECTILINEAR FORMULAS • Rectangle (l × b): P = 2(l + b) • A = l × b Diagonal = √(l2 + b2) • Square (Side s): P = 4s • A = s2 = ½ d2 • Parallelogram & Triangle: Parallelogram: A = b × h Triangle: A = ½ × b × h • 1 Hectare = 10,000 m2 PATHWAYS & BORDERS • Outer Path (Width w): Outer dims: (l + 2w) × (b + 2w) Area = (l+2w)(b+2w) - lb • Inner Path (Width w): Inner dims: (l - 2w) × (b - 2w) Area = lb - (l-2w)(b-2w) • Crossroads in Park: Area = lw + bw - w2 (Subtract center square overlap w2) CIRCULAR MEASURES • Circle (Radius r): Circumference C = 2πr = πd Area A = πr2 • Semi-Circle Boundary Trap: Perimeter = πr + 2r (MUST add straight diameter 2r!) Area = ½ πr2 • Circular Ring (Annulus): Area = π(R2 - r2) FENCING COST = PERIMETER × RATE • TURFING/PAVING COST = AREA × RATE

Chapter Summary & 10 Key Takeaways

Takeaway 1
Perimeter is the total boundary distance; Area is the two-dimensional surface region enclosed.
Takeaway 2
Rectangle: $P = 2(l + b)$ and $A = l \times b$; Square: $P = 4s$ and $A = s^2$.
Takeaway 3
Area of a parallelogram equals $\text{Base} \times \text{Height}$.
Takeaway 4
Area of any triangle equals $\frac{1}{2} \times \text{Base} \times \text{Height}$.
Takeaway 5
Unit conversions: $1\text{ m}^2 = 10,000\text{ cm}^2$; $1\text{ hectare} = 10,000\text{ m}^2$.
Takeaway 6
Outer pathway area around a rectangle: $(l + 2w)(b + 2w) - lb$.
Takeaway 7
Crossroads area running through a park: $lw + bw - w^2$ (subtract central square intersection).
Takeaway 8
Circumference of a circle: $C = 2\pi r = \pi d$; Area: $A = \pi r^2$ (where $\pi \approx 22/7$).
Takeaway 9
Perimeter of a semi-circle is $\pi r + 2r$ (includes the straight diameter).
Takeaway 10
Area of an annular circular ring: $\pi(R^2 - r^2) = \pi(R + r)(R - r)$.

Check Your Understanding (Diagnostic Practice Questions)

Diagnostic questions testing core conceptual clarity. Answers are hidden initially — solve each problem first, then click to reveal the step-by-step verified solution.

1
A rectangular park is $65\text{ m}$ long and $40\text{ m}$ wide. A path $2.5\text{ m}$ wide is constructed all around the outside of the park. Find the area of the path and the cost of gravelling it at $\text{Rs. } 20\text{ per m}^2$.
Reveal Answer & Explanation
Answer: Inner dimensions of park: $l = 65\text{ m}, b = 40\text{ m}$.
$$\text{Area of park} = 65 \times 40 = 2,600\text{ m}^2$$
Width of path $w = 2.5\text{ m}$.
Outer dimensions including path:
$$\text{Outer length } L = 65 + 2(2.5) = 65 + 5 = 70\text{ m}$$
$$\text{Outer breadth } B = 40 + 2(2.5) = 40 + 5 = 45\text{ m}$$
$$\text{Total outer area} = 70 \times 45 = 3,150\text{ m}^2$$
Step 1: Calculate Area of Path:
$$\text{Area of path} = 3,150 - 2,600 = \mathbf{550\text{ m}^2}$$
Step 2: Calculate Cost of Gravelling:
$$\text{Total Cost} = 550 \times 20 = \mathbf{\text{Rs. } 11,000}$$.
Outer dimensions: $70 \times 45 = 3150$. Inner area: $65 \times 40 = 2600$. Path area: $550\text{ m}^2$. Multiply by 20.
2
The circumference of a circular plot is $132\text{ meters}$. Find its diameter and its area (use $\pi = \frac{22}{7}$).
Reveal Answer & Explanation
Answer:

Circumference $C = 2\pi r = 132\text{ m}$.

$$2 \times \frac{22}{7} \times r = 132$$


$$\frac{44}{7} r = 132 \implies r = \frac{132 \times 7}{44} = 3 \times 7 = 21\text{ meters}$$


• Diameter: $d = 2r = 2 \times 21 = \mathbf{42\text{ meters}}$.
• Area: $A = \pi r^2 = \frac{22}{7} \times 21 \times 21 = 22 \times 3 \times 21 = 66 \times 21 = \mathbf{1,386\text{ m}^2}$.


Find radius: $r = (132 \times 7) / 44 = 21\text{ m}$. Diameter is $42\text{ m}$. Area is $(22/7) \times 21^2 = 1,386\text{ m}^2$.
3
A wire is in the shape of a rectangle of length $40\text{ cm}$ and breadth $22\text{ cm}$. If the same wire is rebent into the shape of a square, what will be the measure of each side? Which shape encloses more area?
Reveal Answer & Explanation
Answer:

Length of wire equals the perimeter of the rectangle:

$$\text{Perimeter of rectangle} = 2(l + b) = 2(40 + 22) = 2(62) = 124\text{ cm}$$


When rebent into a square, Perimeter of square $= 124\text{ cm}$:

$$4s = 124 \implies s = \frac{124}{4} = \mathbf{31\text{ cm}}$$


Each side of the square is $31\text{ cm}$.
• Compare Areas:

$$\text{Area of rectangle} = l \times b = 40 \times 22 = 880\text{ cm}^2$$


$$\text{Area of square} = s^2 = 31 \times 31 = 961\text{ cm}^2$$


Since $961 > 880$, the Square encloses more area (by $81\text{ cm}^2$).


Perimeter $= 124\text{ cm}$. Square side $= 124/4 = 31\text{ cm}$. Square area ($961$) is greater than rectangle area ($880$).
4
Find the perimeter of a semi-circular plate of radius $14\text{ cm}$ (take $\pi = \frac{22}{7}$).
Reveal Answer & Explanation
Answer:

Perimeter of a semi-circle consists of the curved arc plus the straight diameter:

$$\text{Perimeter} = \pi r + 2r$$


$$= \left( \frac{22}{7} \times 14 \right) + (2 \times 14)$$


$$= (22 \times 2) + 28 = 44 + 28 = \mathbf{72\text{ cm}}$$

.
The perimeter of the semi-circular plate is $72\text{ cm}$.


Do not forget the diameter! Formula is $\pi r + 2r = 44 + 28 = 72\text{ cm}$.
5
Two crossroads, each of width $3\text{ m}$, run at right angles through the center of a rectangular park of length $70\text{ m}$ and breadth $45\text{ m}$, parallel to its sides. Find the area of the crossroads.
Reveal Answer & Explanation
Answer: Length $l = 70\text{ m}, b = 45\text{ m}$, road width $w = 3\text{ m}$.
• Area of road parallel to length $= l \times w = 70 \times 3 = 210\text{ m}^2$
• Area of road parallel to breadth $= b \times w = 45 \times 3 = 135\text{ m}^2$
• Area of common central intersection $= w \times w = 3 \times 3 = 9\text{ m}^2$
$$\text{Total Area of Crossroads} = 210 + 135 - 9 = 345 - 9 = \mathbf{336\text{ m}^2}$$.
Apply $lw + bw - w^2 = (70 \times 3) + (45 \times 3) - 3^2 = 210 + 135 - 9 = 336\text{ m}^2$.
6
A circular track has an inner radius of $28\text{ m}$ and an outer radius of $35\text{ m}$. Find the area of the track.
Reveal Answer & Explanation
Answer:

The track forms an Annular Circular Ring with $R = 35\text{ m}$ and $r = 28\text{ m}$:

$$\text{Area of track} = \pi(R^2 - r^2) = \pi(R + r)(R - r)$$


$$= \frac{22}{7} \times (35 + 28) \times (35 - 28)$$


$$= \frac{22}{7} \times 63 \times 7 = 22 \times 63 = \mathbf{1,386\text{ m}^2}$$

.
The area of the track is $1,386\text{ m}^2$.


Use difference of squares: $\pi(R+r)(R-r) = (22/7) \times 63 \times 7 = 22 \times 63 = 1,386\text{ m}^2$.
7
The area of a parallelogram is $338\text{ cm}^2$. If its altitude is twice the corresponding base, find the base and the altitude.
Reveal Answer & Explanation
Answer: Let the base be $b\text{ cm}$.
Then altitude $h = 2b\text{ cm}$.
Area of a parallelogram $= b \times h$:
$$b \times 2b = 338$$
$$2b^2 = 338 \implies b^2 = \frac{338}{2} = 169$$
$$b = \sqrt{169} = \mathbf{13\text{ cm}}$$
$$\text{Altitude } h = 2(13) = \mathbf{26\text{ cm}}$$.
Check: $13 \times 26 = 338\text{ cm}^2$. Correct!
Set $b \times 2b = 338 \implies 2b^2 = 338 \implies b^2 = 169 \implies b = 13\text{ cm}$. Height is $26\text{ cm}$.
8
How many times will the wheel of a car rotate in a journey of $11\text{ km}$, if the diameter of the wheel is $70\text{ cm}$?
Reveal Answer & Explanation
Answer: Step 1: Distance covered in one single rotation equals the Circumference of the wheel:
$$d = 70\text{ cm} \implies r = 35\text{ cm}$$
$$\text{Circumference } C = 2\pi r = 2 \times \frac{22}{7} \times 35 = 2 \times 22 \times 5 = 220\text{ cm} = 2.2\text{ m}$$
Step 2: Total journey distance in meters:
$$\text{Total Distance } D = 11\text{ km} = 11 \times 1,000\text{ m} = 11,000\text{ m}$$
Step 3: Number of rotations:
$$\text{Rotations} = \frac{\text{Total Distance}}{\text{Circumference}} = \frac{11,000}{2.2} = \frac{110,000}{22} = \mathbf{5,000 \text{ rotations}}$$.
Circumference is $220\text{ cm} = 2.2\text{ m}$. Distance is $11,000\text{ m}$. Rotations $= 11,000 / 2.2 = 5,000$.
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