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Mensuration for Composite Shapes

General Mathematics

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Area Of Study

Geometry and measurement

Unit

Unit 4: General Mathematics Unit 4

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Mensuration for Composite Shapes

General Mathematics
01 May 2026

Mensuration Formulas for Composite Shapes and Solids

What is Mensuration?

Mensuration is the branch of mathematics concerned with measuring geometric figures — lengths, areas, and volumes. For composite shapes, the strategy is to decompose the figure into standard shapes, compute each component, then add (or subtract).

Composite Area Strategies

Addition: when the shape is a combination of non-overlapping standard shapes.

Subtraction: when a region is removed from a larger shape (e.g., a hole, a cutout).

Common Standard Shapes

Shape Area formula
Rectangle \(A = lw\)
Triangle \(A = \tfrac{1}{2}bh\)
Circle \(A = \pi r^2\)
Semicircle \(A = \tfrac{1}{2}\pi r^2\)
Trapezium \(A = \tfrac{1}{2}(a+b)h\)
Sector \(A = \tfrac{\theta}{360}\pi r^2\)

Worked Example — Composite Area

A running track consists of a rectangle (100 m × 60 m) with semicircles on each short end.

\[A_{\text{rect}} = 100 \times 60 = 6000 \text{ m}^2\]
\[A_{\text{two semicircles}} = \pi r^2 = \pi (30)^2 = 900\pi \approx 2827 \text{ m}^2\]
\[A_{\text{total}} \approx 6000 + 2827 = 8827 \text{ m}^2\]

Perimeter (track length):

\[P = 2 \times 100 + 2\pi(30) = 200 + 60\pi \approx 388.5 \text{ m}\]

Composite Volume Strategies

Decompose into prisms, cylinders, pyramids, cones, hemispheres.

Worked Example — Composite Solid

A swimming pool is 25 m long, 10 m wide, with a shallow end depth of 1.0 m and deep end depth of 2.2 m.

Model as a trapezoidal prism (trapezium cross-section):

\[V = A_{\text{trapezium}} \times \text{width} = \tfrac{1}{2}(1.0 + 2.2)(25) \times 10 = \tfrac{1}{2}(3.2)(25)(10) = 400 \text{ m}^3\]

Capacity: \(400 \times 1000 = 400{,}000 \text{ L}\).

Surface Area of Composite Solids

For surface area: sum the areas of all exposed faces. Internal faces (where parts join) are not included.

Worked Example — Closed Cylinder Topped by Hemisphere

Radius \(r = 3\) m, cylinder height \(h = 5\) m.

\[SA = \underbrace{2\pi r h}_{\text{cylinder wall}} + \underbrace{\pi r^2}_{\text{base}} + \underbrace{2\pi r^2}_{\text{hemisphere}}\]
\[= 2\pi(3)(5) + \pi(9) + 2\pi(9) = 30\pi + 9\pi + 18\pi = 57\pi \approx 179.1 \text{ m}^2\]

(The top circle of the cylinder is replaced by the hemisphere — do not count it.)

Approach Checklist

  1. Sketch and label the composite figure.
  2. Identify each component shape.
  3. Note which faces are internal (not part of SA).
  4. Apply formulas systematically.
  5. Sum components; state answer with units.

REMEMBER: The circular top of a cylinder that meets a hemisphere is an internal face — it is not part of the surface area. Only external faces contribute.

EXAM TIP: Show the decomposition explicitly: write each component’s formula and calculated value before adding. This earns method marks even if arithmetic errors occur.

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