Mensuration: Areas and Volumes

🟢 Lite — Quick Review (1h–1d)

Rapid summary for last-minute revision before your exam.

Mensuration is the branch of mathematics concerned with calculating geometric measurements: lengths, areas, and volumes.

2D Shapes — Area Formulas:

Shape	Formula
Rectangle	$A = l \times w$
Square	$A = s^2$
Triangle	$A = \frac{1}{2} \times base \times height$
Parallelogram	$A = b \times h$
Trapezium	$A = \frac{1}{2}(a+b) \times h$
Circle	$A = \pi r^2$
Ellipse	$A = \pi ab$

Perimeter Formulas:

• Rectangle: $P = 2(l + w)$
• Square: $P = 4s$
• Triangle: $P = a + b + c$
• Circle (Circumference): $C = 2\pi r = \pi d$

3D Shapes — Volume Formulas:

Shape	Formula
Cube	$V = s^3$
Cuboid	$V = l \times w \times h$
Cylinder	$V = \pi r^2 h$
Cone	$V = \frac{1}{3}\pi r^2 h$
Sphere	$V = \frac{4}{3}\pi r^3$
Pyramid	$V = \frac{1}{3} \times \text{Base Area} \times h$
Frustum	$V = \frac{1}{3}\pi h(a^2 + ab + b^2)$

⚡ WAEC Tip: Units matter! Area is in square units (cm², m²). Volume is in cubic units (cm³, m³). When comparing areas of different shapes, make sure you’re using the same units.

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🟡 Standard — Regular Study (2d–2mo)

For students who want genuine understanding.

Surface Area of 3D Shapes:

Right Prism (uniform cross-section): $$SA = 2(\text{Base Area}) + (\text{Perimeter of base}) \times \text{height}$$

Cylinder: $$SA = 2\pi r^2 + 2\pi r h = 2\pi r(r + h)$$

Cone: $$SA = \pi r^2 + \pi r l \quad \text{where } l = \sqrt{r^2 + h^2} \text{ (slant height)}$$

Sphere: $$SA = 4\pi r^2$$

Frustum of a Cone: $$SA = \pi (R + r) \sqrt{(R-r)^2 + h^2} + \pi R^2 + \pi r^2$$ Where R = larger radius, r = smaller radius, h = perpendicular height

Practical Applications:

Problem 1: Find the total surface area of a cylinder with radius 7 cm and height 10 cm. $$SA = 2\pi(7)^2 + 2\pi(7)(10) = 98\pi + 140\pi = 238\pi \approx 747.7 \text{ cm}^2$$

Problem 2: Find the volume of a cone with base radius 5 cm and vertical height 12 cm. $$V = \frac{1}{3}\pi(5)^2(12) = \frac{1}{3}\pi(25)(12) = 100\pi \approx 314.2 \text{ cm}^3$$

Problem 3: A frustum has radii 8 cm and 3 cm, and height 7 cm. Find its volume. $$V = \frac{1}{3}\pi(7)[8^2 + (8)(3) + 3^2] = \frac{1}{3}\pi(7)[64 + 24 + 9] = \frac{1}{3}\pi(7)(97) = \frac{679\pi}{3} \approx 711.4 \text{ cm}^3$$

Problem-Solving Strategies:

1. Convert all measurements to the same unit before calculating
2. Identify the shape — sometimes a complex shape can be broken into simpler ones
3. For combined shapes — add or subtract volumes as needed
4. For similar shapes — areas scale as square of ratio, volumes as cube

Scale Factor Problems:

If linear scale factor is $k$:

• Area scale factor = $k^2$
• Volume scale factor = $k^3$

Problem: Two similar cones have heights 6 cm and 9 cm. If the smaller has volume 24 cm³, find the volume of the larger. $$k = \frac{9}{6} = \frac{3}{2}$$ $$\text{Volume scale factor} = \left(\frac{3}{2}\right)^3 = \frac{27}{8}$$ $$V_{\text{large}} = 24 \times \frac{27}{8} = 81 \text{ cm}^3$$

⚡ Common Student Mistakes: Forgetting to use the slant height (not vertical height) when calculating curved surface area of a cone. Mixing up radius and diameter. Forgetting to halve the height for the frustum formula.

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🔴 Extended — Deep Study (3mo+)

Comprehensive theory for thorough preparation.

Circle-Related Formulas:

Arc Length: $$l = \frac{\theta}{360°} \times 2\pi r \quad \text{(where } \theta \text{ is in degrees)}$$

Sector Area: $$A = \frac{\theta}{360°} \times \pi r^2 = \frac{1}{2}r^2\theta \quad \text{(where } \theta \text{ is in radians)}$$

Segment Area (area bounded by chord and arc): $$A_{\text{segment}} = A_{\text{sector}} - A_{\text{triangle}}$$

Problem: Find the area of a sector with radius 6 cm and angle 60°. $$A = \frac{60}{360} \times \pi \times 36 = \frac{1}{6} \times 36\pi = 6\pi \approx 18.85 \text{ cm}^2$$

Problem: Find the area of the segment formed by a chord in a circle of radius 10 cm, subtending 90° at the centre.

Sector area = $\frac{90}{360} \times \pi \times 100 = 25\pi$ Triangle area = $\frac{1}{2} \times 10 \times 10 = 50$ (right isosceles triangle) Segment area = $25\pi - 50 \approx 28.54$ cm²

Hemispheres:

A hemisphere is half a sphere: $$V = \frac{2}{3}\pi r^3$$ $$SA = 3\pi r^2 \quad \text{(including base)}$$ $$SA = 2\pi r^2 \quad \text{(curved surface only)}$$

Problem: A hemispherical bowl has radius 14 cm. Find its volume and total surface area. $$V = \frac{2}{3}\pi(14)^3 = \frac{2}{3}\pi \times 2744 = \frac{5488\pi}{3} \approx 5749.3 \text{ cm}^3$$ $$SA = 3\pi(14)^2 = 3\pi \times 196 = 588\pi \approx 1847.3 \text{ cm}^2$$

Composite Solids:

Break complex shapes into simpler ones:

Problem: A solid consists of a cylindrical body with a conical top. The cylinder has radius 5 cm and height 12 cm. The cone has the same radius and height 8 cm. Find the total volume.

$$V_{\text{cyl}} = \pi(5)^2(12) = 300\pi$$ $$V_{\text{cone}} = \frac{1}{3}\pi(5)^2(8) = \frac{200\pi}{3}$$ $$V_{\text{total}} = 300\pi + \frac{200\pi}{3} = \frac{900\pi + 200\pi}{3} = \frac{1100\pi}{3} \approx 1151.9 \text{ cm}^3$$

Density Problems:

$$\text{Density} = \frac{\text{Mass}}{\text{Volume}}$$

Problem: A metal sphere has radius 6 cm and density 8.5 g/cm³. Find its mass. $$V = \frac{4}{3}\pi(6)^3 = \frac{4}{3}\pi \times 216 = 288\pi \text{ cm}^3$$ $$m = 8.5 \times 288\pi \approx 7696.9 \text{ g} \approx 7.7 \text{ kg}$$

Conversion Between Units:

Conversion	Factor
1 m = 100 cm	1 m² = 10,000 cm²
1 km = 1000 m	1 hectare = 10,000 m²
1 L = 1000 mL	1 L = 1000 cm³
1 kg = 1000 g	1 tonne = 1000 kg

Problem: Convert 2.5 m³ to cm³. $$2.5 \times 1,000,000 = 2,500,000 \text{ cm}^3$$

Practical WAEC Problems:

Problem: A water tank is in the shape of a cylinder with a cone on the bottom. The cylinder has diameter 4 m and height 3 m. The cone has height 1.5 m. Find the total volume in litres.

Cylinder radius = 2 m $$V_{\text{cyl}} = \pi(2)^2(3) = 12\pi \text{ m}^3$$ $$V_{\text{cone}} = \frac{1}{3}\pi(2)^2(1.5) = 2\pi \text{ m}^3$$ $$V_{\text{total}} = 14\pi \text{ m}^3 = 14\pi \times 1000 \text{ L} \approx 43,982 \text{ L}$$

Problem: A rectangular tank is 80 cm long, 60 cm wide, and 40 cm deep. If 48 litres of water are poured in, what is the depth of water?

Tank base area = $80 \times 60 = 4800$ cm² = 0.48 m² $$48 \text{ L} = 0.048 \text{ m}^3$$ $$\text{Depth} = \frac{0.048}{0.48} = 0.1 \text{ m} = 10 \text{ cm}$$

Earthwork and Trench Problems:

Problem: A trench is 50 m long, 2 m wide, and 1.5 m deep. Earth is removed and spread to form a heap with a uniform depth of 0.2 m over an area of 100 m². Find the dimensions of the heap.

Volume of trench = $50 \times 2 \times 1.5 = 150$ m³ This volume is spread to depth 0.2 m: $$\text{Area} = \frac{150}{0.2} = 750 \text{ m}^2$$ But we already have 100 m²… wait, this means it would be 7.5 m deep over the 100 m² area, which doesn’t make sense.

Actually, the problem might be asking for the dimensions of the heap assuming it forms a cone or something similar. More likely: 150 m³ spread over 100 m² gives: $$\text{New depth} = \frac{150}{100} = 1.5 \text{ m}$$

⚡ WAEC Examination Patterns: Calculate areas and volumes of standard shapes. Solve problems involving combined shapes. Apply similarity (scale factors for area and volume). Convert between different units. Solve density problems. Find surface areas of cones, spheres, and cylinders. Solve practical problems involving water tanks, earthworks, and storage.