Area and Volume Formulas for Every Shape

Mensuration questions are some of the most reliable marks in Matric, O Level and FSc papers, provided you know the formulas and use units carefully. This guide collects the area and volume formulas you will actually need, explains where they come from, and shows how to apply them step by step.

Area Formulas for 2D Shapes

Area measures the space inside a flat shape, in square units such as \( \text{cm}^2 \) or \( \text{m}^2 \).

Rectangles, squares and parallelograms

• Rectangle: \( A = l \times w \)
• Square: \( A = s^2 \)
• Parallelogram: \( A = b \times h \), where \( h \) is the perpendicular height, not the slanted side.

Triangles

• Basic formula: \( A = \frac{1}{2} b h \)
• Using two sides and the included angle: \( A = \frac{1}{2} ab \sin C \)
• Heron's formula, when all three sides are known:

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

Trapezium and rhombus

• Trapezium: \( A = \frac{1}{2}(a + b)h \), where \( a \) and \( b \) are the parallel sides.
• Rhombus (or kite): \( A = \frac{1}{2} d_1 d_2 \), using the two diagonals.

Circles and sectors

• Circle area: \( A = \pi r^2 \); circumference: \( C = 2\pi r \)
• Sector with angle \( \theta \) in degrees: \( A = \frac{\theta}{360^\circ} \pi r^2 \)
• Sector with angle \( \theta \) in radians: \( A = \frac{1}{2} r^2 \theta \), arc length \( s = r\theta \)

Volume and Surface Area of 3D Shapes

Volume measures the space a solid occupies (in \( \text{cm}^3 \), \( \text{m}^3 \)). Surface area is the total area of all its faces.

Prisms: cuboid, cube and cylinder

Any prism has volume equal to cross-sectional area × length.

• Cuboid: \( V = lwh \), \( \text{SA} = 2(lw + lh + wh) \)
• Cube: \( V = s^3 \), \( \text{SA} = 6s^2 \)
• Cylinder: \( V = \pi r^2 h \), curved surface \( = 2\pi r h \), total \( \text{SA} = 2\pi r(r + h) \)

Pyramids and cones

A pyramid or cone holds exactly one third of the matching prism or cylinder.

• Pyramid: \( V = \frac{1}{3} \times \text{base area} \times h \)
• Cone: \( V = \frac{1}{3}\pi r^2 h \), curved surface \( = \pi r l \), total \( \text{SA} = \pi r(r + l) \)

Here \( l \) is the slant height, found using Pythagoras: \( l = \sqrt{r^2 + h^2} \).

Spheres and hemispheres

• Sphere: \( V = \frac{4}{3}\pi r^3 \), \( \text{SA} = 4\pi r^2 \)
• Hemisphere: \( V = \frac{2}{3}\pi r^3 \), curved surface \( = 2\pi r^2 \), total \( \text{SA} = 3\pi r^2 \) (including the flat circular face)

Worked Example 1: A Cylindrical Water Tank

Question: A cylindrical tank has radius 7 m and height 10 m. Find its volume and total surface area. Use \( \pi = \frac{22}{7} \).

Solution:

• Step 1: Volume formula.

$$ V = \pi r^2 h = \frac{22}{7} \times 7^2 \times 10 $$

• Step 2: Simplify.

$$ V = \frac{22}{7} \times 49 \times 10 = 22 \times 7 \times 10 = 1540 \text{ m}^3 $$

• Step 3: Total surface area.

$$ \text{SA} = 2\pi r(r + h) = 2 \times \frac{22}{7} \times 7 \times (7 + 10) $$

• Step 4: Simplify.

$$ \text{SA} = 2 \times 22 \times 17 = 748 \text{ m}^2 $$

So the tank holds \( 1540 \text{ m}^3 \) (that is, 1,540,000 litres, since \( 1 \text{ m}^3 = 1000 \) L) and has a surface area of \( 748 \text{ m}^2 \).

Worked Example 2: A Cone

Question: A cone has base radius 3 cm and vertical height 4 cm. Find its volume and total surface area in terms of \( \pi \).

Solution:

• Step 1: Volume.

$$ V = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi (3)^2(4) = \frac{1}{3}\pi(36) = 12\pi \text{ cm}^3 $$

• Step 2: Slant height using Pythagoras.

$$ l = \sqrt{r^2 + h^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \text{ cm} $$

• Step 3: Total surface area.

$$ \text{SA} = \pi r(r + l) = \pi(3)(3 + 5) = 24\pi \text{ cm}^2 $$

So \( V = 12\pi \approx 37.70 \text{ cm}^3 \) and \( \text{SA} = 24\pi \approx 75.40 \text{ cm}^2 \).

Worked Example 3: Area of a Triangle with Heron's Formula

Question: Find the area of a triangle with sides 13 cm, 14 cm and 15 cm.

Solution:

• Step 1: Semi-perimeter.

$$ s = \frac{13 + 14 + 15}{2} = 21 $$

• Step 2: Substitute.

$$ A = \sqrt{21(21-13)(21-14)(21-15)} = \sqrt{21 \times 8 \times 7 \times 6} $$

• Step 3: Simplify.

$$ A = \sqrt{7056} = 84 \text{ cm}^2 $$

Common Mistakes

• Using slant height instead of vertical height in the volume of a cone or pyramid.
• Using the diameter as the radius. Always halve the diameter first.
• Mixing units. Convert everything to the same unit before calculating. Remember \( 1 \text{ m}^2 = 10{,}000 \text{ cm}^2 \) and \( 1 \text{ m}^3 = 1{,}000{,}000 \text{ cm}^3 \).
• Forgetting the flat face of a hemisphere or the base of a cone when the question asks for total surface area.
• Rounding too early. Keep \( \pi \) exact until the last step.

Tips

• Learn the prism rule: volume = cross-section × length. It covers cuboids, cylinders and triangular prisms in one idea.
• Remember "one third" for anything that comes to a point (pyramids and cones).
• For composite solids, such as a cylinder topped with a hemisphere, split the shape into parts, find each volume, and add.
• Check whether the paper wants \( \pi = 3.142 \), \( \frac{22}{7} \) or an answer "in terms of \( \pi \)".

Key Takeaways

• Area is measured in square units; volume in cubic units.
• Prism volume = cross-sectional area × length; pyramid and cone volume = one third of that.
• Sphere: \( V = \frac{4}{3}\pi r^3 \) and \( \text{SA} = 4\pi r^2 \).
• Use Pythagoras to find the slant height of a cone when only \( r \) and \( h \) are given.
• Consistent units and late rounding prevent most mensuration errors.

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