Right Pyramids, Right Cones and Spheres Revision Notes for Grade 11 NSC Matric Mathematics

Right Pyramids, Right Cones and Spheres

Understanding the basic shapes

Pyramids are three-dimensional geometric solids that have a polygon as their base and triangular faces that meet at a single point called the apex. The key feature of pyramids is that their sides are not perpendicular to the base.

A pyramid becomes a right pyramid when the line connecting the apex to the centre of the base is perpendicular to the base. This perpendicular line represents the true height of the pyramid.

chatImportant

A "right" pyramid doesn't mean it's standing upright - it means the height line from the apex to the base center is perpendicular to the base. This is a crucial distinction for calculations.

Cones work similarly to pyramids, but instead of having a polygon as their base, they have a circular base. The curved surface of a cone connects the circular base to the apex point.

Spheres are perfectly round three-dimensional solids that look identical from any direction. Every point on the surface of a sphere is exactly the same distance from the centre.

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The names of pyramids come from the shape of their base - triangular pyramids have triangular bases, while square pyramids have square bases.

Surface area calculations

Surface area represents the total area of all the surfaces that make up a three-dimensional shape. For pyramids and cones, this includes the base area plus the area of all the triangular faces or curved surface.

Square pyramid surface area

For a square pyramid with base side length b and slant height hs:

Surface area = area of base + area of triangular sides

Surface area = $b^2 + 4(\frac{1}{2}bh_s)$

This simplifies to: Surface area = $b(b + 2h_s)$

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The base area is simply $b^2$ since it's a square. The four triangular faces each have area $\frac{1}{2}bh_s$ , so we multiply by 4.

Triangular pyramid surface area

For a triangular pyramid with base width b, base height hb, and slant height hs:

Surface area = area of base + area of triangular sides

Surface area = $\frac{1}{2}b \times h_b + 3(\frac{1}{2}b \times h_s)$

This simplifies to: Surface area = $\frac{1}{2}b(h_b + 3h_s)$

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The triangular base has area $\frac{1}{2}bh_b$ , and there are three identical triangular faces each with area $\frac{1}{2}bh_s$ .

Right cone surface area

For a right cone with radius r and slant height h:

Surface area = area of base + area of curved surface

Surface area = $\pi r^2 + \frac{1}{2} \times 2\pi rh$

This simplifies to: Surface area = $\pi r(r + h)$

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The circular base has area $\pi r^2$ , and the curved surface area equals $\pi rh$ (this comes from "unrolling" the curved surface).

Sphere surface area

For a sphere with radius r:

Surface area = $4\pi r^2$

This formula represents the total area of the curved surface that encloses the sphere.

Volume calculations

Volume measures the amount of three-dimensional space contained within a shape. For pyramids and cones, the volume formula follows a consistent pattern.

chatImportant

For volume calculations, you must use the perpendicular height (H), not the slant height. The perpendicular height is the straight-line distance from the apex to the base, measured at a 90° angle to the base.

Square pyramid volume

For a square pyramid with base side length b and height H:

Volume = $\frac{1}{3} \times$ area of base $\times$ height

Volume = $\frac{1}{3} \times b^2 \times H$

The height H is the perpendicular distance from the apex to the base, not the slant height.

Triangular pyramid volume

For a triangular pyramid with base area and height H:

Volume = $\frac{1}{3} \times$ area of base $\times$ height

Volume = $\frac{1}{3} \times \frac{1}{2}bh \times H$

Where $\frac{1}{2}bh$ represents the area of the triangular base.

Right cone volume

For a right cone with radius r and height H:

Volume = $\frac{1}{3} \times$ area of base $\times$ height

Volume = $\frac{1}{3} \times \pi r^2 \times H$

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Notice that all pyramids and cones use the same $\frac{1}{3}$ factor in their volume calculations.

Sphere volume

For a sphere with radius r:

Volume = $\frac{4}{3}\pi r^3$

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The sphere is the only shape in this group that doesn't use the $\frac{1}{3}$ factor - instead it uses $\frac{4}{3}$ .

Worked example: SALT telescope building

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Worked Example: SALT Telescope Building Calculations

Question: The Southern African Large Telescope (SALT) is housed in a cylindrical building with a domed roof in the shape of a hemisphere. The height of the building wall is 17 m and the diameter is 26 m.

Part 1: Calculate the total surface area

The building consists of a cylinder with a hemisphere on top. The radius is 13 m (half the diameter).

Total surface area = area of dome + area of cylinder walls

For the hemisphere: Surface area = $\frac{1}{2}(4\pi r^2) = 2\pi r^2$

For the cylinder walls: Surface area = $2\pi rh$ (we don't include the top since it's covered by the dome)

Surface area = $2\pi(13)^2 + 2\pi(13)(17)$
Surface area = $2\pi(169) + 2\pi(221)$
Surface area = $2\pi(169 + 221) = 2\pi(390) = 780\pi$
Surface area ≈ 2450 m²

Part 2: Calculate the total volume

Total volume = volume of hemisphere + volume of cylinder

Hemisphere volume = $\frac{1}{2} \times \frac{4}{3}\pi r^3 = \frac{2}{3}\pi r^3$

Cylinder volume = $\pi r^2 h$

Volume = $\frac{2}{3}\pi(13)^3 + \pi(13)^2(17)$
Volume = $\frac{2}{3}\pi(2197) + \pi(169)(17)$
Volume = $\frac{2}{3}\pi(2197) + \pi(2873)$
Volume ≈ 9543 m³

Key formulas summary

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Surface area formulas:

Square pyramid: $b(b + 2h_s)$
Triangular pyramid: $\frac{1}{2}b(h_b + 3h_s)$
Right cone: $\pi r(r + h)$
Sphere: $4\pi r^2$

Volume formulas:

Square pyramid: $\frac{1}{3}b^2H$
Triangular pyramid: $\frac{1}{3} \times \frac{1}{2}bhH$
Right cone: $\frac{1}{3}\pi r^2H$
Sphere: $\frac{4}{3}\pi r^3$

Remember!

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Key Points to Remember:

All pyramids and cones use $\frac{1}{3}$ in their volume formula, while spheres use $\frac{4}{3}$
Surface area always equals base area plus lateral (side) surface area
A "right" pyramid means the height is perpendicular to the base
For surface area calculations, you need the slant height, but for volume you need the perpendicular height
When working with composite shapes, break them down into familiar shapes and add their areas or volumes together

Right Pyramids, Right Cones and Spheres (Grade 11 NSC Matric Mathematics): Revision Notes