The pyramid: its parts, its height and surface area

Space geometry and vectors, grade 12 (4 units) Lesson 5 of 20Learning

The pyramid: its parts, its height and surface area

The pyramid

A pyramid is a solid whose base is a polygon, and a point off the base plane – the apex – is joined by an edge to every vertex of the base.

The lateral faces – triangles, such as $\displaystyle SAB$ .
The edges – the base edges ( $\displaystyle AB$ , $\displaystyle BC$ …) and the side edges ( $\displaystyle SA$ , $\displaystyle SB$ …).
The height – a perpendicular segment from the apex to the base plane, such as $\displaystyle SH$ . Its foot $\displaystyle H$ may also be outside the base.
A pyramid is named by its base: triangular, quadrilateral …

The right pyramid

A right pyramid is a pyramid whose side edges are all equal.

The height of a right pyramid: its foot $\displaystyle O$ is the centre of the circle through the base’s vertices.

The reason: $\displaystyle SO$ is perpendicular to the base, so the triangles $\displaystyle SOA$ , $\displaystyle SOB$ , $\displaystyle SOC$ … are right-angled at $\displaystyle O$ . They have equal hypotenuses ( $\displaystyle {SA}={SB}={SC}$ …) and the common leg $\displaystyle SO$ , so they are congruent, and $\displaystyle {OA}={OB}={OC}$ …
A base that is a rectangle or a square: $\displaystyle O$ is where the diagonals meet.
A base that is an equilateral triangle: $\displaystyle O$ is where the medians meet, and $\displaystyle OA$ is $\displaystyle \frac{2}{3}$ of a median.
A base that is a right-angled triangle: $\displaystyle O$ is the midpoint of the hypotenuse.

The lateral area and the surface area

The lateral area $\displaystyle M$ is the sum of the areas of the lateral faces, and the surface area is the lateral area and the base:

\displaystyle S

\displaystyle =

\displaystyle M+B

A face’s area is worked out from its height, such as $\displaystyle SM$ in the face $\displaystyle SBC$ . In a right pyramid, $\displaystyle SM$ is the hypotenuse of the right-angled triangle $\displaystyle SOM$ .

Example. In the right pyramid $\displaystyle SABCD$ the base is a square of side 6, and the height is $\displaystyle {SO}={4}$ .

$\displaystyle {OM}={3}$ , so $\displaystyle {SM}={\sqrt{4^2+3^2}}={5}$ .
$\displaystyle {M}={4\cdot\frac{6\cdot5}{2}}={60}$ , and $\displaystyle {S}={60+36}={96}$ .
The side edge: $\displaystyle OB$ is half the square’s diagonal, $\displaystyle {OB}={3\sqrt{2}}$ , so $\displaystyle {SB}={\sqrt{16+18}}={\sqrt{34}}$ .

Exercise 1

The base of the right pyramid $\displaystyle SABCD$ is a rectangle: $\displaystyle {AB}={8}$ , $\displaystyle {BC}={6}$ .
The height is $\displaystyle {SO}={12}$ .

Why is $\displaystyle O$ where the base’s diagonals meet? Explain.

\displaystyle OA

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