L2 · Surface area of right pyramids

Learning goal

“I can find the surface area of a right pyramid by adding its base to its triangular lateral faces - using the SLANT HEIGHT as the height of each triangular face, and adjusting the census when a real problem covers only some of the faces.”

Pyramid — a polyhedron with one polygonal base and triangular lateral faces that meet at a single point. (A cylinder is not a pyramid and not a polyhedron; a pyramid is.)

Apex — the single point at the top of a pyramid where all the triangular faces meet.

Lateral face — one of the triangular side faces of a pyramid, running from a base edge up to the apex.

Slant height — the height of a triangular lateral face: the distance from the midpoint of a base edge straight up the face to the apex. It is NOT the pyramid's vertical height (the perpendicular distance from the apex to the base) - the slant height is the one that lives ON the face, so it is the one the area formula needs.

Worked example

A right square pyramid has base side 6 cm and slant height 5 cm. Find its surface area. Step 1: Census from the net: 1 square base + 4 identical triangular lateral faces. Step 2: Base: 6 x 6 = 36 cm2. Each triangle: 1/2 x 6 x 5 = 15 cm2 - the triangle's base is the base edge (6) and its height is the SLANT height (5), because the slant height is measured on the face itself. Step 3: SA = 36 + 4 x 15 = 36 + 60 = 96 cm2.

Check yourself — explain it

The lateral faces are computed with the slant height rather than the vertical height because...

Paired problem — example still visible

Square pyramid, base side 4 cm, slant height 3 cm -> base 16; triangles 4 x (1/2 x 4 x 3) = 24; SA = 40 cm2

Faded completion — last step is yours

"Square pyramid, base side 10 cm, slant height 6 cm: base = 100; each triangle = 1/2 x 10 x 6 = 30; SA = 100 + 4 x ___ = ___ cm2" -> 30; 220 cm2

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