L2 · Surface Area of Right Pyramids

Learning goal

“I can find the surface area of a right pyramid: area of the base plus one triangle for each base edge, using the SLANT height for the triangles - and I can find the slant height by Pythagoras when only the vertical height is given.”

Right pyramid — a solid with one polygon base; its apex sits directly above the base's centre, and its lateral faces are congruent-or-paired triangles.

Apex — the top point of the pyramid, where all the triangular faces meet.

Slant height — the distance from the apex down the middle of a triangular face to the midpoint of a base edge. It is the height OF THE TRIANGLE, and it is what surface area uses.

Vertical height — the straight-down distance from the apex to the centre of the base, INSIDE the pyramid. Volume uses it (Lesson L6); surface area does not.

Worked example

A right square pyramid has base edge 6 cm and slant height 5 cm. Find its surface area. Step 1: Base: 6 x 6 = 36 cm2. Step 2: One triangular face: 1/2 x base edge x slant height = 1/2 x 6 x 5 = 15 cm2. Step 3: Four faces: 4 x 15 = 60 cm2. Step 4: SA = 36 + 60 = 96 cm2. The slant height 5 is the height of each triangular face - measured ON the face, from apex to the midpoint of a base edge.

Check yourself — explain it

A student computes the triangular faces of the pyramid above using the VERTICAL height (4 cm) instead of the slant height (5 cm). His triangles come out...

Paired problem — example still visible

Square pyramid, base edge 8 cm, slant height 7 cm -> base 64 + 4 x (1/2 x 8 x 7) = 64 + 112 = 176 cm2

Faded completion — last step is yours

"Base edge 10 cm, slant height 9 cm: SA = 100 + 4 x (1/2 x 10 x ___) = 100 + 180 = ___ cm2" -> 9; 280 cm2

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10 scored questions with a 3-rung hint ladder. Mastery needs 8/10 plus both gate questions.

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Grade 10 — Mathematics 10C › C10U02 Surface Area & Volume

L2 · Surface Area of Right Pyramids