Two heights live inside every pyramid. Volume wants one of them and surface area wants the other, and telling them apart is the whole lesson.
Find the volume
A square pyramid has a base edge of 10 cm and a height of 12 cm.
B = 10² = 100 cm², so V = ⅓(100)(12) = 400 cm³. B forgets the one third and gives the prism on the same base — the single most common slip in this topic, and it is always exactly three times too big. C halves instead of thirding. D uses 6 as the height, halving it first.
V = ⅓ B h
Same base, same height, one third the volume. Three identical pyramids pack exactly into the prism they came from — pour rice from one into the other and it takes three fills, which is worth doing once if you have a set of solids to hand.
The height h drops straight down from the apex to the centre of the base. The slant height ℓ runs down the middle of a triangular face to the midpoint of a base edge. They meet in a right triangle whose third side is half the base edge.
ℓ² = 12² + 5² = 144 + 25
half of 10 is 5, not 10
ℓ = √169 = 13 cm
longer than the height ✓
Using 10 instead of 5 gives ℓ = √244 ≈ 15.6, and the answer stops being a whole number — which is usually the first sign. The triangle runs from the centre of the base to the middle of one edge, and that distance is half the edge.
LSA = ½ p ℓ TSA = ½ p ℓ + B
Each face is a triangle of base s and height ℓ, so each has area ½sℓ. Four of them give 4(½sℓ) = ½(4s)ℓ = ½pℓ. And a pyramid has only one base to add, not two.
p = 4(10) = 40 cm
round the square base
LSA = ½(40)(13) = 260 cm²
the four triangles
TSA = 260 + 100 = 360 cm²
plus the one square base
A fast sanity check: the slant height must be longer than the vertical height, because it is the hypotenuse of the triangle holding them both. If your slant comes out shorter, you have swapped them.
B = 6(8) = 48 cm²
the rectangular base
96 = ⅓(48)h = 16h
put the number in first
h = 96 ÷ 16 = 6 cm
check: ⅓(48)(6) = 96 ✓
The thirds cause most of the trouble when they are undone rather than applied. Substitute everything you know into V = ⅓Bh first and solve the ordinary equation that is left; do not try to rearrange the formula in your head.
The same right triangle and the same third, met from other directions.
Slant given, height asked: square base edge 18 cm, slant height 15 cm. Find V.
h² = 15² − 9² = 225 − 81 = 144
ℓ is the hypotenuse, so subtract
h = 12 cm
shorter than ℓ ✓
V = ⅓(324)(12) = 1 296 cm³
B = 18² = 324
Going from ℓ to h, the square of the slant comes first and you subtract.
A rectangular base 8 cm by 6 cm, height 4 cm. Find the lateral area.
faces on the 8 cm edges: ℓ = √4² + 3² = 5
half of 6 is 3
faces on the 6 cm edges: ℓ = √4² + 4² = 4√2
half of 8 is 4
LSA = 2(½)(8)(5) + 2(½)(6)(4√2)
two faces of each kind
LSA = 40 + 24√2 cm²
about 73.9 cm²
A rectangular base has two slant heights, so ½ p ℓ no longer applies: add the faces one pair at a time.
A triangular base with legs 5 cm and 9 cm; pyramid 8 cm tall. Find V.
B = ½(5)(9) = 22.5 cm²
a decimal base is fine
V = ⅓(22.5)(8) = 60 cm³
22.5 × 8 = 180, then ÷ 3
The third applies to every base shape, and a cone too: any solid that narrows to a point holds ⅓ B h.
Pyramid items nearly always hide one length: the slant is given and the volume is asked, or the reverse, so the right triangle h, s2, ℓ does the work before any formula. Write which height each formula needs, find the missing one, then apply ⅓ B h or ½ p ℓ + B. The 10-second check: ℓ must be longer than h, and the volume must be a third of the box around the pyramid.
Find the total surface area
A square pyramid has base edge 6 cm and height 4 cm.
Half the base edge is 3, so ℓ = √4² + 3² = 5 cm. Then p = 24, LSA = ½(24)(5) = 60, and B = 36, giving TSA = 96 cm². A is the lateral area with the height used in place of the slant, ½(24)(4) = 48. B uses the whole base edge in the triangle, ℓ = √52 ≈ 7.2. C is the lateral area alone plus a second base.
Hint: nothing here can start until you have the slant height, and the triangle that gives it uses half of 6. A pyramid has one base, so the final line adds B once.
