Mensuration (3D)

Prisms and pyramids: the one-third rule

A prism has a constant cross-section: $V = (\text{area of base}) \times (\text{length/height})$. A pyramid with the same base and apex over the centre holds exactly one-third: $V = \frac13 \times \text{base area} \times h$. This is the cone/cylinder relation in flat-base form — the fraction $\frac13$ is the whole topic. $V_{\text{prism}} = A_{\text{base}} \times h, \qquad V_{\text{pyramid}} = \frac{1}{3} A_{\text{base}} \times h$

Frustums (cut pyramids/cones)

Slice a cone or pyramid parallel to its base and remove the top: what remains is a frustum. $V = \frac{\pi h}{3}\left(R^2 + r^2 + Rr\right) \text{ (conical)}, \qquad V = \frac{h}{3}\left(A_1 + A_2 + \sqrt{A_1 A_2}\right) \text{ (any)}$ with $R, r$ the two radii (or $A_1, A_2$ the two face areas) and $h$ the perpendicular height between them. Bucket and tumbler questions are frustums.

Lateral surfaces

• Prism lateral area $=$ (perimeter of cross-section) $\times$ (length) — the tube unrolls flat.
• Pyramid lateral area $= \frac12 \times$ (perimeter of base) $\times$ (slant height) — sum of triangles. The total adds the two base/cross-section ends.

Cube cutting and melting

A cube of side $a$ cut into unit cubes yields $a^3$ of them. Melting several cubes into one: sum the volumes and take the cube root — edges $3, 4, 5$ combine into a cube of edge $\sqrt[3]{27+64+125} = 6$. Both facts are volume bookkeeping; nothing else changes on melting.

Worked frustum

Bucket with $R = 5$, $r = 3$, $h = 6$: the bracket $R^2 + r^2 + Rr = 25 + 9 + 15 = 49$, and $\frac{\pi h}{3} = \frac{22}{7} \times 2$, so $V = \frac{22}{7} \times 2 \times 49 = 308$ cu cm. Exam frustums are built so the bracket is a perfect square times $\frac{22}{7}$-friendly numbers — if your bracket is ugly, recheck the radii (diameters halved?) and the terms ($Rr$ included?). For the cut-off top piece instead, either subtract the small cone from the big one or use the similarity ratio: cutting at half the height leaves $\frac18$ of the volume below-cut... rather, the removed top cone has $\frac18$ of the whole cone's volume, so the frustum is $\frac78$ of it.

Cube numbers worth carrying

$a^3$: 1, 8, 27, 64, 125, 216, 343, 512, 729, 1000, 1331, 1728 — the melting answers live here ($216 \to 6$, $729 \to 9$, $1728 \to 12$). Two cubes of edge $a$ melted give $\sqrt[3]{2a^3} = a\sqrt[3]{2}$ — NOT $2a$; only cube-number sums recast into clean cubes, which is why exams pick $3^3 + 4^3 + 5^3 = 6^3$ (the only consecutive-cube identity).

Quick revision

• Prism: $V = A \times h$; pyramid: $\frac13 A h$; lateral prism $= P \times h$, pyramid $= \frac12 P l$.
• Frustum (cone form): $\frac{\pi h}{3}(R^2 + r^2 + Rr)$.
• Cut cube of side $a$ into unit cubes: $a^3$ pieces.
• Melted cubes: $\sqrt[3]{\sum a_i^3}$.