Mensuration (3D)

A triangular prism with cross-section $20$ sq cm and length $12$ cm holds $240$ cu cm. A rectangular base $7\times5$ with length $10$ holds $350$ cu cm.

The base can be a triangle, square, hexagon, anything regular. Find its area first, multiply by the length second. A cuboid is just a prism with a rectangle base, so the tank formula lives here too.

Square base $10$, height $12$: $V=\dfrac13\times100\times12=400$ cu cm. Rectangular base $9\times4$ with height $7$: $V=\dfrac13\times36\times7=84$ cu cm.

Same base, same height: the pyramid always holds one-third of the prism's volume. The cone's one-third is the same idea on a circular base.

When a face slant of $13$ sits on a base of side $10$, the half-base is $5$, so the true height is $\sqrt{169-25}=12$. Slant first, height second.

A cone with its top sliced off. With radii $R$ (bottom) and $r$ (top), height $h$:

Radii $5$ and $3$, height $6$: $\dfrac{22}{7}\times\dfrac{6}{3}\times(25+9+15)=\dfrac{44}{7}\times49=308$ cu cm.

A frustum is a big cone minus the small cone sliced off, and similar triangles link the two radii. Slant height $\sqrt{h^2+(R-r)^2}$: with radii $8$ and $3$ and height $12$ it is $\sqrt{144+25}=13$, the $5$-$12$-$13$ triplet again.

Three cubes of edges $3$, $4$, $5$ melt into one cube. Volume adds: $27+64+125=216=6^3$, so the new edge is $6$ cm.

Melting adds volumes, then one cube root. The cube table turns this into arithmetic. Splitting runs the other way: one cube of edge $12$ gives $8$ cubes of edge $6$, since $\dfrac{1728}{8}=216=6^3$.

• Prism: perimeter of base $\times$ length.
• Pyramid: $\dfrac12\times$ perimeter of base $\times$ slant height.

Square pyramid, base $10$, slant $13$: $\dfrac12\times40\times13=260$ sq cm. The slant height belongs to each triangular face.

Triangular prism with base sides $6$, $8$, $10$ and length $12$: perimeter $24$, so lateral $=24\times12=288$ sq cm. The base is a right triangle by the $6$-$8$-$10$ test.