Surface Area of Pyramids, Cones, and Spheres

This distinction is the source of more marks lost in this topic than anything else. Before any formula — know which height you have and which you need.

Finding Slant Height with Pythagoras:

The right-angled triangle inside the solid has: vertical height $h$ as one leg, half the base dimension as the other leg, and slant height $\ell$ as the hypotenuse.

Surface Area of a Square Pyramid

$$\text{SA} = b^2 + 2b\ell$$

One square base ($b^2$) + four identical triangular lateral faces (each $\frac{1}{2}b\ell$, giving $4 \times \frac{1}{2}b\ell = 2b\ell$ total).

If the pyramid has no base (open at the bottom — like a tent or roof frame): $\text{SA} = 2b\ell$ (lateral faces only).

Surface Area of a Cone

$$\text{SA} = \pi r\ell + \pi r^2$$

Curved lateral surface ($\pi r\ell$) + circular base ($\pi r^2$). Factored: $\text{SA} = \pi r(\ell + r)$. Open cone (no base): $\text{SA} = \pi r\ell$.

Surface Area of a Sphere and Hemisphere

$$\text{SA}_\text{sphere} = 4\pi r^2$$

One of the most elegant results in geometry: the total surface area of a sphere equals exactly four circles of the same radius laid flat.

A hemisphere has two surfaces: curved outer dome $\frac{1}{2} \times 4\pi r^2 = 2\pi r^2$ plus flat circular base $\pi r^2$.

$$\text{SA}_\text{hemisphere} = 2\pi r^2 + \pi r^2 = 3\pi r^2$$

Composite Solids — Hidden Faces at Joins

When two solids are joined, the faces at the join become internal. They are hidden from the outside and are not part of the surface area.

At any join between two solids, the joined faces are not included in the total SA. List which faces are exposed on the outer surface — then calculate only those.