Honors Geometry Unit 12 Review: Surface Area and Volume

Solids split into two families. A polyhedron has only flat polygon faces, straight edges, and vertices, so prisms and pyramids count. Cylinders, cones, and spheres have curved surfaces, so they are not polyhedra.

A prism has two congruent, parallel polygon bases joined by lateral faces that are parallelograms (rectangles in a right prism). A pyramid has one polygon base and triangular lateral faces that meet at the apex. Cylinders and cones are the circular versions of these two. A sphere is the set of all points a fixed distance $r$ from a center. Solids are named by their base, so a hexagonal pyramid has a hexagon base and 6 triangular faces.

For any convex polyhedron, Euler's formula links faces, vertices, and edges with $F + V - E = 2$. A cube has 6 faces, 8 vertices, and 12 edges, and $6 + 8 - 12 = 2$.

A cross-section is the 2D shape you get when a plane slices a solid. Slices of a prism or cylinder parallel to the base are congruent to the base. A right circular cone gives a circle when cut parallel to the base between the base and apex. A vertical cut through the apex gives a triangle, and a tilted cut can give an ellipse when it forms a closed curve on the curved surface. The angle of the cut changes the shape, so a cube can give a square, a rectangle, a triangle, or even a hexagon.

Volume counts the cubic units that fit inside a solid. In a prism or cylinder, every slice parallel to the base is the same shape, so volume is base area times height, $V = Bh$. A cylinder uses the same formula with a circular base, which gives $V = \pi r^2 h$.

Pyramids and cones taper to a point. Each one holds exactly one-third of the prism or cylinder with the same base and height, so $V = \frac{1}{3}Bh$ and $V = \frac{1}{3}\pi r^2 h$. The $h$ here is always the perpendicular height, measured at a right angle from the base to the apex, never the slant along a face.

Cavalieri's principle explains why slanted (oblique) solids use the same formulas. If two solids have the same height and equal cross-sectional areas at every level, they have equal volumes. A leaning stack of coins holds the same volume as a straight stack, so an oblique cylinder still has $V = \pi r^2 h$ with the perpendicular height.

Worked example. A regular square pyramid has base edges of 10 cm and a slant height of 13 cm. Find its volume. The height, half the base edge (5 cm), and the slant height form a right triangle, so $h = \sqrt{13^2 - 5^2} = \sqrt{144} = 12$ cm. The base area is $B = 10^2 = 100$ cm². Then $V = \frac{1}{3}(100)(12) = 400$ cm³.

Surface area is the total area of a solid's outer surfaces, measured in square units. For a solid with a flat net, find the area of each piece and add them up.

It helps to split the total into lateral area (the sides) and base area. For a right prism, the sides unroll into one long rectangle whose length is the base perimeter, so $L = ph$ and $SA = ph + 2B$. A right cylinder's side unrolls into a rectangle that is $2\pi r$ long and $h$ tall, giving $SA = 2\pi rh + 2\pi r^2$.

A regular pyramid has congruent triangular faces, and the height of each triangle is the slant height $\ell$. That gives $L = \frac{1}{2}p\ell$ and $SA = \frac{1}{2}p\ell + B$. A right circular cone has $L = \pi r\ell$ and $SA = \pi r\ell + \pi r^2$. These surface area formulas use slant height, while volume formulas use perpendicular height. For a right cone, use $\ell^2 = r^2 + h^2$ to relate them. For a regular pyramid, use the base apothem, the distance from the base center to the midpoint of an edge, in place of $r$.

Worked example. A right circular cone has radius 6 cm and height 8 cm. Find its total surface area. First find the slant height, $\ell = \sqrt{6^2 + 8^2} = \sqrt{100} = 10$ cm. The lateral area is $\pi(6)(10) = 60\pi$ cm², and the base is $\pi(6)^2 = 36\pi$ cm². So the total is $SA = 96\pi \approx 301.6$ cm².

A sphere depends on one measurement, its radius. Surface area is $SA = 4\pi r^2$, exactly four times the area of a great circle (the circle you get by slicing through the center). Volume is $V = \frac{4}{3}\pi r^3$. The exponent matches the dimension, with $r^2$ for area and $r^3$ for volume.

A hemisphere is half a sphere, so its volume is $\frac{2}{3}\pi r^3$. Its curved surface is $2\pi r^2$. If the hemisphere is a closed solid, add the flat circle on the bottom for a total of $3\pi r^2$.

You'll often work backward. If $SA = 256\pi$, then $4\pi r^2 = 256\pi$, so $r^2 = 64$ and $r = 8$. From a volume, use $r = \sqrt[3]{\frac{3V}{4\pi}}$.

Scaling works the same way for every solid in this unit. Multiply all lengths by $k$, and surface area multiplies by $k^2$ while volume multiplies by $k^3$. Doubling a ball's radius makes its surface 4 times larger and its volume 8 times larger. Spheres also enclose the most volume for a given surface area, which is why soap bubbles are round.

Worked example. A spherical sculpture has a radius of 2 m, and one can of paint covers 15 m². The surface area is $4\pi(2)^2 = 16\pi \approx 50.27$ m². Dividing gives $50.27 \div 15 \approx 3.35$, so you need 4 cans, because 3 cans would leave part of the sculpture unpainted.