Area and Volume

Choosing, decomposing, and running formulas backward

The reference sheet hands you the formulas. The work is choosing the right one, breaking a complicated shape into simple ones, and, when the test runs the formula backward, solving for the piece left blank.

The reference sheet gives area formulas for a circle, rectangle, and triangle and volume formulas for a rectangular prism, cylinder, sphere, cone, and pyramid. Other relationships, including parallelogram and trapezoid area, still need to be recognized or derived. That means three skills: matching the figure to its formula, decomposing a complicated shape into simple pieces you can add or subtract, and running a formula backward to find a dimension when the area or volume is given. In every problem, the main task is the same: choose the correct formula, decompose the figure when necessary, and identify the requested dimension or measure.

Area formulas for triangles, quadrilaterals, and circles

Four area formulas cover almost every plane figure. A rectangle is base times height; a triangle is half of that, $\frac12 bh$, since it fills half its bounding rectangle. A parallelogram is base times height (the perpendicular height, not the slanted side) and a trapezoid, with its two parallel sides, is the average of those sides times the height, $\frac12(b_1 + b_2)h$. In every case the height is the perpendicular distance, which is the one detail worth checking twice.

Figure 1. The core plane-area formulas. In each, the height $h$ is the perpendicular distance, not a slanted side.

Worked: apply an area formula. A trapezoid has parallel sides $6$ and $14$ and height $5$. Average the parallel sides, then multiply by the height: $A = \frac12(6 + 14)(5) = \frac12(20)(5) = 50$. The average of the bases, $10$, times the height $5$, gives the same $50$; the trapezoid behaves like a rectangle of that average width.

When a problem lists a slanted side alongside the height, the slanted side is a distractor. A parallelogram with base $9$, slanted side $5$, and perpendicular height $4$ has area $9 \times 4 = 36$, not $9 \times 5$.

Decompose composite figures to find area

Many figures are not single shapes but combinations: a rectangle with a semicircle on one end, an L-shaped room, a square with a triangle removed. The method never changes: cut the figure into shapes you have formulas for, then add the pieces that are present and subtract the pieces that are cut away. The only skill is seeing the simple shapes inside the complicated one.

Figure 2. A composite figure is a sum of simple ones: this area is a rectangle plus a half-circle.

Worked: build an area from pieces. A region is a $6$-by-$4$ rectangle with a semicircle of radius $2$ on one end. Add the two areas: rectangle $= 6 \times 4 = 24$, half-circle $= \frac12\pi(2^2) = 2\pi$. So the total area is $24 + 2\pi$. For a shape with a piece removed, you would subtract that piece instead of adding it.

Figure 3. A circle inscribed in a square: the region inside the square but outside the circle is the square's area minus the circle's.

Worked: a circle-in-square shaded region. A circle is inscribed in a square of side $10$, as shown. What is the area inside the square but outside the circle? The circle's diameter equals the square's side, so its radius is $5$. Subtract the circle's area from the square's: $100 - 25\pi$. Mixing up radius and diameter is the usual trap in this family; the inscribed circle's diameter, not its radius, is what matches the square's side.

Volume formulas for prisms, cylinders, cones, spheres, and pyramids

Volume formulas split into two families. For solids with the same cross-section all the way up, prisms and cylinders, the volume is simply the base area times the height: a rectangular prism is $\ell wh$, a cylinder is $\pi r^2 h$. For solids that taper to a point, cones and pyramids, the volume is exactly one-third of the matching prism or cylinder: a cone is $\frac13\pi r^2 h$. The sphere stands alone at $\frac43\pi r^3$. Knowing which family a solid belongs to tells you whether the one-third is there.

Figure 4. Prisms and cylinders are base area times height; cones and pyramids are one-third of that; the sphere has its own formula.

Worked: compute and compare volumes. A cylinder with radius $3$ and height $10$ has volume $\pi(3^2)(10) = 90\pi$. A cone with the same radius and height holds one-third as much: $\frac13(90\pi) = 30\pi$. A sphere of radius $3$ has volume $\frac43\pi(3^3) = \frac43\pi(27) = 36\pi$. Recognizing the one-third for the tapering cone saves a full recalculation.

A square pyramid uses the same one-third rule, but its most common trap is a labeling one: a pyramid problem often gives the slant height, the distance along a triangular face, rather than the vertical height the volume formula actually needs.

Figure 5. A square pyramid's vertical height runs straight down to the base's center; its slant height runs along a face to the base's edge. The two are not the same segment.

Worked: do not mistake slant height for vertical height. A square pyramid has a base side of $6$ and a vertical height of $10$. What is its volume? The volume formula wants the vertical height directly; since it is already given here, no conversion is needed: $V = \frac13(\text{base area})(\text{height}) = \frac13(36)(10) = 120$. Had the problem instead given a slant height of $10$, the vertical height would first need to be recovered using the right triangle formed by the vertical height, half the base side, and the slant height, the Pythagorean theorem applied inside the solid, before this same formula could be used.

The same trap appears with cones. A cone with slant height $13$ and radius $5$ has vertical height $\sqrt{13^2 - 5^2} = 12$, and only that $12$ belongs in $\frac13\pi r^2 h$.

Decompose an unfamiliar solid into a known base and height

Not every solid arrives as a named prism, cylinder, cone, sphere, or pyramid. A tetrahedron, a solid with four triangular faces, is not on that list, but it still yields to the same one-third rule once its base and height are identified correctly: volume is one-third of the base area times the height, exactly as for any pyramid, since a tetrahedron is simply a pyramid with a triangular base.

Figure 6. A tetrahedron with three mutually perpendicular edges at $O$. Any two of those edges frame the triangular base; the third is the height above it.

Worked: volume of a right tetrahedron. Three edges of a tetrahedron meet at vertex $O$ at right angles to each other, with lengths $6$, $6$, and $8$. What is the volume? Two of the perpendicular edges, the two of length $6$, frame a right-triangle base; the third edge, length $8$, is then perpendicular to that entire base and serves as the height. Base area $= \frac12(6)(6) = 18$, so $V = \frac13(18)(8) = 48$. The same three edges could be relabeled to use a different pair as the base, and the volume would come out to $48$ either way.

Surface area from the faces of a solid

The surface area of a solid is the total area of all its faces, so you find it by adding up the areas of the flat pieces and curved pieces that wrap the solid. A rectangular prism has six rectangular faces in three matching pairs; a cylinder unrolls into two circles (the ends) plus a rectangle (the wrapped side) whose width is the circumference.

Worked: find a surface area. A cylinder has radius $4$ and height $6$. Add the two circular ends and the wrapped side: two ends $= 2\pi(4^2) = 32\pi$, side $= 2\pi(4)(6) = 48\pi$. So the total surface area is $32\pi + 48\pi = 80\pi$. The side's area is the circumference $2\pi r$ times the height.

Track which measure a question requests

Surface area and volume describe the same solid from two different questions, how much skin wraps it against how much room is inside it, and a problem's wording is the only signal for which one is wanted. "How much material," "how much paint," or "wrapping paper" point to surface area; "how much it holds," "capacity," or "water it can contain" point to volume. The two numbers are rarely close enough that guessing which formula to use is safe.

The same cylinder, radius $4$ and height $6$, makes the gap concrete: its surface area is $80\pi$, but its volume is $\pi(4^2)(6) = 96\pi$. Reaching for the wrong one of these two nearby-looking numbers is a requested-object error, not a computation error, since both formulas would be applied correctly, just to the wrong question.

Solve for dimensions and apply scale factors

Two higher-order moves complete the section. First, the test often gives the area or volume and asks for a dimension, so you set up the formula and solve for the unknown. Second, recall how scaling propagates: enlarging a solid by a factor $k$ multiplies every length by $k$, the surface area by $k^2$, and the volume by $k^3$. Tripling a solid's dimensions multiplies its volume by $27$, not by $3$.

Worked: run a formula backward. A cylinder has volume $100\pi$ and radius $5$. Solve the volume formula for the height: $100\pi = \pi(5^2)h = 25\pi h$, so $h = \frac{100\pi}{25\pi} = 4$. A sphere with volume $288\pi$: from $\frac43\pi r^3 = 288\pi$, divide to get $r^3 = 216$, so $r = 6$; the $\pi$ cancels and a cube root finishes it.

When two similar solids are described by a volume ratio, recover the length ratio with a cube root. Volumes in the ratio $27 : 8$ have lengths in the ratio $3 : 2$, so a side of $4$ on the smaller solid corresponds to $4 \times \frac32 = 6$ on the larger one. Applying the volume ratio straight to a side length skips the cube root and is the error these questions are built to catch.

Summary

Area and volume reward formula judgment over memorization, since the formulas are given. For plane figures, match the shape (triangle as half a rectangle, trapezoid as the average of its bases times height) and decompose composite regions by adding and subtracting simple pieces. For solids, prisms and cylinders are base area times height, cones and pyramids are one-third of that, and the sphere is $\frac43\pi r^3$. Use the vertical height, never the slant height, in a volume formula. Run any formula backward to find an unknown dimension, and remember that volume scales by the cube of a length factor. Pick the right formula, break the figure into parts, and solve for what is asked.