| Primary source | OpenStax College Algebra 2e, Section 2.3: "Models and Applications" (applies geometry formulas to word problems) |
| Direct link | https://openstax.org/books/college-algebra-2e/pages/2-3-models-and-applications |
| Supplementary | OpenStax Prealgebra 2e, Chapter 9: "Math Models and Geometry" |
| Supplementary link | https://openstax.org/books/prealgebra-2e/pages/9-introduction |
Geometric formulas are compiled in most textbook appendices. College Algebra 2e uses them throughout Chapter 2 (word problems) and later chapters.
Most geometry formulas are not arbitrary -- they have derivable geometric explanations. Understanding the explanation tells you which formula to use when a problem does not explicitly name the shape.
Two examples:
Memorizing formulas is unavoidable. But connecting each formula to a geometric picture means you can recover it from reasoning if you forget, and you can adapt it when the problem changes shape.
| Shape | Perimeter / Circumference | Area |
|---|---|---|
| Rectangle | $P = 2l + 2w$ | $A = lw$ |
| Square | $P = 4s$ | $A = s^2$ |
| Triangle | $P = a + b + c$ | $A = \frac{1}{2}bh$ |
| Circle | $C = 2\pi r = \pi d$ | $A = \pi r^2$ |
| Trapezoid | $P = a + b_1 + b_2 + c$ | $A = \frac{1}{2}(b_1 + b_2)h$ |
| Parallelogram | $P = 2a + 2b$ | $A = bh$ |
| Shape | Surface Area | Volume |
|---|---|---|
| Rectangular box | $SA = 2lw + 2lh + 2wh$ | $V = lwh$ |
| Cube | $SA = 6s^2$ | $V = s^3$ |
| Cylinder | $SA = 2\pi r^2 + 2\pi rh$ | $V = \pi r^2 h$ |
| Sphere | $SA = 4\pi r^2$ | $V = \frac{4}{3}\pi r^3$ |
| Cone | $SA = \pi r^2 + \pi r l$ ($l$ = slant height) | $V = \frac{1}{3}\pi r^2 h$ |
| Pyramid (square base) | $SA = s^2 + 2sl$ | $V = \frac{1}{3}s^2 h$ |
For a right triangle with legs $a$, $b$ and hypotenuse $c$: \[ a^2 + b^2 = c^2 \]
Area: $A = lw$ (length times width). Every cell in a $l$-by-$w$ grid is 1 square unit; there are $lw$ cells.
Perimeter: $P = 2l + 2w$. The boundary consists of two lengths and two widths.
Example 1. A rectangular yard has perimeter 48 meters and length twice the width. Find the dimensions and area.
Let $w$ = width. Then $l = 2w$.
$P = 2(2w) + 2w = 6w = 48 \implies w = 8$ m, $l = 16$ m.
$A = 16 \times 8 = 128$ m$^2$.
Area: $A = \frac{1}{2}bh$.
Why. Every triangle fits inside a rectangle with base $b$ and height $h$. The area of the rectangle is $bh$. Any triangle uses exactly half the rectangle's area. (You can verify this by cutting the rectangle diagonally or by noting that the triangle is composed of two right triangles that together fill half the rectangle.)
Perimeter: $P = a + b + c$ (sum of the three side lengths).
Example 2. A triangular sail has base 4 m and height 7 m. Find its area.
$A = \frac{1}{2}(4)(7) = 14$ m$^2$.
The height is perpendicular to the base. For an obtuse triangle, the height may fall outside the triangle. The formula still works: $h$ is the perpendicular distance from the vertex to the line containing the base.
Circumference: $C = 2\pi r$.
Area: $A = \pi r^2$.
Why $\pi$. The number $\pi \approx 3.14159...$ is the ratio of any circle's circumference to its diameter: $\pi = C/d$. This is a property of circles themselves; it does not depend on the size. From this, $C = \pi d = 2\pi r$.
The area formula $A = \pi r^2$ can be derived by cutting the circle into thin wedges, rearranging them into an approximate rectangle of width $\pi r$ and height $r$, giving area $\pi r^2$.
Example 3. A circular fountain has circumference 25 m. Find its radius and area.
$C = 2\pi r = 25 \implies r = \dfrac{25}{2\pi} \approx 3.98$ m.
$A = \pi r^2 = \pi \left(\dfrac{25}{2\pi}\right)^2 = \dfrac{625}{4\pi} \approx 49.7$ m$^2$.
Parallelogram area: $A = bh$. A parallelogram with base $b$ and height $h$ has the same area as the rectangle with those dimensions. (Proof: cut the left triangle off and move it to the right side. The parallelogram becomes a rectangle.)
Trapezoid area: $A = \frac{1}{2}(b_1 + b_2)h$.
Why. A trapezoid with parallel bases $b_1$ and $b_2$ and height $h$ can be viewed as an "average rectangle": the average of the two bases times the height. Equivalently, place two identical trapezoids together (one flipped) to form a parallelogram with base $b_1 + b_2$ and height $h$. Each trapezoid is half of that parallelogram.
Prism principle: For any solid with a uniform cross-section, $V = (\text{cross-sectional area}) \times (\text{height})$.
Pyramid and cone: These taper from a base to a point (apex). Their volume is exactly one-third of the corresponding prism. \[ V_{\text{cone}} = \frac{1}{3}\pi r^2 h \qquad V_{\text{pyramid}} = \frac{1}{3}(\text{base area}) \times h \]
Example 4. A cylindrical can has radius 3 cm and height 10 cm. Find its volume and lateral surface area.
$V = \pi(3)^2(10) = 90\pi \approx 283$ cm$^3$.
Lateral surface area (the curved side, not the ends): unroll it to get a rectangle of width $2\pi r = 6\pi$ and height 10. $SA_{\text{lateral}} = 60\pi \approx 188$ cm$^2$.
Total surface area (including both ends): $SA = 60\pi + 2\pi(9) = 78\pi \approx 245$ cm$^2$.
In a right triangle with legs $a$ and $b$ and hypotenuse $c$ (the side opposite the right angle): \[ a^2 + b^2 = c^2 \]
Example 5. A ladder 13 feet long leans against a wall with its base 5 feet from the wall. How high does it reach?
$5^2 + h^2 = 13^2 \implies 25 + h^2 = 169 \implies h^2 = 144 \implies h = 12$ feet.
Example 6. Find the diagonal of a rectangle with dimensions 7 cm by 24 cm.
$d^2 = 7^2 + 24^2 = 49 + 576 = 625 \implies d = 25$ cm.
Pythagorean triples (integer solutions to $a^2 + b^2 = c^2$): $3$-$4$-$5$, $5$-$12$-$13$, $8$-$15$-$17$. Recognizing these saves computation in timed settings.
Two figures are similar if they have the same shape (all angles equal) but possibly different sizes. Their corresponding sides are proportional.
If the ratio of corresponding lengths is $k$ (the scale factor), then:
Example 7. Two similar triangles have corresponding bases 4 cm and 10 cm. The smaller triangle has area 12 cm$^2$. Find the larger triangle's area.
Scale factor $k = 10/4 = 2.5$. Area scales by $k^2 = 6.25$.
Larger area $= 12 \times 6.25 = 75$ cm$^2$.
| Error | Example | Correction |
|---|---|---|
| Using diameter instead of radius | $A = \pi d^2$ | $A = \pi r^2$; $r = d/2$ |
| Forgetting the $\frac{1}{2}$ for triangles | $A = bh$ | $A = \frac{1}{2}bh$ |
| Using slant height instead of vertical height | $V_{\text{cone}} = \frac{1}{3}\pi r^2 l$ | $V = \frac{1}{3}\pi r^2 h$ ($h$ = perpendicular height) |
| Units: squaring or cubing | Leaving area in cm not cm$^2$ | Area: cm$^2$; Volume: cm$^3$ |
| Scale factors applied linearly to area | Doubling linear size doubles area | Doubling size quadruples area ($k^2 = 4$) |
Problem 1. Find the area and circumference of a circle with diameter 14 cm. Leave answers in terms of $\pi$.
$r = 7$ cm.
$C = 2\pi(7) = 14\pi$ cm.
$A = \pi(7)^2 = 49\pi$ cm$^2$.
Problem 2. A right triangle has legs 9 and 40. Find the hypotenuse.
$c^2 = 81 + 1600 = 1681 = 41^2$
$c = 41$
Problem 3. A sphere has radius 6 cm. Find its volume and surface area.
$V = \frac{4}{3}\pi(6)^3 = \frac{4}{3}\pi(216) = 288\pi \approx 904.8$ cm$^3$.
$SA = 4\pi(6)^2 = 144\pi \approx 452.4$ cm$^2$.
Problem 4. A cylinder has volume $100\pi$ cm$^3$ and height 4 cm. Find the radius.
$\pi r^2 (4) = 100\pi \implies r^2 = 25 \implies r = 5$ cm.
Problem 5. A right triangle has hypotenuse 17 and one leg 8. Find the area.
Other leg: $a^2 = 17^2 - 8^2 = 289 - 64 = 225 \implies a = 15$.
$A = \frac{1}{2}(8)(15) = 60$ square units.
Problem 6. A trapezoidal plot of land has parallel sides of length 20 m and 34 m, and a height (perpendicular distance between the parallel sides) of 15 m. Find its area.
$A = \frac{1}{2}(20 + 34)(15) = \frac{1}{2}(54)(15) = 405$ m$^2$.
Problem 7. A rectangular box with a square base has volume 200 cm$^3$. Express the total surface area as a function of the base side length $s$.
Let $s$ = side length of the base, $h$ = height.
Volume: $s^2 h = 200 \implies h = \dfrac{200}{s^2}$.
Surface area: $SA = 2s^2 + 4sh = 2s^2 + 4s \cdot \dfrac{200}{s^2} = 2s^2 + \dfrac{800}{s}$.
This function of $s$ is what you minimize in calculus to find the optimal box dimensions.
Problem 8. Two similar cones have heights 5 cm and 20 cm. If the smaller cone has volume 30$\pi$ cm$^3$, find the volume of the larger cone.
Scale factor $k = 20/5 = 4$.
Volume scales by $k^3 = 64$.
Larger volume $= 30\pi \times 64 = 1920\pi$ cm$^3$.
Problem 9. A cylindrical water tank has radius 3 m and height 8 m. Water is draining at 2 m$^3$/min. At what rate is the water level falling?
Volume of water when height is $h$: $V = \pi(3)^2 h = 9\pi h$.
As $V$ decreases at 2 m$^3$/min:
$\dfrac{dV}{dt} = 9\pi \dfrac{dh}{dt} = -2$
$\dfrac{dh}{dt} = \dfrac{-2}{9\pi} \approx -0.071$ m/min.
(This problem previews related rates from MATH161 Chapter 3. The geometry formula $V = 9\pi h$ is the essential setup step.)
Every area formula counts unit squares. Every volume formula counts unit cubes.
A rectangle $l \times w$ fits exactly $lw$ unit squares: one per grid cell. A triangle is half a rectangle: $\frac{1}{2}lw = \frac{1}{2}bh$. A circle of radius $r$ fits inside a square of side $2r$ (area $4r^2$) and contains approximately $\pi r^2$ of them, where $\pi \approx 3.14$ is the "fraction of the square" a circle occupies.
Volume works the same way: a box $l \times w \times h$ fits $lwh$ unit cubes. A cylinder is a stack of circular slabs: $V = (\pi r^2) \cdot h$. A cone tapers: fill it three times to fill the cylinder, so $V_{\text{cone}} = \frac{1}{3}V_{\text{cylinder}}$.
When you forget a formula, ask: "What shape is this, and how do I count squares or cubes in it?" The answer usually leads you back to the formula.
For students who find math intimidating: The Pythagorean theorem and the area of a circle are the two most widely used formulas in this lesson. If you can recall those two and know how to solve for a missing dimension, you can handle most geometry calculations in a calculus course.
For students interested in proof: The formula $A = \pi r^2$ can be derived rigorously using the definite integral $\int_{-r}^{r} 2\sqrt{r^2 - x^2}\,dx = \pi r^2$, which evaluates the area of a circle as twice the area above the $x$-axis. This is a MATH162 computation; it uses every geometry formula in this lesson plus the techniques of integration by trigonometric substitution.
For students interested in careers: Geometric formulas appear in architecture (floor plans, material estimates), civil engineering (road cross-sections, drainage basins), manufacturing (material usage, container optimization), and medicine (organ volumes, drug dosing by body surface area).