Digital SAT Suite Math Reference Sheet

1. What You Need to Know

You’ll be given a Digital SAT Suite Math Reference Sheet during PSAT 10. Your job isn’t to memorize these formulas, but to:

Recognize which formula applies fast.
Substitute correctly (especially radius vs diameter, height vs slant height, and which side is opposite/adjacent).
Solve for the requested variable (often the problem hides what you need to isolate).

What’s on the reference sheet (high-yield categories):

Area (rectangle, triangle, circle)
Volume (rectangular solid, cylinder, cone, sphere)
Coordinate geometry (slope, equation of a line)
Right-triangle relationships (Pythagorean theorem, special right triangles, trig definitions)

Critical reminder: The sheet is a starting point. Many problems require 1–3 extra algebra steps after you plug in.

2. Step-by-Step Breakdown

How to use the reference sheet efficiently (every time)

Identify the topic from the question: area, volume, line, right triangle, trig.
Sketch and label (even a quick doodle): mark given values, unknowns, and units.
Choose the matching reference formula and write it down.
Substitute carefully:
- Check whether the problem gives diameter but the formula uses radius.
- Check whether height is vertical (for cone/cylinder volume).
- For trig, confirm which side is opposite vs adjacent relative to the angle.
Solve algebraically for the variable asked.
Sanity check:
- Units: area should be square units; volume cubic units.
- Magnitude: does your answer seem too big or too small?

Mini worked walkthrough (line equation)

You’re told a line has slope $m=3$ and passes through the point $(-2,5)$ . Find its equation.

Use point-slope form:

$y-y_1=m(x-x_1)$

Substitute $m=3$ , $x_1=-2$ , $y_1=5$ :

$y-5=3(x+2)$

(If needed) convert to slope-intercept:

$y-5=3x+6$

$y=3x+11$

Decision point: If the problem asks for an equation, point-slope is often enough; if it asks for y-intercept or to compare lines, rewrite as $y=mx+b$ .

3. Key Formulas, Rules & Facts

A) Area formulas

Formula	When to use	Notes / traps
$A=lw$	Rectangle	$l$ and $w$ must be in same units
$A=\frac{1}{2}bh$	Triangle	$h$ is perpendicular to $b$
$A=\pi r^2$	Circle area	If given diameter $d$ , use $r=\frac{d}{2}$
$C=2\pi r$	Circle circumference	If given diameter $d$ , also true: $C=\pi d$ (derive from $r=\frac{d}{2}$ )

B) Volume formulas

Formula	When to use	Notes / traps
$V=lwh$	Rectangular solid	Volume is cubic units
$V=\pi r^2h$	Cylinder	$h$ is the vertical height
$V=\frac{1}{3}\pi r^2h$	Cone	Uses vertical height, not slant height
$V=\frac{4}{3}\pi r^3$	Sphere	Don’t confuse with surface area (not on sheet)

C) Coordinate geometry

Formula	When to use	Notes / traps
$m=\frac{y_2-y_1}{x_2-x_1}$	Slope between two points	If $x_2=x_1$ , slope is undefined (vertical line)
$y-y_1=m(x-x_1)$	Equation of a line	Great when you know $m$ and one point

Extra slope facts (not a formula, but test-critical):

Horizontal line: $y=c$ and slope $m=0$ .
Vertical line: $x=c$ and slope is undefined.

D) Right triangles (the big 3)

Relationship	When to use	Notes / traps
$a^2+b^2=c^2$	Right triangle	$c$ is the hypotenuse (longest side, opposite the right angle)
$\sin(\theta)=\frac{\text{opposite}}{\text{hypotenuse}}$	Right-triangle trig	“Opposite” depends on which angle you’re using
$\cos(\theta)=\frac{\text{adjacent}}{\text{hypotenuse}}$	Right-triangle trig	Adjacent is the non-hypotenuse side touching the angle
$\tan(\theta)=\frac{\text{opposite}}{\text{adjacent}}$	Right-triangle trig	Only uses legs (no hypotenuse)

E) Special right triangles

Triangle	Side ratios (relative)	How to use fast
$45^\circ\text{-}45^\circ\text{-}90^\circ$	$x,\,x,\,x\sqrt{2}$	Hypotenuse is $\sqrt{2}$ times a leg
$30^\circ\text{-}60^\circ\text{-}90^\circ$	$x,\,x\sqrt{3},\,2x$	Short leg opposite $30^\circ$ ; hypotenuse is double short leg

4. Examples & Applications

Example 1: Circle area from diameter

A circle has diameter $14$ . Find its area.

Convert to radius:

$r=\frac{14}{2}=7$

Area:

$A=\pi r^2=\pi(7^2)=49\pi$

Key insight: Most circle traps are just forgetting to halve the diameter.

Example 2: Cylinder volume with units

A cylinder has radius $3\,cm$ and height $10\,cm$ . Find volume.

$V=\pi r^2h=\pi(3^2)(10)=90\pi$

Units:

$90\pi\,cm^3$

Key insight: If the answer choices mix $cm^2$ and $cm^3$ , the question is testing whether you know area vs volume units.

Example 3: Slope and parallel lines

Line $\ell_1$ passes through $(2,1)$ and $(8,4)$ . Find its slope.

$m=\frac{4-1}{8-2}=\frac{3}{6}=\frac{1}{2}$

If a line $\ell_2$ is **parallel** to $\ell_1$ , then:

$m_2=\frac{1}{2}$

Key insight: Parallel lines have equal slope; perpendicular lines have negative reciprocal slope (you must know this even though it’s not a reference-sheet formula).

Example 4: Special triangle speed solve

A right triangle is $30^\circ\text{-}60^\circ\text{-}90^\circ$ with hypotenuse $18$ . Find the short leg.
In a $30^\circ\text{-}60^\circ\text{-}90^\circ$ triangle, hypotenuse is $2x$ :

$2x=18$

$x=9$

Key insight: Don’t reach for trig if the angles scream “special triangle.”

5. Common Mistakes & Traps

Radius vs diameter mix-up
- Wrong: plugging diameter into $A=\pi r^2$ or $C=2\pi r$ .
- Why wrong: the formulas use $r$ .
- Fix: convert first using $r=\frac{d}{2}$ .
Using slant height in cone volume
- Wrong: using the slanted side as $h$ in $V=\frac{1}{3}\pi r^2h$ .
- Why wrong: volume uses perpendicular height.
- Fix: if slant height is given, you may need the Pythagorean theorem to find vertical height.
Forgetting the triangle height must be perpendicular
- Wrong: using a side length as $h$ when it’s not perpendicular to the base.
- Why wrong: $A=\frac{1}{2}bh$ assumes a right angle between $b$ and $h$ .
- Fix: look for a right angle mark or drop an altitude in your sketch.
Slope sign errors (order mismatch)
- Wrong: computing $\frac{y_2-y_1}{x_1-x_2}$ or mixing point order.
- Why wrong: inconsistent subtraction flips the sign.
- Fix: pick an order and stick to it in both numerator and denominator.
Division by zero on vertical lines
- Wrong: trying to compute $m$ when $x_2=x_1$ and forcing a number.
- Why wrong: vertical lines have undefined slope.
- Fix: recognize vertical lines have equation $x=c$ .
Opposite vs adjacent relative to the wrong angle
- Wrong: labeling opposite/adjacent without referencing the specific $\theta$ .
- Why wrong: those labels change when the angle changes.
- Fix: circle the angle $\theta$ first, then label sides.
Messing up special triangle side placement
- Wrong: putting $x\sqrt{3}$ opposite $30^\circ$ .
- Why wrong: in $30^\circ\text{-}60^\circ\text{-}90^\circ$ , the **short leg** is opposite $30^\circ$ .
- Fix: memorize “short opposite $30^\circ$ .”
Unit mismatch
- Wrong: using $r$ in meters and $h$ in centimeters.
- Why wrong: formulas assume consistent units.
- Fix: convert before substituting.

6. Memory Aids & Quick Tricks

Trick / mnemonic	What it helps you remember	When to use
SOHCAHTOA	$\sin$ , $\cos$ , $\tan$ definitions	Any right-triangle trig question
“Hypotenuse is across from the right angle”	Identifies $c$ in $a^2+b^2=c^2$	Pythagorean problems
$45^\circ\text{-}45^\circ\text{-}90^\circ$ is $x,x,x\sqrt{2}$	Fast side lengths	Isosceles right triangle
$30^\circ\text{-}60^\circ\text{-}90^\circ$ is $x,x\sqrt{3},2x$	Fast side lengths	Triangle has $30^\circ$ or $60^\circ$
“Parallel: same slope”	Line relationship	Comparing two lines
“Perpendicular: negative reciprocal”	Line relationship	Right-angle line problems

7. Quick Review Checklist

You can instantly match the situation to the correct formula: area, volume, slope/line, right triangles.
You always check radius vs diameter before using circle formulas.
For cones/cylinders, you use vertical height in the volume formula.
You know volume answers must be in cubic units and area in square units.
You can compute slope with

$m=\frac{y_2-y_1}{x_2-x_1}$

You can write a line equation with

$y-y_1=m(x-x_1)$

You label right triangles correctly and can use

$a^2+b^2=c^2$

You recognize special triangles and their ratios:

$x,x,x\sqrt{2}$

$x,x\sqrt{3},2x$

You can use trig definitions without mixing up opposite/adjacent:

$\sin(\theta)=\frac{\text{opposite}}{\text{hypotenuse}}$

$\cos(\theta)=\frac{\text{adjacent}}{\text{hypotenuse}}$

$\tan(\theta)=\frac{\text{opposite}}{\text{adjacent}}$

You’ve got this: use the sheet to set up quickly, then let your algebra finish the job.