SAT Math Geometry and Trigonometry: Complete Study Notes

Area and Volume

Geometry questions often look like “plug into a formula,” but the real SAT skill is choosing the right measurement (area vs. surface area vs. volume), using consistent units, and breaking complicated shapes into simpler ones. A helpful mental model is: area is “paint the face,” surface area is “wrap the object,” and volume is “fill the object.” Many SAT questions hide that choice inside context.

What area and volume mean (and why units matter)

Area measures how much 2D space a shape covers, like counting how many square tiles fit inside a boundary. Area units are square units (for example, square inches, cm^2).

Volume measures how much 3D space a solid occupies, like counting cubic blocks. Volume units are cubic units (for example, cubic feet, m^3).

On the SAT, unit mistakes are common because diagrams may omit units or mix dimensions. Build the habit of writing units next to your final answer and sanity-checking: if you found area, units must be squared; if volume, cubed. If the problem mixes units (like centimeters and meters), convert before computing—especially for volume, where conversion factors cube.

Perimeter vs. area (a frequent SAT trap)

Perimeter is total boundary length, so it uses linear units. Area uses square units. Changing a figure can increase perimeter while decreasing area (and vice versa), so do not assume they move together.

Core area ideas: decomposition and “missing piece” strategy

Many SAT figures are composite: they can be split into rectangles, triangles, semicircles, and other familiar shapes. Decomposition works because area is additive: if two regions don’t overlap, the total area is the sum of their areas. Many problems are really “add and subtract” questions—compute a large area and subtract cutouts, or assemble several smaller areas.

A reliable process:

  1. Mark all known lengths.
  2. Draw helpful lines to create rectangles and right triangles.
  3. Compute areas of parts.
  4. Add/subtract carefully.

Common traps include double-counting an overlapping region and assuming a diagonal creates two equal-area triangles when that’s only guaranteed in special cases (like a diagonal of a parallelogram/rectangle).

Area of common plane figures

You’re usually given enough information to compute area directly, but sometimes you must rebuild missing measures using angle facts, the Pythagorean Theorem, or similarity.

Rectangles and parallelograms

A rectangle is a parallelogram with right angles. A parallelogram has opposite sides parallel. In both, area is “base times height,” where the height is the perpendicular distance between parallel sides (not necessarily a side length).

A_{\text{rectangle}}=lw

A_{\text{parallelogram}}=bh

Common trap: using a slanted side as the height in a non-rectangle parallelogram. The height must be perpendicular to the base.

Triangles

A triangle’s area is half the area of a parallelogram with the same base and height.

A_{\triangle}=\frac{1}{2}bh

The height must be perpendicular to the chosen base; a frequent SAT mistake is using a slanted side as “height” when it isn’t perpendicular. If you’re given two sides and the included angle, you might be tempted to use a trig area formula, but SAT problems more often expect you to drop an altitude and use right-triangle relationships.

Trapezoids

A trapezoid has at least one pair of parallel sides (the bases). Its area is the average of the bases times the height.

A_{\text{trapezoid}}=\frac{1}{2}(b_1+b_2)h

The height is the perpendicular distance between the bases; do not use a slanted leg unless it’s perpendicular.

Circles (area)

Circle area depends on the square of the radius.

A_{\text{circle}}=\pi r^2

Worked example (composite area)

A rectangle is 10 by 6. A semicircle with diameter 6 is removed from one side (so radius 3). Find remaining area.

1) Rectangle area:

A_R=10\cdot 6=60

2) Semicircle area (half a circle):

A_S=\frac{1}{2}\pi r^2=\frac{1}{2}\pi\cdot 3^2=\frac{9\pi}{2}

3) Remaining area:

A=60-\frac{9\pi}{2}

The main thinking step is recognizing the parts that make up the figure, not algebra.

Volume of common solids (and a unifying idea)

A powerful way to reduce mistakes is to remember:

Volume = (area of cross-section) × (height)

This viewpoint makes prisms and cylinders feel consistent.

Rectangular prisms

V_{\text{rect prism}}=lwh

Triangular prisms

If the triangular base has area %%LATEX9%% and the prism length is %%LATEX10%%:

V_{\text{tri prism}}=A_{\triangle}L

Cylinders

A cylinder is a “circular prism,” so base area is \pi r^2.

V_{\text{cylinder}}=\pi r^2h

Cones and pyramids

A cone/pyramid has one-third the volume of the corresponding cylinder/prism with the same base and height.

V_{\text{cone}}=\frac{1}{3}\pi r^2h

V_{\text{pyramid}}=\frac{1}{3}Bh

Here %%LATEX16%% is the area of the base (any shape), and %%LATEX17%% is the perpendicular height.

Spheres

A sphere’s volume depends on the cube of the radius.

V_{\text{sphere}}=\frac{4}{3}\pi r^3

Surface area (when it appears)

Surface area is the total area of the outer faces of a 3D solid. SAT problems often test whether you can identify which faces exist and avoid counting internal faces.

Rectangular prism surface area

SA_{\text{rect prism}}=2(lw+lh+wh)

Cylinder surface area

A cylinder’s surface area includes two circles plus the “wrapped” rectangle (circumference times height).

SA_{\text{cylinder}}=2\pi r^2+2\pi rh

Cone surface area (slant height matters)

Cone surface area uses slant height %%LATEX21%%, not the vertical height. Sometimes %%LATEX22%% is given; other times you find it using the Pythagorean Theorem.

SA_{\text{cone}}=\pi r^2+\pi r\ell

Sphere surface area

SA_{\text{sphere}}=4\pi r^2

Scaling: how dimensions change area and volume

Scaling is a high-yield SAT idea. If every length in a figure is multiplied by a scale factor %%LATEX25%%, then perimeter scales by %%LATEX26%%, area scales by %%LATEX27%%, and volume scales by %%LATEX28%%. This isn’t random: area is built from two independent length dimensions and volume from three.

\text{Area scales by }k^2

\text{Volume scales by }k^3

Worked example (scaling)

A cube has side length 3. A similar cube has side length 6.

1) Scale factor:

k=\frac{6}{3}=2

2) Volume scales by:

k^3=8

Original volume:

V_1=3^3=27

New volume:

V_2=6^3=216

Check ratio:

\frac{216}{27}=8

Worked example (volume and unit conversion)

A cylindrical water tank has radius 2 meters and height 3 meters. What is its volume in cubic meters?

Use cylinder volume:

V=\pi r^2h=\pi(2^2)(3)=12\pi

So the volume is 12\pi cubic meters. If a problem mixes units, convert first to avoid power-of-ten mistakes—especially because volume conversions cube.

Exam Focus
  • Typical question patterns:
    • Composite figures: “Find the shaded area” by adding/subtracting rectangles, triangles, semicircles.
    • Solids: identify whether it’s asking for volume vs. surface area (often disguised in context).
    • Volume via base area: prisms/cylinders where you must compute the base area first.
    • Scale changes: “If dimensions double, what happens to area/volume?” (or “If the radius doubles…”)
  • Common mistakes:
    • Using a non-perpendicular side as triangle/parallelogram/trapezoid height.
    • Mixing radius and diameter (especially in circle-based solids).
    • Mixing units (feet with inches, cm with m), especially for volume where conversion factors cube.
    • Confusing surface area with volume (wrapping vs. filling).
    • Scaling area by %%LATEX38%% instead of %%LATEX39%%, or volume by %%LATEX40%% instead of %%LATEX41%%.

Lines, Angles, and Triangles

This topic is the language of geometry. Lines and angles provide structure; triangles are the “building blocks” because many polygons and real-world structures can be split into triangles. On the SAT, you’re usually asked to infer missing measures using a small set of angle facts and triangle properties.

Lines and angles: the basic relationships

An angle measures rotation. On the SAT, angles are almost always in degrees.

When lines intersect, several relationships immediately give equations:

  • Vertical angles (opposite angles formed by intersecting lines) are equal.
  • A linear pair sums to 180 degrees because it forms a straight line.
  • Angles around a point sum to 360 degrees.

A useful translation into algebra is: if two angles form a straight line, then:

a+b=180

These facts matter because they let you convert a diagram into equations.

Parallel lines and transversals

When a transversal cuts two parallel lines:

  • Corresponding angles are equal.
  • Alternate interior angles are equal.
  • Same-side interior angles are supplementary (sum to 180 degrees).

A common SAT move is to give one angle and ask for another far away. Your job is to “walk” the angle through the diagram using equality and supplementary relationships. If you’re unsure, sketch quickly and mark which angles are equal versus supplementary before writing an equation.

Worked example (parallel lines: supplementary)

Two parallel lines are cut by a transversal. One angle is 65 degrees. Find the same-side interior angle on the other intersection.

Same-side interior angles are supplementary:

x+65=180

x=115

Worked example (parallel lines with expressions)

Two parallel lines are cut by a transversal. A corresponding angle is labeled %%LATEX45%% degrees and its corresponding partner is labeled %%LATEX46%% degrees. Find x.

Corresponding angles are equal:

3x+10=5x-30

Solve:

40=2x

x=20

A frequent error is choosing the wrong relationship (for example, adding to 180 when the angles should be equal).

Coordinate geometry: slope and distance (as line tools)

SAT geometry often happens on the coordinate plane.

Slope measures steepness (rise over run). For points %%LATEX51%% and %%LATEX52%%:

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

Interpretation matters: positive slope rises left to right, negative slope falls left to right, slope 0 is horizontal, and undefined slope is vertical.

Two key line facts:

  • Parallel lines have equal slopes.
  • Perpendicular (non-vertical) lines have slopes that are negative reciprocals, which can be written as:

m_1m_2=-1

Distance between two points comes from the Pythagorean Theorem:

d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}

Common traps include forgetting to square differences before adding, swapping coordinate order inconsistently, or dropping the square root too early.

Worked example (distance on the coordinate plane)

Find the distance between %%LATEX56%% and %%LATEX57%%.

d=\sqrt{(3-(-1))^2+(5-2)^2}=\sqrt{4^2+3^2}=\sqrt{25}=5

Triangle angle facts

Two foundational facts drive most triangle questions.

1) Sum of interior angles:

A+B+C=180

2) Exterior angle theorem: an exterior angle equals the sum of the two remote interior angles.

\text{Exterior angle}=A+B

These are powerful because they create equations even when a triangle is embedded in a larger figure.

Special triangle types and what they guarantee

  • Isosceles triangle: at least two equal sides, and the angles opposite those equal sides are equal.
  • Equilateral triangle: all sides equal, so all angles equal; each angle is 60 degrees.

The SAT often gives an isosceles marking and expects you to set base angles equal, then use the 180-sum.

Congruence vs. similarity (don’t blur them)

Two triangles are congruent if they are exactly the same size and shape. Two triangles are similar if they have the same shape but may be different sizes (equal corresponding angles, proportional corresponding sides). SAT problems frequently rely on similarity to set up proportions.

Common congruence criteria you may use to justify correspondences:

  • SSS
  • SAS
  • ASA

For similarity, a very common SAT criterion is:

  • AA similarity: if two angles match, the triangles are similar.

Similar triangles and scale factor

Once triangles are similar, corresponding side ratios are equal. If the scale factor from triangle 1 to triangle 2 is %%LATEX61%%, each corresponding side is multiplied by %%LATEX62%%.

Example idea: if a small triangle has sides 3, 4, 5 and the scale factor is 2, the larger triangle’s sides are 6, 8, 10.

Worked example (similar triangles: 3-4-5 scaled)

Triangle 1 has sides 3, 4, 5. A similar triangle has the side corresponding to 3 equal to 12. Find its side corresponding to 5.

Scale factor:

k=\frac{12}{3}=4

So the corresponding side is:

5\cdot 4=20

Worked example (using similarity for missing length)

Two triangles are similar. In the smaller triangle, a side is 6 and the corresponding side in the larger triangle is 15. Another side in the smaller triangle is 10. Find the corresponding side in the larger triangle.

1) Scale factor:

k=\frac{15}{6}=\frac{5}{2}

2) Multiply the corresponding side:

10\cdot \frac{5}{2}=25

A common mistake is flipping the scale factor and shrinking when you should grow.

Perimeter and triangle inequality

Perimeter is the sum of side lengths.

In any triangle with sides %%LATEX67%%, %%LATEX68%%, c, the triangle inequality must hold:

a+b>c

a+c>b

b+c>a

This appears when the SAT asks for possible side lengths or ranges.

Exam Focus
  • Typical question patterns:
    • Parallel lines: “Find x” using corresponding/alternate interior (equal) or same-side interior (supplementary) relationships.
    • Triangle angles: use the 180-sum or exterior angle theorem to find missing angles.
    • Similar triangles: set up proportions to find an unknown side length.
    • Coordinate geometry: use slope for parallel/perpendicular or distance for lengths.
  • Common mistakes:
    • Confusing corresponding/alternate interior (equal) with same-side interior (sum to 180).
    • Assuming “looks isosceles” without given equal sides or equal angles.
    • Setting up a similarity ratio with mismatched corresponding sides, or flipping “small-to-large” versus “large-to-small.”
    • Computing slope with swapped coordinates inconsistently (keep point order consistent).

Right Triangles and Trigonometry

Right triangles are a major SAT tool because they connect geometry (shapes) to algebra (equations). Trigonometry on the SAT is almost always right-triangle trig, not advanced identities.

Why right triangles are the bridge to many SAT problems

A right triangle has one 90-degree angle. Right triangles matter because they turn geometry into arithmetic: lengths relate through the Pythagorean Theorem, and angles relate to side ratios through trigonometry.

Many “non-right” problems become right-triangle problems when you draw an altitude (a perpendicular segment). Being willing to add that helpful line is often the whole trick.

Pythagorean Theorem

In a right triangle with legs %%LATEX74%% and %%LATEX75%% and hypotenuse c (opposite the right angle):

a^2+b^2=c^2

Use it to find a missing side when you know the other two. It’s also a recognition tool: if three side lengths satisfy the equation, the triangle is right. A key habit is labeling the hypotenuse correctly: it is always the longest side and always opposite the 90-degree angle.

Worked example (Pythagorean)

A right triangle has legs 9 and 12. Find the hypotenuse.

c^2=9^2+12^2=81+144=225

c=15

Common trap: treating c as any side. It must be the longest side.

Special right triangles (high-yield)

Special right triangles save time and prevent mistakes.

45-45-90 triangle

This comes from cutting a square along its diagonal. The legs are equal. If each leg is x, then:

\text{hypotenuse}=x\sqrt{2}

Side ratio:

1:1:\sqrt{2}

30-60-90 triangle

This comes from splitting an equilateral triangle in half. If the shortest leg (opposite 30 degrees) is %%LATEX84%%, then the hypotenuse is %%LATEX85%% and the longer leg is x\sqrt{3}.

Ratio:

1:\sqrt{3}:2

Memory aid: in 30-60-90, the hypotenuse is double the short; the remaining side uses \sqrt{3}.

Common trap: mixing up which side is opposite 30 degrees. The shortest side is always opposite the smallest angle.

Trigonometric ratios (SOH-CAH-TOA)

Trigonometry connects an acute angle in a right triangle to ratios of side lengths. For an acute angle \theta:

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

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

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

“Opposite” means across from the angle %%LATEX93%%. “Adjacent” means next to %%LATEX94%% but not the hypotenuse. A frequent confusion is “adjacent to what?” Adjacent is defined relative to the specific angle you’re using.

A useful memory aid is SOH-CAH-TOA (Sine Opposite Hypotenuse, Cosine Adjacent Hypotenuse, Tangent Opposite Adjacent).

Using trig ratios to find missing sides

The workflow is consistent:

  1. Identify the angle \theta you’re using.
  2. Label opposite, adjacent, hypotenuse relative to \theta.
  3. Choose the ratio involving the known and unknown sides.
  4. Solve the equation.
Worked example (trig for a missing side)

In a right triangle, %%LATEX97%% degrees. The hypotenuse is 10. Find the side opposite %%LATEX98%%.

Use sine:

\sin(30)=\frac{\text{opposite}}{10}

Use the known value:

\sin(30)=\frac{1}{2}

So:

\frac{1}{2}=\frac{\text{opposite}}{10}

\text{opposite}=5

Trig values from special triangles (a conceptual way to remember)

Rather than memorizing a table, derive key values:

  • In a 30-60-90 triangle with hypotenuse 2, the legs are 1 and \sqrt{3}.
  • In a 45-45-90 triangle with legs 1, the hypotenuse is \sqrt{2}.

From these:

\sin(30)=\frac{1}{2}

\cos(30)=\frac{\sqrt{3}}{2}

\sin(60)=\frac{\sqrt{3}}{2}

\cos(60)=\frac{1}{2}

\sin(45)=\frac{\sqrt{2}}{2}

\cos(45)=\frac{\sqrt{2}}{2}

Then tangent is sine divided by cosine, for example:

\tan(45)=1

Common trap: confusing which functions use the hypotenuse (sine and cosine do; tangent does not).

Angles of elevation and depression (real-world trig setup)

An angle of elevation is measured up from a horizontal line of sight; an angle of depression is measured down from a horizontal line. Both create right triangles with horizontal and vertical legs.

A common setup is standing %%LATEX112%% units from a building and looking up at angle %%LATEX113%% to the top:

\tan(\theta)=\frac{\text{height}}{d}

SAT often keeps numbers friendly so you can use special angles or simple algebra.

Worked example (elevation)

You are 20 feet from a flagpole. The angle of elevation to the top is 45 degrees. Find the height of the pole.

\tan(45)=\frac{h}{20}

Since:

\tan(45)=1

We get:

1=\frac{h}{20}

h=20

When trig is unnecessary

Some problems hand you a right triangle with angles 30, 60, or 45. In those cases, special triangles are often faster and more exact than approximating trig values. If the problem expects an exact radical answer, that’s usually your clue to use special-triangle ratios.

Exam Focus
  • Typical question patterns:
    • Find a missing side with the Pythagorean Theorem or special right triangle ratios.
    • Use %%LATEX119%%, %%LATEX120%%, or \tan to relate an angle to side lengths.
    • Word problems with elevation/depression or ladder-against-a-wall setups.
  • Common mistakes:
    • Mislabeling the hypotenuse or mixing up opposite vs. adjacent relative to \theta.
    • Using the wrong trig function (for example, sine when you need tangent).
    • Mixing up 30-60-90 side assignments (short leg opposite 30 degrees).
    • Rounding too early when the test expects an exact radical form.

Circles

Circles show up in SAT geometry both as measurement problems and as relationship problems involving chords, tangents, arcs, and angles. The core idea is that a circle is the set of all points in a plane at a fixed distance from a center.

The basic circle vocabulary

A circle is the set of points in a plane that are all the same distance from a fixed point (the center). That distance is the radius %%LATEX123%%. The diameter %%LATEX124%% is twice the radius:

d=2r

A chord connects two points on the circle (a diameter is a chord through the center). A tangent touches the circle at exactly one point. A secant cuts through the circle, intersecting it twice.

Circumference and area

The circumference is the distance around the circle:

C=2\pi r

The area is:

A=\pi r^2

A key insight for comparison questions: doubling r doubles circumference but quadruples area.

Arcs and sectors: fractions of a circle

An arc is part of the circumference. A sector is a “slice” of the circle. For SAT problems in degrees, use the fraction \theta/360.

Arc length:

L=\frac{\theta}{360}\cdot 2\pi r

Sector area:

A_{\text{sector}}=\frac{\theta}{360}\cdot \pi r^2

Common traps include using 180 instead of 360, mixing radius and diameter, or using the wrong fraction.

Worked example (arc length)

A circle has radius 6 and central angle 120 degrees. Find the arc length.

L=\frac{120}{360}\cdot 2\pi\cdot 6

L=\frac{1}{3}\cdot 12\pi

L=4\pi

Worked example (sector area)

A circle has radius 6. Find the area of a 120 degree sector.

A_{\text{sector}}=\frac{120}{360}\cdot \pi(6^2)

A_{\text{sector}}=\frac{1}{3}\cdot 36\pi=12\pi

Central angles vs. inscribed angles (the “half” relationship)

A central angle has its vertex at the center. An inscribed angle has its vertex on the circle. If they intercept the same arc, then the inscribed angle is half the central angle.

\text{inscribed angle}=\frac{1}{2}(\text{central angle})

Equivalently, if an arc has measure %%LATEX138%% degrees, the inscribed angle intercepting that arc measures %%LATEX139%%. A common trap is mixing up which arc is intercepted; it’s the arc between the points where the angle’s sides meet the circle, on the inside of the angle.

Worked example (inscribed angle)

A central angle intercepting an arc measures 80 degrees. An inscribed angle intercepts the same arc. Find the inscribed angle.

\text{inscribed}=\frac{1}{2}\cdot 80=40

Chords and symmetry (distance from center matters)

Two high-utility chord facts:

  • Equal chords are the same distance from the center.
  • A radius (or diameter) drawn perpendicular to a chord bisects the chord.

This is useful because it creates right triangles you can solve with the Pythagorean Theorem.

Worked example (chord length from center distance)

A circle has radius 10. A chord is 6 units from the center (perpendicular distance). Find the chord length.

Draw a radius to the chord’s midpoint; it’s perpendicular and bisects the chord. Let half the chord be x. Then:

x^2+6^2=10^2

x^2+36=100

x^2=64

x=8

Chord length is 2x=16.

Tangents: a right angle you can trust (and equal tangents)

A tangent touches the circle at exactly one point. The key SAT theorem is: a tangent is perpendicular to the radius drawn to the point of tangency.

Written conceptually:

\text{radius}\perp\text{tangent}

That gives you an immediate right triangle whenever you see a tangent and a radius.

Another common SAT fact: tangent segments from the same external point are equal in length. If two tangents are drawn from point P to a circle, then:

PA=PB

Worked example (equal tangents)

From external point %%LATEX150%%, two tangents touch a circle at %%LATEX151%% and %%LATEX152%%. If %%LATEX153%% and %%LATEX154%%, find %%LATEX155%%.

Set them equal:

3x+1=2x+9

x=8

Circle equations on the coordinate plane

On the SAT, circle equations usually appear in standard form. A circle with center %%LATEX158%% and radius %%LATEX159%% has equation:

(x-h)^2+(y-k)^2=r^2

Special case (center at the origin):

x^2+y^2=r^2

This comes from the distance formula: points on the circle are exactly distance r from the center.

Worked example (equation of a circle)

A circle has center (2,-1) and radius 5. Write its equation.

(x-2)^2+(y+1)^2=25

Common trap: sign errors. The equation uses %%LATEX165%% and %%LATEX166%%, so if %%LATEX167%% then %%LATEX168%% becomes y+1.

Worked example (center and radius from an equation)

Given:

(x-2)^2+(y+1)^2=25

The center is (2,-1) and the radius is 5 because:

r^2=25

r=5

A common mistake is claiming the center is %%LATEX174%% by “flipping signs” without using the structure %%LATEX175%% and y-k.

Exam Focus
  • Typical question patterns:
    • Compute circumference/area, or arc length/sector area from a central angle using the \theta/360 proportion.
    • Use relationships: inscribed angle is half the corresponding central angle; tangent-radius perpendicular; equal tangents from one external point.
    • Use chord facts (perpendicular from center bisects chord) to create right triangles.
    • Interpret or rewrite a circle equation to find center and radius.
  • Common mistakes:
    • Mixing radius and diameter in circumference/area/arc formulas.
    • Using the wrong fraction for sectors (forgetting denominator 360 for degrees).
    • Forgetting inscribed angles are half the intercepted arc (not equal to it).
    • Sign errors when reading centers from equations, especially with negatives.