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SAT Math

SAT Math: Geometry & Trigonometry — Complete Study Guide

Master area, volume, lines, angles, triangles, circles, and right triangle trigonometry for the Digital SAT. Includes all reference formulas provided on the test.

Last updated: February 10, 2026Estimated reading time: 1.5 hours

Important Note: Reference Sheet

The Digital SAT gives you a math reference sheet in the testing app. You should still memorize these formulas because speed matters, but it helps to know the formulas are available during the test.

These are the key formulas from the official reference sheet:

  • Area of circle: A=πr2A = \pi r^2
  • Circumference: C=2πrC = 2\pi r
  • Area of rectangle: A=lwA = lw
  • Area of triangle: A=12bhA = \frac{1}{2}bh
  • Pythagorean theorem: a2+b2=c2a^2 + b^2 = c^2
  • Special right triangles: 3030-6060-9090 has sides (x, x3, 2x)(x,\ x\sqrt{3},\ 2x) and 4545-4545-9090 has sides (x, x, x2)(x,\ x,\ x\sqrt{2})
  • Volume of rectangular prism: V=lwhV = lwh
  • Volume of cylinder: V=πr2hV = \pi r^2h
  • Volume of sphere: V=43πr3V = \frac{4}{3}\pi r^3
  • Volume of cone: V=13πr2hV = \frac{1}{3}\pi r^2h
  • Volume of pyramid: V=13lwhV = \frac{1}{3}lwh

Geometry and trigonometry usually contribute about 5 to 7 questions, but these are high-leverage points because many students lose easy points to diagram mistakes, unit confusion, and rushed angle logic. This guide focuses on the exact question patterns the SAT repeats.

Strategy tip: For every geometry question, write the target quantity first (area, angle, side length, volume, or expression). That one habit prevents many wrong turns.


1. Area and Volume

Geometry questions often mix basic formulas with context. You may need to break a shape into parts, convert units, or interpret a word problem that hides a familiar formula.

2D area calculations

Common 2D area formulas on SAT:

  • Rectangle: A=lwA = lw
  • Triangle: A=12bhA = \frac{1}{2}bh
  • Circle: A=πr2A = \pi r^2
  • Trapezoid: A=12(b1+b2)hA = \frac{1}{2}(b_1+b_2)h

For composite figures, split into simple pieces (rectangles, triangles, semicircles), calculate each area, then add or subtract.

3D volume calculations

Most SAT volume questions use direct substitution:

  • Rectangular prism: V=lwhV = lwh
  • Cylinder: V=πr2hV = \pi r^2h
  • Cone: V=13πr2hV = \frac{1}{3}\pi r^2h
  • Sphere: V=43πr3V = \frac{4}{3}\pi r^3
  • Pyramid: V=13lwhV = \frac{1}{3}lwh (for rectangular base dimensions ll and ww)

Watch whether the problem gives radius or diameter.

Surface area and density

Surface area sometimes appears as context (paint needed, wrapping material, exposed faces). For rectangular prisms:

SA=2(lw+lh+wh)SA = 2(lw + lh + wh)

For cylinders:

SA=2πr2+2πrhSA = 2\pi r^2 + 2\pi rh

Density model:

density=massvolume\text{density} = \frac{\text{mass}}{\text{volume}}

So you can also use:

mass=densityvolume\text{mass} = \text{density} \cdot \text{volume}

Worked Example 1: Composite area

Worked Example 1: Composite floor plan area

An L-shaped room is made from a 14 m×10 m14\text{ m} \times 10\text{ m} rectangle with a 6 m×4 m6\text{ m} \times 4\text{ m} rectangular corner cut out. What is the room's area?

  1. Area of large rectangle:

    1410=140 m214\cdot10 = 140\text{ m}^2
  2. Area of cut-out rectangle:

    64=24 m26\cdot4 = 24\text{ m}^2
  3. Subtract to get L-shape area:

    14024=116 m2140 - 24 = 116\text{ m}^2

Final answer:

116 m2\boxed{116\text{ m}^2}

Worked Example 2: Volume with context

Worked Example 2: Water tank volume

A cylindrical water tank has radius 33 feet and height 1212 feet. How many cubic feet of water does it hold when full? Use π3.14\pi \approx 3.14.

  1. Use cylinder volume formula:

    V=πr2hV = \pi r^2h
  2. Substitute values:

    V=3.143212V = 3.14\cdot 3^2 \cdot 12
  3. Simplify:

    32=9,912=1083^2 = 9,\quad 9\cdot 12 = 108 V=3.14108=339.12V = 3.14\cdot108 = 339.12

So the tank holds:

339.12 ft3\boxed{339.12\text{ ft}^3}

If asked in gallons, you would then apply a conversion factor.

Worked Example 3: Density

Worked Example 3: Density of a metal block

A metal block is a rectangular prism with dimensions 5 cm5\text{ cm} by 4 cm4\text{ cm} by 3 cm3\text{ cm}. Its mass is 480 g480\text{ g}. What is its density in g/cm3^3?

  1. Find volume:

    V=lwh=543=60 cm3V = lwh = 5\cdot4\cdot3 = 60\text{ cm}^3
  2. Use density formula:

    d=mV=48060=8d = \frac{m}{V} = \frac{480}{60} = 8

Density is:

8 g/cm3\boxed{8\text{ g/cm}^3}

Strategy tip: In geometry contexts, always include units in intermediate steps. Unit tracking catches many setup errors before they cost points.


2. Lines, Angles, and Polygons

Angle relationships are a frequent SAT geometry target because they test structure recognition more than calculation.

Core angle relationships

  • Complementary angles sum to 9090^\circ.
  • Supplementary angles sum to 180180^\circ.
  • Vertical angles are equal.

When parallel lines are cut by a transversal:

  • Corresponding angles are equal.
  • Alternate interior angles are equal.
  • Same-side interior (co-interior) angles are supplementary.

Polygon angle sums

Interior angle sum of an nn-gon:

(n2)180(n-2)\cdot180^\circ

For regular polygons (all equal angles), each interior angle is:

(n2)180n\frac{(n-2)\cdot180^\circ}{n}

Exterior angle theorem (triangle)

An exterior angle of a triangle equals the sum of the two remote interior angles.

If exterior angle is EE and remote interior angles are aa and bb:

E=a+bE = a+b

Worked Example 4: Parallel lines and transversal

Worked Example 4: Find x from angle relationships

Lines mm and nn are parallel. A transversal crosses them. One angle is labeled (4x+10)(4x+10)^\circ and its alternate interior partner is (6x20)(6x-20)^\circ. Find xx.

  1. Alternate interior angles are equal (because lines are parallel):

    4x+10=6x204x+10 = 6x-20
  2. Solve:

    30=2x30 = 2x x=15x = 15

Check angle measure:

4(15)+10=70,6(15)20=704(15)+10 = 70^\circ,\quad 6(15)-20 = 70^\circ

Answer:

x=15\boxed{x=15}

Worked Example 5: Polygon angles

Worked Example 5: Interior angle of regular nonagon

A regular nonagon has n=9n=9 sides. Find each interior angle.

  1. Interior angle sum:

    (n2)180=(92)180=7180=1260(n-2)\cdot180 = (9-2)\cdot180 = 7\cdot180 = 1260^\circ
  2. Since the nonagon is regular, divide equally among 9 angles:

    12609=140\frac{1260}{9} = 140^\circ

Each interior angle is:

140\boxed{140^\circ}

Strategy tip: If you forget a polygon formula, triangulate from one vertex. You can see why an nn-gon splits into (n2)(n-2) triangles.


3. Triangles

Triangles are central in SAT geometry. Most questions use one of a small number of repeatable tools.

Triangle inequality theorem

For side lengths aa, bb, and cc:

a+b>c,a+c>b,b+c>aa+b>c,\quad a+c>b,\quad b+c>a

Equivalent shortcut: longest side must be less than sum of other two.

Pythagorean theorem and converse

For right triangle with legs a,ba,b and hypotenuse cc:

a2+b2=c2a^2+b^2=c^2

Converse: if side lengths satisfy this equation, the triangle is right.

Special right triangles

  • 4545-4545-9090: sides x,x,x2x, x, x\sqrt{2}
  • 3030-6060-9090: sides x,x3,2xx, x\sqrt{3}, 2x (short leg, long leg, hypotenuse)

Similar triangles

Triangles are similar if corresponding angles are equal, and side lengths are proportional.

Common similarity criteria:

  • AA similarity
  • SAS similarity
  • SSS similarity

Proportion setup example:

a1a2=b1b2=c1c2\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}

Congruent triangles

Congruence means same shape and same size (all corresponding sides and angles equal). SAT may test recognition in diagrams.

Triangle area beyond base-height

Standard formula:

A=12bhA=\frac{1}{2}bh

Some contexts require finding height from other given values first.

Worked Example 6: Pythagorean theorem application

Worked Example 6: Ladder against a wall

A ladder reaches a point 1212 feet high on a wall. The base of the ladder is 55 feet from the wall. How long is the ladder?

  1. Model as right triangle:

    • Legs: 1212 and 55
    • Hypotenuse: ladder length cc
  2. Apply Pythagorean theorem:

    122+52=c212^2 + 5^2 = c^2 144+25=c2144 + 25 = c^2 169=c2169 = c^2
  3. Take square root:

    c=13c=13

Ladder length:

13 feet\boxed{13\text{ feet}}

Worked Example 7: Special right triangle

Worked Example 7: 30-60-90 triangle

In a 3030-6060-9090 triangle, the shorter leg is 77. Find the longer leg and hypotenuse.

For 3030-6060-9090, side ratio is:

x:x3:2xx : x\sqrt{3} : 2x

Given shorter leg x=7x=7:

  • Longer leg =73= 7\sqrt{3}
  • Hypotenuse =14= 14

Answer:

longer leg =73, hypotenuse =14\boxed{\text{longer leg }=7\sqrt{3},\ \text{hypotenuse }=14}

Worked Example 8: Similar triangle proportions

Worked Example 8: Solve side in similar triangles

Two triangles are similar. In the smaller triangle, corresponding sides are 66, 88, and 1010. In the larger triangle, the side corresponding to 66 is 1515. Find the side corresponding to 1010 in the larger triangle.

  1. Find scale factor from small to large:

    k=156=52k = \frac{15}{6} = \frac{5}{2}
  2. Apply to side 10:

    1052=2510\cdot\frac{5}{2}=25

Answer:

25\boxed{25}

Strategy tip: For similarity questions, choose one clear corresponding pair first to get the scale factor. Then apply that same factor consistently.


4. Right Triangle Trigonometry

SAT trigonometry is practical and focused. You mainly need SOH-CAH-TOA with degree mode.

SOH-CAH-TOA definitions

For an acute angle θ\theta in a right triangle:

sin(θ)=oppositehypotenuse\sin(\theta)=\frac{\text{opposite}}{\text{hypotenuse}} cos(θ)=adjacenthypotenuse\cos(\theta)=\frac{\text{adjacent}}{\text{hypotenuse}} tan(θ)=oppositeadjacent\tan(\theta)=\frac{\text{opposite}}{\text{adjacent}}

Complementary angle relationship

In right triangles, acute angles are complementary:

sin(x)=cos(90x)\sin(x)=\cos(90^\circ-x)

and similarly:

cos(x)=sin(90x)\cos(x)=\sin(90^\circ-x)

Use cases

  • Find missing side given angle and one side.
  • Find angle given side ratios.
  • Model real contexts (height, slope, distance, elevation).

Angles of elevation and depression

  • Elevation: angle measured upward from horizontal.
  • Depression: angle measured downward from horizontal.

SAT usually gives enough info to form a right triangle and apply one trig ratio.

Degree mode note

The SAT uses degrees, not radians, for trig in standard questions. Ensure calculator is in degree mode.

Worked Example 9: Find a side using trig

Worked Example 9: Using sine to find opposite side

In a right triangle, θ=35\theta=35^\circ and hypotenuse is 1818. Find the side opposite θ\theta.

  1. Use sine ratio:

    sin(35)=opposite18\sin(35^\circ)=\frac{\text{opposite}}{18}
  2. Solve for opposite side oo:

    o=18sin(35)o = 18\sin(35^\circ)
  3. Approximate with calculator:

    o18(0.5736)=10.3248o \approx 18(0.5736)=10.3248

So opposite side is about:

10.3\boxed{10.3}

(to nearest tenth)

Worked Example 10: Angle of elevation

Worked Example 10: Building height from angle

From a point on the ground, the angle of elevation to the top of a building is 4040^\circ. The observer is 3030 meters from the building base. Find the building height.

  1. Draw right triangle:

    • Adjacent side: 3030
    • Opposite side: building height hh
    • Angle at observer: 4040^\circ
  2. Use tangent:

    tan(40)=h30\tan(40^\circ)=\frac{h}{30}
  3. Solve:

    h=30tan(40)h=30\tan(40^\circ)
  4. Approximate:

    h30(0.8391)=25.173h\approx 30(0.8391)=25.173

Building height is about:

25.2 m\boxed{25.2\text{ m}}

Worked Example 11: Complementary angle relationship

Worked Example 11: Use sin(x)=cos(90-x)

If sin(x)=0.8\sin(x)=0.8 for an acute angle xx, find cos(90x)\cos(90^\circ-x).

Using complementary identity:

sin(x)=cos(90x)\sin(x)=\cos(90^\circ-x)

So directly:

cos(90x)=0.8\cos(90^\circ-x)=0.8

Answer:

0.8\boxed{0.8}

Strategy tip: Many SAT trig questions become one-step problems if you choose the ratio that uses known and target sides immediately.


5. Circles

Circle questions combine algebra and geometry and appear in both modules.

Circle equation

Standard form of a circle:

(xh)2+(yk)2=r2(x-h)^2 + (y-k)^2 = r^2
  • Center: (h,k)(h,k)
  • Radius: rr

Converting from general form

General quadratic form in xx and yy:

x2+y2+Dx+Ey+F=0x^2 + y^2 + Dx + Ey + F = 0

Convert by grouping xx and yy terms and completing the square.

Arc length and sector area

For central angle θ\theta in degrees:

L=θ3602πrL = \frac{\theta}{360}\cdot 2\pi r Asector=θ360πr2A_{\text{sector}} = \frac{\theta}{360}\cdot \pi r^2

Central and inscribed angles

  • Central angle intercepts arc with same measure as arc.
  • Inscribed angle intercepting same arc is half the central angle.

Tangent lines

A tangent line is perpendicular to radius at point of tangency.

If radius to tangent point forms one side and tangent line forms another, the angle is 9090^\circ.

Radian basics

SAT is mostly degree-based, but conversion may appear:

180=π radians180^\circ = \pi\text{ radians}

So:

radians=degreesπ180\text{radians} = \text{degrees}\cdot\frac{\pi}{180}

Worked Example 12: Circle equation from conditions

Worked Example 12: Build equation from center and point

A circle has center (2,3)(2,-3) and passes through point (6,0)(6,0). Find its equation.

  1. Use standard form:

    (xh)2+(yk)2=r2(x-h)^2 + (y-k)^2 = r^2
  2. Plug center (h,k)=(2,3)(h,k)=(2,-3):

    (x2)2+(y+3)2=r2(x-2)^2 + (y+3)^2 = r^2
  3. Find r2r^2 from distance to point (6,0)(6,0):

    r2=(62)2+(0+3)2=42+32=16+9=25r^2 = (6-2)^2 + (0+3)^2 = 4^2 + 3^2 = 16+9=25
  4. Final equation:

    (x2)2+(y+3)2=25(x-2)^2 + (y+3)^2 = 25

Answer:

(x2)2+(y+3)2=25\boxed{(x-2)^2 + (y+3)^2 = 25}

Worked Example 13: Sector area in context

Worked Example 13: Pizza slice area

A pizza has radius 1010 inches. A slice has central angle 5454^\circ. What is the area of the slice?

  1. Use sector area formula:

    A=θ360πr2A = \frac{\theta}{360}\cdot\pi r^2
  2. Substitute:

    A=54360π102A = \frac{54}{360}\cdot\pi\cdot10^2
  3. Simplify:

    54360=320,102=100\frac{54}{360}=\frac{3}{20},\quad 10^2=100 A=320100π=15πA=\frac{3}{20}\cdot100\pi=15\pi

Area of slice:

15π in2\boxed{15\pi\text{ in}^2}

Worked Example 14: Tangent line problem

Worked Example 14: Radius to tangent is perpendicular

Circle center is OO. Point TT is on the circle, and line \ell is tangent at TT. If radius OT=9OT=9 and point PP lies on tangent line such that PT=12PT=12, find OPOP.

  1. Radius to tangent point is perpendicular to tangent line, so triangle OTPOTP is right at TT.

  2. Legs are:

    • OT=9OT=9
    • PT=12PT=12
  3. Use Pythagorean theorem:

    OP2=92+122=81+144=225OP^2 = 9^2 + 12^2 = 81+144=225
  4. Square root:

    OP=15OP=15

Answer:

15\boxed{15}

Strategy tip: In tangent problems, explicitly mark the right angle at tangency. That mark often unlocks the whole question.


Quick Reference

Geometry & Trigonometry Quick Formula Box

Area

Arectangle=lw,Atriangle=12bh,Acircle=πr2A_{\text{rectangle}}=lw,\quad A_{\text{triangle}}=\frac{1}{2}bh,\quad A_{\text{circle}}=\pi r^2

Circumference

C=2πrC=2\pi r

Pythagorean theorem

a2+b2=c2a^2+b^2=c^2

Special right triangles

30-60-90:(x, x3, 2x),45-45-90:(x, x, x2)30\text{-}60\text{-}90: (x,\ x\sqrt3,\ 2x),\quad 45\text{-}45\text{-}90: (x,\ x,\ x\sqrt2)

Volumes

Vprism=lwh,Vcylinder=πr2hV_{\text{prism}}=lwh,\quad V_{\text{cylinder}}=\pi r^2hVcone=13πr2h,Vsphere=43πr3,Vpyramid=13lwhV_{\text{cone}}=\frac13\pi r^2h,\quad V_{\text{sphere}}=\frac43\pi r^3,\quad V_{\text{pyramid}}=\frac13 lwh

Trig ratios

sinθ=opphyp,cosθ=adjhyp,tanθ=oppadj\sin\theta=\frac{\text{opp}}{\text{hyp}},\quad \cos\theta=\frac{\text{adj}}{\text{hyp}},\quad \tan\theta=\frac{\text{opp}}{\text{adj}}

Circle equation

(xh)2+(yk)2=r2(x-h)^2+(y-k)^2=r^2

Arc/sector

L=θ3602πr,Asector=θ360πr2L=\frac{\theta}{360}\cdot2\pi r,\quad A_{\text{sector}}=\frac{\theta}{360}\cdot\pi r^2

Desmos Tips (Graphing Circles)

Desmos can save time on coordinate-geometry circle questions.

  1. To graph a circle directly, type standard form:

    (xh)2+(yk)2=r2(x-h)^2 + (y-k)^2 = r^2
  2. To inspect center and radius quickly:

    • Rewrite equation into standard form first.
    • Then graph it and verify geometry visually.
  3. For intersection problems (line and circle):

    • Enter both equations.
    • Click intersection points to read coordinates.
  4. For equation-matching questions:

    • Graph answer choices one by one and compare center/intercepts/size.

Digital SAT Tips

  1. Geometry and trig are usually fewer questions, but these are highly score-efficient if mastered.
  2. Draw your own clean sketch when diagram is crowded.
  3. Label known values directly on the diagram.
  4. Check whether answers need exact form (like 15π15\pi) or decimal approximation.
  5. For trig problems, confirm degree mode on calculator.
  6. For circle equations, extract center and radius before doing anything else.
  7. In word problems, convert units before formula substitution.
  8. When possible, eliminate answer choices by estimation before full computation.

Strategy tip: If the question asks for an expression (for example, x+yx+y), do not stop at finding just one variable.


Practice Problems

Problem 1 (composite area)

A shape is made of a 12×912\times9 rectangle with a 4×34\times3 rectangle removed from one corner. What is its area?

A) 8484
B) 9090
C) 9696
D) 108108

Problem 2 (parallel lines)

Two parallel lines are cut by a transversal. Corresponding angles are labeled (3x+12)(3x+12)^\circ and (5x18)(5x-18)^\circ. What is xx?

A) 99
B) 1212
C) 1515
D) 1818

Problem 3 (special triangle)

In a 4545-4545-9090 triangle, the hypotenuse is 10210\sqrt2. What is each leg?

A) 55
B) 1010
C) 10210\sqrt2
D) 2020

Problem 4 (trig)

In a right triangle, angle θ=50\theta=50^\circ and adjacent side to θ\theta is 7. Which expression gives the hypotenuse hh?

A) h=7sin50h=7\sin50^\circ
B) h=7cos50h=7\cos50^\circ
C) h=7cos50h=\frac{7}{\cos50^\circ}
D) h=7tan50h=\frac{7}{\tan50^\circ}

Problem 5 (circle sector)

A circle has radius 12 cm. What is the area of a 6060^\circ sector?

A) 12π12\pi
B) 24π24\pi
C) 36π36\pi
D) 48π48\pi

Answers and Brief Explanations

1) C
Large rectangle area =129=108=12\cdot9=108. Removed area =43=12=4\cdot3=12. Composite area =10812=96=108-12=96.

2) C
Corresponding angles are equal:

3x+12=5x183x+12=5x-1830=2xx=1530=2x\Rightarrow x=15

3) B
In 4545-4545-9090, hypotenuse =x2=x\sqrt2. So x2=102x=10x\sqrt2=10\sqrt2\Rightarrow x=10.

4) C

cosθ=adjhyp=7h\cos\theta=\frac{\text{adj}}{\text{hyp}}=\frac{7}{h}

So h=7cos50h=\frac{7}{\cos50^\circ}.

5) B
Sector area:

60360π(12)2=16144π=24π\frac{60}{360}\cdot\pi(12)^2=\frac16\cdot144\pi=24\pi

Geometry and trigonometry on the SAT are predictable when you keep diagrams organized, formulas deliberate, and units consistent. Aim to turn these 5-7 questions into reliable points by practicing setup and interpretation, not just arithmetic.