Skip to content
🎯⭐ INTERACTIVE LESSON

Geometry and Trigonometry

Learn step-by-step with interactive practice!

Geometry and Trigonometry - Complete Interactive Lesson

Part 1: Lines & Angles

Geometry: Lines, Angles, and Triangles

Part 1 of 7 — Angle Relationships

Geometry accounts for roughly 10-15% of SAT Math questions. Mastering angle relationships gives you quick points.

Fundamental Angle Rules

  • Supplementary angles: Sum to 180°180°
  • Complementary angles: Sum to 90°90°
  • Vertical angles: Equal (formed by intersecting lines)
  • Angles on a straight line: Sum to 180°180°

Parallel Lines Cut by a Transversal

When a line crosses two parallel lines, it creates 8 angles with key relationships:

  • Corresponding angles are equal (same position at each intersection)
  • Alternate interior angles are equal (opposite sides, between parallels)
  • Alternate exterior angles are equal (opposite sides, outside parallels)
  • Co-interior (same-side interior) angles sum to 180°180°

Triangle Angle Sum

The angles in any triangle sum to 180°180°.

If a triangle has angles 55°55° and 70°70°: Third angle=180°−55°−70°=55°\text{Third angle} = 180° - 55° - 70° = 55°

This is an isosceles triangle (two equal angles → two equal sides).

Exterior Angle Theorem

An exterior angle of a triangle equals the sum of the two non-adjacent interior angles.

Exterior angle=Remote interior1+Remote interior2\text{Exterior angle} = \text{Remote interior}_1 + \text{Remote interior}_2

SAT Trap ⚠️

When the SAT shows a figure with parallel lines, check if they actually SAY the lines are parallel. "Looks parallel" ≠ IS parallel. Look for arrows or explicit statements.

Angle Relationships Practice 🎯

Deep Dive: Multi-Step Angle Problems

Worked Example 1: Parallel Lines with Algebra

StepWork
ProblemLines l∥ml \parallel m are cut by a transversal. One angle is (3x+10)°(3x + 10)° and its alternate interior angle is (5x−30)°(5x - 30)°. Find xx and the angle measure.
Set upAlternate interior angles are equal: 3x+10=5x−303x + 10 = 5x - 30
Solve40=2x40 = 2x → x=20x = 20
AnswerEach angle =3(20)+10=70°= 3(20) + 10 = 70° ✓

Worked Example 2: Exterior Angle with Algebra

StepWork
ProblemIn △ABC, ∠A=(2x+5)°\angle A = (2x + 5)°, ∠B=(x+10)°\angle B = (x + 10)°, exterior angle at C=(4x−5)°C = (4x - 5)°. Find all angles.
Apply theoremExterior == sum of remotes: 4x−5=(2x+5)+(x+10)4x - 5 = (2x + 5) + (x + 10)
Solve4x−5=3x+154x - 5 = 3x + 15 → x=20x = 20
AnglesA=45°A = 45°, B=30°B = 30°, C=180°−45°−30°=105°C = 180° - 45° - 30° = 105°
CheckExterior at CC: 4(20)−5=75°4(20) - 5 = 75° = 45°+30°45° + 30° ✓

Angle Relationship Quick Reference

RelationshipRuleHow to Spot
Vertical anglesEqualX shape at intersection
SupplementarySum to 180°180°Adjacent on straight line
CorrespondingEqualSame position at each parallel crossing
Alternate interiorEqualZ or S pattern between parallels
Co-interiorSum to 180°180°U or C pattern between parallels
Exterior angleSum of two remotesOutside vertex of triangle

Advanced Angle Problems 🎯

Identify the Angle Relationship — Name each angle pair.

Part 1 Summary: Angle Relationships

RuleFormula/FactWhen to Use
Supplementarya+b=180°a + b = 180°Angles on a straight line
Complementarya+b=90°a + b = 90°Corner/right angle split
Vertical anglesa=ba = bIntersecting lines
Triangle sumA+B+C=180°A + B + C = 180°Any triangle
Exterior angle=remote1+remote2= \text{remote}_1 + \text{remote}_2Angle outside triangle vertex
Parallel + transversalCorresponding & alternate = equal; co-interior sum to 180°180°Arrows on lines or stated parallel

SAT Strategy

  • If angles involve variables, set up an equation using the appropriate rule.
  • Always check: do the angles sum correctly?
  • Angles at a point sum to 360°360° — don't confuse with straight line (180°180°).

Next: Triangle properties, special right triangles, and similarity →

Part 2: Triangle Properties

Triangle Properties & Theorems

Part 2 of 7 — Special Triangles, Similarity, Congruence

Special Right Triangles

The SAT provides these in the reference sheet, but memorizing them saves time:

45-45-90 Triangle:

  • Legs: xx, xx
  • Hypotenuse: x2x\sqrt{2}

30-60-90 Triangle:

  • Short leg: xx (opposite 30°)
  • Long leg: x3x\sqrt{3} (opposite 60°)
  • Hypotenuse: 2x2x (opposite 90°)

Example

A 30-60-90 triangle has a hypotenuse of 10. Find the legs.

  • Hypotenuse =2x=10= 2x = 10 → x=5x = 5
  • Short leg =5= 5
  • Long leg =53≈8.66= 5\sqrt{3} \approx 8.66

Triangle Inequality Theorem

For any triangle with sides aa, bb, cc: a+b>ca + b > c

The sum of any two sides must exceed the third.

Example: Can a triangle have sides 3, 5, and 9?
3+5=8<93 + 5 = 8 < 9 → No!

Similar Triangles (AA Similarity)

If two angles of one triangle equal two angles of another, the triangles are similar (same shape, proportional sides).

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

SAT Trap ⚠️

In 30-60-90 triangles, students often mix up which leg is which. Remember: the shortest side is opposite the smallest angle (30°).

Triangle Properties Practice 🎯

Deep Dive: Applying Triangle Properties

Worked Example 1: Special Triangle → Area

StepWork
ProblemAn equilateral triangle has side length 8. Find its area.
StrategySplit into two 30-60-90 triangles by drawing the height.
Find heightHalf-base =4= 4 (short leg). Height =43= 4\sqrt{3} (long leg of 30-60-90).
AreaA=12(8)(43)=163≈27.7A = \frac{1}{2}(8)(4\sqrt{3}) = 16\sqrt{3} \approx 27.7

SAT shortcut: Area of equilateral triangle =s234= \frac{s^2\sqrt{3}}{4}. Plug in: 6434=163\frac{64\sqrt{3}}{4} = 16\sqrt{3} ✓

Worked Example 2: Similar Triangles with Algebra

StepWork
Problem△ABC ~ △DEF. In △ABC: AB=6AB = 6, BC=9BC = 9. In △DEF: DE=10DE = 10, EF=?EF = ?
Set up proportionABDE=BCEF\frac{AB}{DE} = \frac{BC}{EF} → 610=9EF\frac{6}{10} = \frac{9}{EF}
Cross multiply6⋅EF=906 \cdot EF = 90 → EF=15EF = 15

Pythagorean Triples to Memorize

TripleMultiples You'll See
3,4,53, 4, 56,8,106, 8, 10; 9,12,159, 12, 15; 15,20,2515, 20, 25
5,12,135, 12, 1310,24,2610, 24, 26
8,15,178, 15, 17Less common but appears
7,24,257, 24, 25Rare on SAT

Recognizing these saves you from using the Pythagorean theorem every time.

Triangle Inequality: Finding the Range

If two sides are 55 and 1111, the third side xx must satisfy: 11−5<x<11+511 - 5 < x < 11 + 5 6<x<166 < x < 16

Advanced Triangle Problems 🎯

Match the Triangle Property — Select the correct value for each scenario.

Part 2 Summary: Triangle Properties

PropertyKey Facts
45-45-90Legs =x= x, hyp =x2= x\sqrt{2}
30-60-90Short =x= x, long =x3= x\sqrt{3}, hyp =2x= 2x
Pythagorean theorema2+b2=c2a^2 + b^2 = c^2 (right triangles only)
Triangle inequalitySum of two sides >> third side
Similar trianglesEqual angles → proportional sides
Area ratio (similar)== (side ratio)2^2

SAT Strategy

  • Spot Pythagorean triples (3-4-5, 5-12-13) before computing.
  • For special right triangles, identify which angle or side you're given first, then find xx.
  • In similar triangle problems, match corresponding sides carefully — order matters.

Next: Area, perimeter, and quadrilateral properties →

Part 3: Circle Properties

Area, Perimeter, and Quadrilaterals

Part 3 of 7 — Polygons and Their Properties

Essential Area Formulas

ShapeAreaPerimeter
RectangleA=lwA = lwP=2l+2wP = 2l + 2w
SquareA=s2A = s^2P=4sP = 4s
TriangleA=12bhA = \frac{1}{2}bhP=a+b+cP = a + b + c
ParallelogramA=bhA = bhP=2a+2bP = 2a + 2b
TrapezoidA=12(b1+b2)hA = \frac{1}{2}(b_1 + b_2)hSum of all sides

Key Insight: Height ≠ Side Length

The height (altitude) is the perpendicular distance from base to top. In non-right triangles and parallelograms, the height is NOT the same as a side length.

Coordinate Geometry Areas

For a rectangle or right triangle on the coordinate plane:

  • Find the lengths of the sides using the distance formula or by counting grid units
  • Apply the appropriate area formula

Shaded Region Problems

Strategy: Shaded area=Total area−Unshaded area\text{Shaded area} = \text{Total area} - \text{Unshaded area}

Example: A circle of radius 5 is inscribed in a square. Find the shaded area (corners).

  • Square area: (2×5)2=100(2 \times 5)^2 = 100
  • Circle area: π(5)2=25π≈78.54\pi(5)^2 = 25\pi \approx 78.54
  • Shaded area: 100−25π≈21.46100 - 25\pi \approx 21.46

SAT Trap ⚠️

In "shaded region" problems, make sure you subtract the RIGHT shape. Draw the overlapping shapes clearly and label dimensions.

Area & Perimeter Practice 🎯

Deep Dive: Complex Area Problems

Worked Example 1: Multi-Shape Shaded Region

StepWork
ProblemA square with side 10 contains an inscribed circle. A smaller square is inscribed inside the circle. Find the area between the two squares.
Large squareArea =102=100= 10^2 = 100
CircleDiameter =10= 10, so r=5r = 5. Area =25π= 25\pi (for reference)
Small squareIts diagonal == circle diameter =10= 10. Side =102=52= \frac{10}{\sqrt{2}} = 5\sqrt{2}. Area =(52)2=50= (5\sqrt{2})^2 = 50
Between squares100−50=50100 - 50 = 50 square units

Worked Example 2: Coordinate Plane Area

StepWork
ProblemFind the area of the triangle with vertices A(1,2)A(1, 2), B(7,2)B(7, 2), C(4,8)C(4, 8).
Find baseABAB is horizontal: length =7−1=6= 7 - 1 = 6
Find heightCC is above ABAB: height =8−2=6= 8 - 2 = 6
Area12(6)(6)=18\frac{1}{2}(6)(6) = 18

Coordinate area shortcut: When one side is horizontal or vertical, use it as the base — the height is just the perpendicular distance.

Height vs. Slant Side — The #1 Trap

ShapeHeight is...NOT the height
TrianglePerpendicular from base to opposite vertexA non-perpendicular side
ParallelogramPerpendicular distance between parallel sidesThe slanted side
TrapezoidPerpendicular between the two parallel basesThe slanted legs

SAT trap: A parallelogram has sides 8 and 5 with a height of 4. Area =8×4=32= 8 \times 4 = 32 (NOT 8×5=408 \times 5 = 40).

Advanced Area & Perimeter Problems 🎯

Choose the Right Formula — Select the correct area formula for each shape.

Part 3 Summary: Area & Perimeter

ShapeAreaPerimeter
Triangle12bh\frac{1}{2}bha+b+ca + b + c
Rectanglelwlw2l+2w2l + 2w
Squares2s^24s4s; diagonal =s2= s\sqrt{2}
Parallelogrambhbh (NOT side \times side)2a+2b2a + 2b
Trapezoid12(b1+b2)h\frac{1}{2}(b_1 + b_2)hSum of all sides

Key Strategies

  • Shaded regions: Total area − unshaded area. Draw and label clearly.
  • Coordinate plane: Use horizontal/vertical sides as base when possible.
  • Height trap: Always use the PERPENDICULAR height, not the slant side.

Next: Circle geometry — arcs, sectors, and central angles →

Part 4: Area & Volume

Circles: Arc Length, Sector Area, Central Angles

Part 4 of 7 — Circle Geometry

Circle Fundamentals

PropertyFormula
CircumferenceC=2πr=πdC = 2\pi r = \pi d
AreaA=πr2A = \pi r^2
Arc lengthL=θ360°×2πrL = \frac{\theta}{360°} \times 2\pi r
Sector areaAsector=θ360°×πr2A_{\text{sector}} = \frac{\theta}{360°} \times \pi r^2

Where θ\theta is the central angle in degrees.

The Proportion Rule

A central angle of θ°\theta° creates an arc that is θ360\frac{\theta}{360} of the full circle. This fraction applies to BOTH arc length AND sector area.

Example: A circle with radius 10 has a central angle of 72°72°.

  • Arc length =72360×2π(10)=15×20π=4π= \frac{72}{360} \times 2\pi(10) = \frac{1}{5} \times 20\pi = 4\pi
  • Sector area =72360×π(10)2=15×100π=20π= \frac{72}{360} \times \pi(10)^2 = \frac{1}{5} \times 100\pi = 20\pi

Inscribed Angle Theorem

An inscribed angle is HALF the central angle that subtends the same arc.

Inscribed angle=12×Central angle\text{Inscribed angle} = \frac{1}{2} \times \text{Central angle}

Special case: An inscribed angle that subtends a diameter (semicircle) is always 90°90°.

Tangent Lines

A tangent to a circle is perpendicular to the radius at the point of tangency, so the radius and the tangent line form a right angle there.

Circle Geometry Practice 🎯

Deep Dive: Multi-Step Circle Problems

Worked Example 1: Arc Length from Context

StepWork
ProblemA clock's minute hand is 6 inches long. How far does the tip travel in 20 minutes?
Central angle20 min =2060=13= \frac{20}{60} = \frac{1}{3} of full rotation =13×360°=120°= \frac{1}{3} \times 360° = 120°
Arc lengthL=120360×2π(6)=13×12π=4π≈12.57L = \frac{120}{360} \times 2\pi(6) = \frac{1}{3} \times 12\pi = 4\pi \approx 12.57 inches

Worked Example 2: Sector Area to Find Radius

StepWork
ProblemA sector with central angle 60°60° has area 24π24\pi. Find the radius.
Set up60360×πr2=24π\frac{60}{360} \times \pi r^2 = 24\pi
Simplify16πr2=24π\frac{1}{6}\pi r^2 = 24\pi → r2=144r^2 = 144
Answerr=12r = 12

Key Circle Relationships

GivenFindMethod
RadiusCircumferenceC=2πrC = 2\pi r
CircumferenceRadiusr=C2πr = \frac{C}{2\pi}
AreaRadiusr=Aπr = \sqrt{\frac{A}{\pi}}
Arc length + angleRadiusr=L×3602π×θr = \frac{L \times 360}{2\pi \times \theta}
Sector area + angleRadiusr=Asector×360π×θr = \sqrt{\frac{A_{\text{sector}} \times 360}{\pi \times \theta}}

Radians on the SAT

Some SAT questions use radians instead of degrees:

  • Full circle =2π= 2\pi radians =360°= 360°
  • Arc length in radians: L=rθL = r\theta
  • Sector area in radians: A=12r2θA = \frac{1}{2}r^2\theta

Conversion: θrad=θdeg×π180\theta_{\text{rad}} = \theta_{\text{deg}} \times \frac{\pi}{180}

Advanced Circle Problems 🎯

Circle Calculations — Select the correct result.

Part 4 Summary: Circle Geometry

PropertyFormulaKey Relationship
CircumferenceC=2πrC = 2\pi rr=C/(2π)r = C/(2\pi)
AreaA=πr2A = \pi r^2r=A/πr = \sqrt{A/\pi}
Arc length (deg)θ360×2πr\frac{\theta}{360} \times 2\pi rFraction of circumference
Sector area (deg)θ360×πr2\frac{\theta}{360} \times \pi r^2Same fraction of area
Arc length (rad)L=rθL = r\thetaSimpler in radians
Inscribed angle=12×= \frac{1}{2} \times central angleInscribed in semicircle =90°= 90°
Tangent line⊥\perp to radiusCreates right angle at tangent point

SAT Strategy

  • The fraction θ360\frac{\theta}{360} is the same for both arc length and sector area.
  • If the SAT gives you arc length, work backward to find radius or angle.
  • Watch for radian vs. degree — the formulas change.

Next: Volume and surface area of 3D figures →

Part 5: Coordinate Geometry

Volume and Surface Area

Part 5 of 7 — 3D Figures

The SAT reference sheet includes these formulas, but knowing them cold saves time.

Volume Formulas

ShapeVolume
Rectangular prismV=lwhV = lwh
CylinderV=πr2hV = \pi r^2 h
ConeV=13πr2hV = \frac{1}{3}\pi r^2 h
SphereV=43πr3V = \frac{4}{3}\pi r^3
PyramidV=13BhV = \frac{1}{3}Bh (where BB = base area)

Surface Area

ShapeSurface Area
Rectangular prismSA=2(lw+lh+wh)SA = 2(lw + lh + wh)
CylinderSA=2πr2+2πrhSA = 2\pi r^2 + 2\pi rh
SphereSA=4πr2SA = 4\pi r^2

Common SAT Problem: Filling and Draining

"A cylindrical tank has radius 3 ft and height 10 ft. Water fills it at 2 cubic feet per minute. How long until it's full?"

V=π(3)2(10)=90π≈282.7 ft3V = \pi(3)^2(10) = 90\pi \approx 282.7 \text{ ft}^3 Time=90π2=45π≈141.4 minutes\text{Time} = \frac{90\pi}{2} = 45\pi \approx 141.4 \text{ minutes}

Scaling Rule for 3D

If dimensions are scaled by factor kk:

  • Lengths scale by kk
  • Areas scale by k2k^2
  • Volumes scale by k3k^3

Example: If you double all dimensions of a box, its volume increases by 23=82^3 = 8 times.

Volume & Surface Area Practice 🎯

Deep Dive: 3D Problem-Solving Strategies

Worked Example 1: Transferring Between Shapes

StepWork
ProblemWater from a full cylinder (radius 3, height 12) is poured into a cone (radius 6, height hh). The cone is filled exactly. Find hh.
Cylinder volumeV=π(3)2(12)=108πV = \pi(3)^2(12) = 108\pi
Cone volumeV=13π(6)2h=12πhV = \frac{1}{3}\pi(6)^2 h = 12\pi h
Set equal12πh=108π12\pi h = 108\pi → h=9h = 9

Worked Example 2: Surface Area in Context

StepWork
ProblemA rectangular box (4 \times 6 \times 3) needs to be wrapped with no overlap. How much wrapping paper is needed?
Surface areaSA=2(4⋅6+4⋅3+6⋅3)=2(24+12+18)=2(54)=108SA = 2(4 \cdot 6 + 4 \cdot 3 + 6 \cdot 3) = 2(24 + 12 + 18) = 2(54) = 108 sq units

Scaling Rules — Complete Table

DimensionScale FactorExample (original → doubled)
Lengthkk5→105 → 10
Perimeterkk20→4020 → 40
Area / Surface areak2k^225→10025 → 100
Volumek3k^3125→1000125 → 1000

Common SAT 3D Question Types

  1. "How much fits inside?" → Volume
  2. "How much material to cover?" → Surface area
  3. "Pour from one to another" → Set volumes equal
  4. "What happens when dimensions change?" → Scaling rules
  5. "How long to fill/drain?" → Volume \div rate

Advanced Volume & Surface Area 🎯

3D Figure Identification — Match the description to the correct formula or value.

Part 5 Summary: Volume & Surface Area

ShapeVolumeSurface Area
Rectangular prismlwhlwh2(lw+lh+wh)2(lw + lh + wh)
Cubes3s^36s26s^2
Cylinderπr2h\pi r^2 h2πr2+2πrh2\pi r^2 + 2\pi rh
Cone13πr2h\frac{1}{3}\pi r^2 hπr2+πrl\pi r^2 + \pi r l (ll = slant)
Sphere43πr3\frac{4}{3}\pi r^34πr24\pi r^2

Scaling Rules

  • Lengths ×k\times k, Areas ×k2\times k^2, Volumes ×k3\times k^3

SAT Strategy

  • "Pour from shape A to shape B" → Set volumes equal and solve.
  • Watch units — if radius is in cm and height in m, convert first.
  • Cone = 13\frac{1}{3} cylinder; hemisphere = 12×43πr3=23πr3\frac{1}{2} \times \frac{4}{3}\pi r^3 = \frac{2}{3}\pi r^3.

Next: Coordinate geometry — distance, midpoint, and circle equations →

Part 6: Problem-Solving Workshop

Coordinate Geometry

Part 6 of 7 — Distance, Midpoint, and Equations of Lines/Circles

Distance Formula

d=(x2−x1)2+(y2−y1)2d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

This is just the Pythagorean theorem applied to the coordinate plane.

Midpoint Formula

M=(x1+x22, y1+y22)M = \left(\frac{x_1 + x_2}{2},\, \frac{y_1 + y_2}{2}\right)

Slope

m=y2−y1x2−x1=riserunm = \frac{y_2 - y_1}{x_2 - x_1} = \frac{\text{rise}}{\text{run}}

Parallel lines: Same slope (m1=m2m_1 = m_2)
Perpendicular lines: Negative reciprocal slopes (m1×m2=−1m_1 \times m_2 = -1)

Equation of a Circle

Standard form: (x−h)2+(y−k)2=r2(x - h)^2 + (y - k)^2 = r^2

  • Center: (h,k)(h, k)
  • Radius: rr

Example: (x−3)2+(y+2)2=25(x - 3)^2 + (y + 2)^2 = 25

  • Center: (3,−2)(3, -2) ← note: y+2y + 2 means k=−2k = -2
  • Radius: 25=5\sqrt{25} = 5

Converting General Form to Standard Form (Completing the Square)

x2+y2−6x+4y−12=0x^2 + y^2 - 6x + 4y - 12 = 0

Group and complete the square: (x2−6x+9)+(y2+4y+4)=12+9+4(x^2 - 6x + 9) + (y^2 + 4y + 4) = 12 + 9 + 4 (x−3)2+(y+2)2=25(x - 3)^2 + (y + 2)^2 = 25

SAT Trap ⚠️

When reading circle equations, remember: (x−h)2(x - h)^2 means the center's x-coordinate is +h+h, and (y+k)2(y + k)^2 means the center's y-coordinate is −k-k. The signs flip!

Coordinate Geometry Practice 🎯

Deep Dive: Coordinate Geometry Problem Solving

Worked Example 1: Finding a Missing Vertex

StepWork
ProblemA rectangle has three vertices at A(1,1)A(1, 1), B(7,1)B(7, 1), C(7,5)C(7, 5). Find vertex DD.
StrategyOpposite sides of a rectangle are equal and parallel.
ReasoningDD has the same xx as AA and same yy as CC: D(1,5)D(1, 5).
VerifyAB=6AB = 6, CD=6CD = 6 ✓. BC=4BC = 4, AD=4AD = 4 ✓.

Worked Example 2: Is It a Right Triangle?

StepWork
ProblemTriangle with vertices P(0,0)P(0, 0), Q(4,0)Q(4, 0), R(0,3)R(0, 3). Is it a right triangle?
SlopesPQPQ: slope =0= 0 (horizontal). PRPR: slope undefined (vertical).
CheckHorizontal ⊥ vertical → YES, right angle at PP.
Confirm with lengthsPQ=4PQ = 4, PR=3PR = 3, QR=16+9=5QR = \sqrt{16 + 9} = 5. Since 32+42=523^2 + 4^2 = 5^2, it's a 3-4-5 right triangle ✓

Equation of a Line — Forms You Need

FormEquationWhen to Use
Slope-intercepty=mx+by = mx + bKnow slope and y-intercept
Point-slopey−y1=m(x−x1)y - y_1 = m(x - x_1)Know slope and a point
StandardAx+By=CAx + By = CSAT often gives this form

Perpendicular Bisector Strategy

To find the perpendicular bisector of segment AB‾\overline{AB}:

  1. Find the midpoint of ABAB
  2. Find the slope of ABAB
  3. Take the negative reciprocal for the perpendicular slope
  4. Write the line through the midpoint with that slope

Advanced Coordinate Geometry 🎯

Coordinate Geometry Quick Checks — Select the correct answer.

Part 6 Summary: Coordinate Geometry

ToolFormulaKey Fact
Distance(x2−x1)2+(y2−y1)2\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}Same as Pythagorean theorem
Midpoint(x1+x22,y1+y22)\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)Average the coordinates
Slopey2−y1x2−x1\frac{y_2-y_1}{x_2-x_1}Rise over run
Parallel linesm1=m2m_1 = m_2Same slope
Perpendicular linesm1⋅m2=−1m_1 \cdot m_2 = -1Negative reciprocals
Circle (standard)(x−h)2+(y−k)2=r2(x-h)^2 + (y-k)^2 = r^2Signs flip for center

SAT Strategy

  • Know your Pythagorean triples — saves time on distance problems.
  • Completing the square converts general form circles to standard form.
  • Watch the sign flip in circle equations: (x+3)2(x + 3)^2 means center x=−3x = -3.

Next: Comprehensive geometry review and SAT strategy →

Part 7: Review & Applications

Geometry Review & SAT Strategy

Part 7 of 7 — Comprehensive Review

Formula Quick Reference

CategoryKey Formulas
AnglesTriangle sum =180°= 180°, exterior angle == sum of remotes
TrianglesA=12bhA = \frac{1}{2}bh, Pythagorean theorem: a2+b2=c2a^2 + b^2 = c^2
Special △30-60-90: x,x3,2xx, x\sqrt{3}, 2x; 45-45-90: x,x,x2x, x, x\sqrt{2}
CirclesC=2πrC = 2\pi r, A=πr2A = \pi r^2, sector =θ360= \frac{\theta}{360} of full
VolumeCylinder =πr2h= \pi r^2 h, Cone =13πr2h= \frac{1}{3}\pi r^2 h, Sphere =43πr3= \frac{4}{3}\pi r^3
Coordinated=Δx2+Δy2d = \sqrt{\Delta x^2 + \Delta y^2}, circle: (x−h)2+(y−k)2=r2(x-h)^2 + (y-k)^2 = r^2

Common SAT Geometry Question Patterns

  1. "Find the missing angle" → Use angle sum rules
  2. "Find the area of the shaded region" → Total minus unshaded
  3. "Similar triangles" → Set up proportions
  4. "Volume word problem" → Identify the shape, plug into formula
  5. "Coordinate geometry" → Distance, midpoint, or circle equation

Strategy: Draw It

If the SAT doesn't give you a figure, draw one yourself. Even a rough sketch helps you avoid errors.

If they DO give you a figure:

  • "Not drawn to scale" → Don't trust visual proportions
  • "Figure drawn to scale" → You can estimate to eliminate wrong answers

Top 3 Geometry Mistakes

  1. Using the wrong formula (mixing up circumference and area)
  2. Forgetting to take the square root when finding radius from area
  3. Not converting units (e.g., diameter given but formula needs radius)

Geometry Comprehensive Review 🎯

Deep Dive: Multi-Step SAT Geometry Problems

Worked Example 1: Combining Multiple Concepts

StepWork
ProblemA circle is inscribed in an equilateral triangle with side 12. Find the area of the region inside the triangle but outside the circle.
Triangle areaA=s234=14434=363A = \frac{s^2\sqrt{3}}{4} = \frac{144\sqrt{3}}{4} = 36\sqrt{3}
Inscribed circle radiusr=s36=1236=23r = \frac{s\sqrt{3}}{6} = \frac{12\sqrt{3}}{6} = 2\sqrt{3}
Circle areaA=π(23)2=12πA = \pi(2\sqrt{3})^2 = 12\pi
Shaded region363−12π≈62.35−37.70≈24.6536\sqrt{3} - 12\pi \approx 62.35 - 37.70 \approx 24.65

Worked Example 2: Coordinate + Geometry Hybrid

StepWork
ProblemA circle has center (3,4)(3, 4) and passes through the origin. Find the circle's area.
RadiusDistance from (3,4)(3,4) to (0,0)(0,0): r=9+16=5r = \sqrt{9 + 16} = 5
AreaA=π(5)2=25πA = \pi(5)^2 = 25\pi

SAT Geometry Decision Framework

Question TypeFirst StepCommon Trap
Missing angleIdentify angle relationship (parallel? triangle? vertical?)Assuming lines are parallel without proof
Shaded regionTotal−Unshaded\text{Total} - \text{Unshaded}Subtracting the wrong shape
Similar trianglesSet up proportion with corresponding sidesMatching sides in wrong order
Volume word problemIdentify 3D shape, plug in valuesConfusing radius with diameter
Circle equationConvert to standard form if neededSign errors in center coordinates
ScalingApply kk, k2k^2, or k3k^3 depending on dimensionUsing k2k^2 for volume

Common SAT Geometry Mistakes — Quick Check

MistakeCorrect Approach
Area of circle with diameter 10 → π(10)2\pi(10)^2r=5r = 5, so A=25πA = 25\pi
30-60-90 short leg = hypotenuseShort leg =hyp2= \frac{\text{hyp}}{2}
Using slant height as heightHeight is perpendicular
Forgetting \sqrt{} for radius from arear=A/πr = \sqrt{A/\pi}, not A/πA/\pi

SAT Geometry Challenge 🎯

Geometry Concept Quick Check — Select the correct answer for each scenario.

Full Topic Summary: Geometry & Angles

PartTopicKey Formulas & Facts
1Angle RelationshipsSupplementary (180°180°), complementary (90°90°), vertical (equal), exterior angle theorem
2Triangle Properties30-60-90 (x,x3,2xx, x\sqrt{3}, 2x), 45-45-90 (x,x,x2x, x, x\sqrt{2}), similarity, inequality
3Area & Perimeter12bh\frac{1}{2}bh, bhbh (parallelogram), 12(b1+b2)h\frac{1}{2}(b_1+b_2)h (trapezoid), shaded == total −- unshaded
4Circle GeometryC=2πrC = 2\pi r, A=πr2A = \pi r^2, arc/sector =θ360= \frac{\theta}{360} of whole, inscribed =12= \frac{1}{2} central
5Volume & SACylinder πr2h\pi r^2 h, cone 13πr2h\frac{1}{3}\pi r^2 h, sphere 43πr3\frac{4}{3}\pi r^3, scaling k3k^3
6Coordinate GeometryDistance, midpoint, slope, parallel/perpendicular, circle equations
7Review & StrategyMulti-step problems, decision framework, common mistakes

Top SAT Geometry Strategies

  1. Draw and label — if no figure given, sketch one
  2. Know your triples — 3-4-5, 5-12-13, 8-15-17
  3. Height ≠ slant side — always perpendicular
  4. Diameter vs. radius — read carefully, divide by 2 if needed
  5. "Not drawn to scale" — don't trust the picture
  6. Complete the square for circle equations in general form
  7. Scaling: lengths kk, areas k2k^2, volumes k3k^3

🎉 Geometry & Angles complete! You're ready for SAT geometry questions.