Geometry Basics - 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 $180°$
Complementary angles: Sum to $90°$
Vertical angles: Equal (formed by intersecting lines)
Angles on a straight line: Sum to $180°$

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 $180°$

Triangle Angle Sum

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

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

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.

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

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

Step	Work
Problem	Lines $l \parallel m$ are cut by a transversal. One angle is $(3x + 10)°$ and its alternate interior angle is $(5x - 30)°$ . Find $x$ and the angle measure.
Set up	Alternate interior angles are equal: $3x + 10 = 5x - 30$
Solve	$40 = 2x$ → $x = 20$
Answer	Each angle $= 3(20) + 10 = 70°$ ✓

Worked Example 2: Exterior Angle with Algebra

Step	Work
Problem	In △ABC, $\angle A = (2x + 5)°$ , $\angle B = (x + 10)°$ , exterior angle at $C = (4x - 5)°$ . Find all angles.
Apply theorem	Exterior $=$ sum of remotes: $4x - 5 = (2x + 5) + (x + 10)$
Solve	$4x - 5 = 3x + 15$ → $x = 20$
Angles	$A = 45°$ , $B = 30°$ , $C = 180° - 45° - 30° = 105°$
Check	Exterior at $C$ : $4(20) - 5 = 75°$ = $45° + 30°$ ✓

Angle Relationship Quick Reference

Relationship	Rule	How to Spot
Vertical angles	Equal	X shape at intersection
Supplementary	Sum to $180°$	Adjacent on straight line
Corresponding	Equal	Same position at each parallel crossing
Alternate interior	Equal	Z or S pattern between parallels
Co-interior	Sum to $180°$	U or C pattern between parallels
Exterior angle	Sum of two remotes	Outside vertex of triangle

Advanced Angle Problems 🎯

Identify the Angle Relationship — Name each angle pair.

Part 1 Summary: Angle Relationships

Rule	Formula/Fact	When to Use
Supplementary	$a + b = 180°$	Angles on a straight line
Complementary	$a + b = 90°$	Corner/right angle split
Vertical angles	$a = b$	Intersecting lines
Triangle sum	$A + B + C = 180°$	Any triangle
Exterior angle	$= \text{remote}_1 + \text{remote}_2$	Angle outside triangle vertex
Parallel + transversal	Corresponding & alternate = equal; co-interior sum to $180°$	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 $360°$ — don't confuse with straight line ( $180°$ ).

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

30-60-90 Triangle:

Short leg: $x$ (opposite 30°)
Long leg: $x\sqrt{3}$ (opposite 60°)
Hypotenuse: $2x$ (opposite 90°)

Example

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

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

Triangle Inequality Theorem

For any triangle with sides $a$ , $b$ , $c$ : $a + b > c$

The sum of any two sides must exceed the third.

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

Similar Triangles (AA Similarity)

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

$\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$

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

Step	Work
Problem	An equilateral triangle has side length 8. Find its area.
Strategy	Split into two 30-60-90 triangles by drawing the height.
Find height	Half-base $= 4$ (short leg). Height $= 4\sqrt{3}$ (long leg of 30-60-90).
Area	$A = \frac{1}{2}(8)(4\sqrt{3}) = 16\sqrt{3} \approx 27.7$

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

Worked Example 2: Similar Triangles with Algebra

Step	Work
Problem	△ABC ~ △DEF. In △ABC: $AB = 6$ , $BC = 9$ . In △DEF: $DE = 10$ , $EF = ?$
Set up proportion	$\frac{AB}{DE} = \frac{BC}{EF}$ → $\frac{6}{10} = \frac{9}{EF}$
Cross multiply	$6 \cdot EF = 90$ → $EF = 15$

Pythagorean Triples to Memorize

Triple	Multiples You'll See
$3, 4, 5$	$6, 8, 10$ ; $9, 12, 15$ ; $15, 20, 25$
$5, 12, 13$	$10, 24, 26$
$8, 15, 17$	Less common but appears
$7, 24, 25$	Rare on SAT

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

Triangle Inequality: Finding the Range

If two sides are $5$ and $11$ , the third side $x$ must satisfy: $11 - 5 < x < 11 + 5$ $6 < x < 16$

Advanced Triangle Problems 🎯

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

Part 2 Summary: Triangle Properties

Property	Key Facts
45-45-90	Legs $= x$ , hyp $= x\sqrt{2}$
30-60-90	Short $= x$ , long $= x\sqrt{3}$ , hyp $= 2x$
Pythagorean theorem	$a^2 + b^2 = c^2$ (right triangles only)
Triangle inequality	Sum of two sides $>$ third side
Similar triangles	Equal angles → proportional sides
Area ratio (similar)	$=$ (side ratio) $^2$

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 $x$ .
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

Shape	Area	Perimeter
Rectangle	$A = lw$	$P = 2l + 2w$
Square	$A = s^2$	$P = 4s$
Triangle	$A = \frac{1}{2}bh$	$P = a + b + c$
Parallelogram	$A = bh$	$P = 2a + 2b$
Trapezoid	$A = \frac{1}{2}(b_1 + b_2)h$	Sum 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: $\text{Shaded area} = \text{Total area} - \text{Unshaded area}$

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

Square area: $(2 \times 5)^2 = 100$
Circle area: $\pi(5)^2 = 25\pi \approx 78.54$
Shaded area: $100 - 25\pi \approx 21.46$

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

Step	Work
Problem	A square with side 10 contains an inscribed circle. A smaller square is inscribed inside the circle. Find the area between the two squares.
Large square	Area $= 10^2 = 100$
Circle	Diameter $= 10$ , so $r = 5$ . Area $= 25\pi$ (for reference)
Small square	Its diagonal $=$ circle diameter $= 10$ . Side $= \frac{10}{\sqrt{2}} = 5\sqrt{2}$ . Area $= (5\sqrt{2})^2 = 50$
Between squares	$100 - 50 = 50$ square units

Worked Example 2: Coordinate Plane Area

Step	Work
Problem	Find the area of the triangle with vertices $A(1, 2)$ , $B(7, 2)$ , $C(4, 8)$ .
Find base	$AB$ is horizontal: length $= 7 - 1 = 6$
Find height	$C$ is above $AB$ : height $= 8 - 2 = 6$
Area	$\frac{1}{2}(6)(6) = 18$

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

Shape	Height is...	NOT the height
Triangle	Perpendicular from base to opposite vertex	A non-perpendicular side
Parallelogram	Perpendicular distance between parallel sides	The slanted side
Trapezoid	Perpendicular between the two parallel bases	The slanted legs

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

Advanced Area & Perimeter Problems 🎯

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

Part 3 Summary: Area & Perimeter

Shape	Area	Perimeter
Triangle	$\frac{1}{2}bh$	$a + b + c$
Rectangle	$lw$	$2l + 2w$
Square	$s^2$	$4s$ ; diagonal $= s\sqrt{2}$
Parallelogram	$bh$ (NOT side × side)	$2a + 2b$
Trapezoid	$\frac{1}{2}(b_1 + b_2)h$	Sum 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

Property	Formula
Circumference	$C = 2\pi r = \pi d$
Area	$A = \pi r^2$
Arc length	$L = \frac{\theta}{360°} \times 2\pi r$
Sector area	$A_{\text{sector}} = \frac{\theta}{360°} \times \pi r^2$

Where $\theta$ is the central angle in degrees.

The Proportion Rule

A central angle of $\theta°$ creates an arc that is $\frac{\theta}{360}$ 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 $72°$ .

Arc length $= \frac{72}{360} \times 2\pi(10) = \frac{1}{5} \times 20\pi = 4\pi$
Sector area $= \frac{72}{360} \times \pi(10)^2 = \frac{1}{5} \times 100\pi = 20\pi$

Inscribed Angle Theorem

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

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

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

Tangent Lines

A tangent to a circle is perpendicular to the radius at the point of tangency (

Circle Geometry Practice 🎯

Deep Dive: Multi-Step Circle Problems

Worked Example 1: Arc Length from Context

Step	Work
Problem	A clock's minute hand is 6 inches long. How far does the tip travel in 20 minutes?
Central angle	20 min $= \frac{20}{60} = \frac{1}{3}$ of full rotation $= \frac{1}{3} \times 360° = 120°$
Arc length	$L = \frac{120}{360} \times 2\pi(6) = \frac{1}{3} \times 12\pi = 4\pi \approx 12.57$ inches

Worked Example 2: Sector Area to Find Radius

Step	Work
Problem	A sector with central angle $60°$ has area $24\pi$ . Find the radius.
Set up	$\frac{60}{360} \times \pi r^2 = 24\pi$
Simplify	$\frac{1}{6}\pi r^2 = 24\pi$ → $r^2 = 144$
Answer	$r = 12$

Key Circle Relationships

Given	Find	Method
Radius	Circumference	$C = 2\pi r$
Circumference	Radius	$r = \frac{C}{2\pi}$
Area	Radius	$r = \sqrt{\frac{A}{\pi}}$
Arc length + angle	Radius	$r = \frac{L \times 360}{2\pi \times \theta}$
Sector area + angle	Radius	$r = \sqrt{\frac{A_{\text{sector}} \times 360}{\pi \times \theta}}$

Radians on the SAT

Some SAT questions use radians instead of degrees:

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

Conversion: $\theta_{\text{rad}} = \theta_{\text{deg}} \times \frac{\pi}{180}$

Advanced Circle Problems 🎯

Circle Calculations — Select the correct result.

Part 4 Summary: Circle Geometry

Property	Formula	Key Relationship
Circumference	$C = 2\pi r$	$r = C/(2\pi)$
Area	$A = \pi r^2$	$r = \sqrt{A/\pi}$
Arc length (deg)	$\frac{\theta}{360} \times 2\pi r$	Fraction of circumference
Sector area (deg)	$\frac{\theta}{360} \times \pi r^2$	Same fraction of area
Arc length (rad)	$L = r\theta$	Simpler in radians
Inscribed angle	$= \frac{1}{2} \times$ central angle	Inscribed in semicircle $= 90°$
Tangent line	$\perp$ to radius	Creates right angle at tangent point

SAT Strategy

The fraction $\frac{\theta}{360}$ 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

Shape	Volume
Rectangular prism	$V = lwh$
Cylinder	$V = \pi r^2 h$
Cone	$V = \frac{1}{3}\pi r^2 h$
Sphere	$V = \frac{4}{3}\pi r^3$
Pyramid	$V = \frac{1}{3}Bh$ (where $B$ = base area)

Surface Area

Shape	Surface Area
Rectangular prism	$SA = 2(lw + lh + wh)$
Cylinder	$SA = 2\pi r^2 + 2\pi rh$
Sphere	$SA = 4\pi r^2$

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?"

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

Scaling Rule for 3D

If dimensions are scaled by factor $k$ :

Lengths scale by $k$
Areas scale by $k^2$
Volumes scale by $k^3$

Example: If you double all dimensions of a box, its volume increases by $2^3 = 8$ times.

Volume & Surface Area Practice 🎯

Deep Dive: 3D Problem-Solving Strategies

Worked Example 1: Transferring Between Shapes

Step	Work
Problem	Water from a full cylinder (radius 3, height 12) is poured into a cone (radius 6, height $h$ ). The cone is filled exactly. Find $h$ .
Cylinder volume	$V = \pi(3)^2(12) = 108\pi$
Cone volume	$V = \frac{1}{3}\pi(6)^2 h = 12\pi h$
Set equal	$12\pi h = 108\pi$ → $h = 9$

Worked Example 2: Surface Area in Context

Step	Work
Problem	A rectangular box (4 × 6 × 3) needs to be wrapped with no overlap. How much wrapping paper is needed?
Surface area	$SA = 2(4 \cdot 6 + 4 \cdot 3 + 6 \cdot 3) = 2(24 + 12 + 18) = 2(54) = 108$ sq units

Scaling Rules — Complete Table

Dimension	Scale Factor	Example (original → doubled)
Length	$k$	$5 → 10$
Perimeter	$k$	$20 → 40$
Area / Surface area	$k^2$	$25 → 100$
Volume	$k^3$	$125 → 1000$

Common SAT 3D Question Types

"How much fits inside?" → Volume
"How much material to cover?" → Surface area
"Pour from one to another" → Set volumes equal
"What happens when dimensions change?" → Scaling rules
"How long to fill/drain?" → Volume ÷ rate

Advanced Volume & Surface Area 🎯

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

Part 5 Summary: Volume & Surface Area

Shape	Volume	Surface Area
Rectangular prism	$lwh$	$2(lw + lh + wh)$
Cube	$s^3$	$6s^2$
Cylinder	$\pi r^2 h$	$2\pi r^2 + 2\pi rh$
Cone	$\frac{1}{3}\pi r^2 h$	$\pi r^2 + \pi r l$ ( $l$ = slant)
Sphere	$\frac{4}{3}\pi r^3$	$4\pi r^2$

Scaling Rules

Lengths $\times k$ , Areas $\times k^2$ , Volumes $\times k^3$

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 = $\frac{1}{3}$ cylinder; hemisphere = $\frac{1}{2} \times \frac{4}{3}\pi r^3 = \frac{2}{3}\pi r^3$ .

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

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

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

Midpoint Formula

$M = \left(\frac{x_1 + x_2}{2},\, \frac{y_1 + y_2}{2}\right)$

Slope

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

Parallel lines: Same slope ( $m_1 = m_2$ )
Perpendicular lines: Negative reciprocal slopes ( $m_1 \times m_2 = -1$ )

Equation of a Circle

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

Center: $(h, k)$
Radius: $r$

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

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

Converting General Form to Standard Form (Completing the Square)

$x^2 + y^2 - 6x + 4y - 12 = 0$

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

SAT Trap ⚠️

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

Coordinate Geometry Practice 🎯

Deep Dive: Coordinate Geometry Problem Solving

Worked Example 1: Finding a Missing Vertex

Step	Work
Problem	A rectangle has three vertices at $A(1, 1)$ , $B(7, 1)$ , $C(7, 5)$ . Find vertex $D$ .
Strategy	Opposite sides of a rectangle are equal and parallel.
Reasoning	$D$ has the same $x$ as $A$ and same $y$ as $C$ : $D(1, 5)$ .
Verify	$AB = 6$ , $CD = 6$ ✓. $BC = 4$ , $AD = 4$ ✓.

Worked Example 2: Is It a Right Triangle?

Step	Work
Problem	Triangle with vertices $P(0, 0)$ , $Q(4, 0)$ , $R(0, 3)$ . Is it a right triangle?
Slopes	$PQ$ : slope $= 0$ (horizontal). $PR$ : slope undefined (vertical).
Check	Horizontal ⊥ vertical → YES, right angle at $P$ .
Confirm with lengths	$PQ = 4$ , $PR = 3$ , $QR = \sqrt{16 + 9} = 5$ . Since $3^2 + 4^2 = 5^2$ , it's a 3-4-5 right triangle ✓

Equation of a Line — Forms You Need

Form	Equation	When to Use
Slope-intercept	$y = mx + b$	Know slope and y-intercept
Point-slope	$y - y_1 = m(x - x_1)$	Know slope and a point
Standard	$Ax + By = C$	SAT often gives this form

Perpendicular Bisector Strategy

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

Find the midpoint of $AB$
Find the slope of $AB$
Take the negative reciprocal for the perpendicular slope
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

Tool	Formula	Key Fact
Distance	$\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$	Same as Pythagorean theorem
Midpoint	$\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)$	Average the coordinates
Slope	$\frac{y_2-y_1}{x_2-x_1}$	Rise over run
Parallel lines	$m_1 = m_2$	Same slope
Perpendicular lines	$m_1 \cdot m_2 = -1$	Negative reciprocals
Circle (standard)	$(x-h)^2 + (y-k)^2 = r^2$	Signs 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: $(x + 3)^2$ means center $x = -3$ .

Next: Comprehensive geometry review and SAT strategy →

Part 7: Review & Applications

Geometry Review & SAT Strategy

Part 7 of 7 — Comprehensive Review

Formula Quick Reference

Category	Key Formulas
Angles	Triangle sum $= 180°$ , exterior angle $=$ sum of remotes
Triangles	$A = \frac{1}{2}bh$ , Pythagorean theorem: $a^2 + b^2 = c^2$
Special △	30-60-90: $x, x\sqrt{3}, 2x$ ; 45-45-90: $x, x, x\sqrt{2}$
Circles	$C = 2\pi r$ , $A = \pi r^2$ , sector $= \frac{\theta}{360}$ of full
Volume	Cylinder $= \pi r^2 h$ , Cone $= \frac{1}{3}\pi r^2 h$ , Sphere $= \frac{4}{3}\pi r^3$
Coordinate	$d = \sqrt{\Delta x^2 + \Delta y^2}$ , circle: $(x-h)^2 + (y-k)^2 = r^2$

Common SAT Geometry Question Patterns

"Find the missing angle" → Use angle sum rules
"Find the area of the shaded region" → Total minus unshaded
"Similar triangles" → Set up proportions
"Volume word problem" → Identify the shape, plug into formula
"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

Using the wrong formula (mixing up circumference and area)
Forgetting to take the square root when finding radius from area
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

Step	Work
Problem	A circle is inscribed in an equilateral triangle with side 12. Find the area of the region inside the triangle but outside the circle.
Triangle area	$A = \frac{s^2\sqrt{3}}{4} = \frac{144\sqrt{3}}{4} = 36\sqrt{3}$
Inscribed circle radius	$r = \frac{s\sqrt{3}}{6} = \frac{12\sqrt{3}}{6} = 2\sqrt{3}$
Circle area	$A = \pi(2\sqrt{3})^2 = 12\pi$
Shaded region	$36\sqrt{3} - 12\pi \approx 62.35 - 37.70 \approx 24.65$

Worked Example 2: Coordinate + Geometry Hybrid

Step	Work
Problem	A circle has center $(3, 4)$ and passes through the origin. Find the circle's area.
Radius	Distance from $(3,4)$ to $(0,0)$ : $r = \sqrt{9 + 16} = 5$
Area	$A = \pi(5)^2 = 25\pi$

SAT Geometry Decision Framework

Question Type	First Step	Common Trap
Missing angle	Identify angle relationship (parallel? triangle? vertical?)	Assuming lines are parallel without proof
Shaded region	$\text{Total} - \text{Unshaded}$	Subtracting the wrong shape
Similar triangles	Set up proportion with corresponding sides	Matching sides in wrong order
Volume word problem	Identify 3D shape, plug in values	Confusing radius with diameter
Circle equation	Convert to standard form if needed	Sign errors in center coordinates
Scaling	Apply $k$ , $k^2$ , or $k^3$ depending on dimension	Using $k^2$ for volume

Common SAT Geometry Mistakes — Quick Check

Mistake	Correct Approach
Area of circle with diameter 10 → $\pi(10)^2$	$r = 5$ , so $A = 25\pi$
30-60-90 short leg = hypotenuse	Short leg $= \frac{\text{hyp}}{2}$
Using slant height as height	Height is perpendicular
Forgetting $\sqrt{}$ for radius from area	$r = \sqrt{A/\pi}$ , not $A/\pi$

SAT Geometry Challenge 🎯

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

Full Topic Summary: Geometry & Angles

Part	Topic	Key Formulas & Facts
1	Angle Relationships	Supplementary ( $180°$ ), complementary ( $90°$ ), vertical (equal), exterior angle theorem
2	Triangle Properties	30-60-90 ( $x, x\sqrt{3}, 2x$ ), 45-45-90 ( $x, x, x\sqrt{2}$ ), similarity, inequality
3	Area & Perimeter	$\frac{1}{2}bh$ , $bh$ (parallelogram), $\frac{1}{2}(b_1+b_2)h$ (trapezoid), shaded $=$ total $-$ unshaded
4	Circle Geometry	$C = 2\pi r$ , $A = \pi r^2$ , arc/sector $= \frac{\theta}{360}$ of whole, inscribed $= \frac{1}{2}$ central
5	Volume & SA	Cylinder $\pi r^2 h$ , cone $\frac{1}{3}\pi r^2 h$ , sphere $\frac{4}{3}\pi r^3$ , scaling $k^3$
6	Coordinate Geometry	Distance, midpoint, slope, parallel/perpendicular, circle equations
7	Review & Strategy	Multi-step problems, decision framework, common mistakes

Top SAT Geometry Strategies

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

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