A lesson plan for each of the textbook's 88 sections, sized to a single class period of roughly 45–55 minutes. The structure stays constant — warm-up, direct instruction, guided practice, independent practice, closure — so students learn what to expect; the content varies as it must. Timings are suggestions, not commandments. A class that gets stuck on the warm-up is telling you something important; honour that signal over the lesson plan.
Each plan ends with the answer key for the corresponding textbook problem set. Materials assume standard issue: notebooks, pencils, compasses, straightedges, protractors, graph paper. The few sections that require something more (mirrors, dice, playing cards) flag it.
Ask: "What's the smallest object in geometry? What's the largest?" Collect answers without commenting. The goal is to surface intuitions about points and planes before defining them.
Introduce point, line, and plane as undefined terms — accepted by intuition, not built from anything simpler. Sketch each on the board with conventional notation. Demonstrate the dot for a point, two arrows for a line, a parallelogram for a plane. Distinguish $\overline{AB}$ (segment), $\overleftrightarrow{AB}$ (line), $\overrightarrow{AB}$ (ray). Discuss "collinear" and "coplanar" with the classroom corner as exemplar.
Work Example 1 from textbook. Then ask volunteers to: name three points, three lines, and any plane visible on the board diagram. Have one student come up and label segment, ray, line on a fresh sketch.
Problem Set 1.1 problems 1–4. Circulate. Common error: confusing line with segment.
Exit ticket: "Draw a plane containing line $\ell$ and a point $P$ not on $\ell$." Collect, scan briefly, return next class.
Problem Set 1.1, problem 5; read §1.2 in advance of next class.
On the board: a number line with points at $-3$ and $8$. Ask: "How far apart are they?" Most students will subtract. Ask: "What if I put them in the other order — does the distance change?" Set up the need for absolute value.
Define distance on a number line: $|a - b|$. Introduce the Segment Addition Postulate: if $B$ is between $A$ and $C$, then $AB + BC = AC$. Sketch examples. Emphasise between — collinear and ordered, not merely "somewhere near."
Work textbook Example 2 (Segment Addition). Then: "If $AB = 3x$, $BC = 5$, $AC = 20$, find $x$." Have students solve at desks; one volunteer at the board.
Problem Set 1.2, problems 1–4.
Pose: "Can a segment have negative length?" Discuss briefly. Conclude with "distance is always non-negative — that's why absolute value matters."
Problem Set 1.2, problem 5; preview §1.3.
Plot $A(1, 1)$ and $B(4, 5)$. "How far apart?" Most students will measure or eyeball. After 90 seconds, ask: "Could we calculate this without a ruler?"
Derive the distance formula from the Pythagorean theorem: $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$. Then midpoint: $M = \left(\dfrac{x_1 + x_2}{2}, \dfrac{y_1 + y_2}{2}\right)$ — the "average of the coordinates." Work textbook Example 3.
Compute $AB$ for the warm-up: $\sqrt{9 + 16} = 5$. Then find its midpoint: $(2.5, 3)$.
Problem Set 1.3, problems 1–4.
Quick poll: "What's the distance from $(0, 0)$ to $(3, 4)$?" Should be unanimous: $5$. The $3$-$4$-$5$ triangle is a useful touchstone — flag it.
Problem Set 1.3, problem 5.
Stand at the front and rotate your arm. Ask "When have I made a right angle? A straight angle?" Have students estimate degrees aloud.
Define an angle: two rays sharing an endpoint (vertex). Measure: a real number from $0°$ to $180°$. Acute ($< 90°$), right ($= 90°$), obtuse ($> 90°$, $< 180°$), straight ($= 180°$). Demonstrate protractor use. Angle Addition Postulate: $m\angle ABC + m\angle CBD = m\angle ABD$ if $C$ is in the interior of $\angle ABD$.
Students measure five angles from the textbook with protractors. Compare results across the class — a useful demonstration of measurement error.
Problem Set 1.4, problems 1–4.
Sketch an angle and ask students to estimate its measure (no protractor). Quick way to build calibrated intuition.
Problem Set 1.4, problem 5.
"If two angles add to $90°$, what's special about them? To $180°$?" Take answers; introduce terminology if not offered.
Complementary: sum $90°$. Supplementary: sum $180°$. Linear pair: two adjacent angles whose non-common sides form a line (supplementary by construction). Vertical angles: two non-adjacent angles formed by intersecting lines (always congruent — prove informally now, formally in Chapter 2).
Sketch two intersecting lines, label one angle $40°$. Have students fill in the other three (40°, 140°, 140°). Walk through reasoning.
Problem Set 1.5, problems 1–4.
"Can an angle be its own complement?" Yes — $45°$. "Its own supplement?" Yes — $90°$. Quick wrap-up.
Problem Set 1.5, problem 5.
Show four shapes: a triangle, a square, a "bow-tie" (self-intersecting quadrilateral), and a concave quadrilateral. Ask: "Which are polygons?" Discuss.
Polygon: a closed plane figure formed by a finite number of segments meeting only at endpoints, with no self-intersections. Name by number of sides (triangle through dodecagon). Regular = equilateral + equiangular. Convex vs concave: in a convex polygon, every interior diagonal lies entirely inside.
Have students draw one example each of: an equilateral but not equiangular polygon (rhombus), an equiangular but not equilateral polygon (rectangle), and a concave pentagon.
Problem Set 1.6, problems 1–4.
"Is a circle a polygon?" No — infinitely many "sides" of zero length, depending on how you frame it; but more precisely, a polygon is bounded by straight segments.
Problem Set 1.6, problem 5.
Hold up a cube and a sphere. "What's the same? What's different?" Note that one is a polyhedron, one isn't.
Polyhedron: a solid with polygonal faces. Vertices, edges, faces. Prism: two parallel congruent bases. Pyramid: one base, triangular faces meeting at apex. Cylinder/cone: curved analogues. Sphere: locus equidistant from a centre. Foreshadow Chapter 12 and Euler's formula.
Count $V$, $E$, $F$ for a cube together: $8, 12, 6$. Verify Euler: $8 - 12 + 6 = 2$.
Problem Set 1.7, problems 1–4.
Show a soccer ball. "How might Euler's formula apply here?" ($V - E + F = 2$ still.) Tease the wider applicability.
Problem Set 1.7, problem 5; review Chapter 1 for quiz next class.
Sequence on the board: $2, 4, 8, 16, \ldots$ "What's next?" Most say $32$. Then offer $2, 4, 8, 16, 31$ — possible too (Moser's circle problem). Make the point: patterns are conjectural until proved.
Define inductive reasoning: drawing a general conclusion from specific cases. Distinguish from deductive (next class). Define conjecture and counterexample. One counterexample suffices to disprove; no number of examples suffices to prove. Work textbook Example 1.
"All prime numbers are odd." Counterexample? ($2$.) "If $n$ is even, $n^2$ is even." Conjecture; can we verify with a few cases? (Yes, but it doesn't prove it.)
Problem Set 2.1, problems 1–4.
"Why isn't checking a million cases a proof?" Discuss briefly — patterns can fail at any case.
Problem Set 2.1, problem 5.
"It is raining and the sun is shining." When is this true? When is it false? Start a truth table on the board.
Define statement (declarative, truth-evaluable). Negation $\sim p$, conjunction $p \wedge q$, disjunction $p \vee q$. Truth tables for each. Note "or" is inclusive in mathematics — at least one true.
Build the truth table for $p \wedge q$ and $p \vee q$ together. Distinguish "exclusive or" (everyday English) from "inclusive or" (math).
Problem Set 2.2, problems 1–4.
"Negate: 'All cats are black.'" Common error: "All cats are not black." Correct: "Some cat is not black." Subtle, important.
Problem Set 2.2, problem 5.
"If it rains, then the ground gets wet." True. "If the ground is wet, then it rained." Always true? Discuss the rogue sprinkler.
Conditional: "If $p$, then $q$" ($p \to q$). Hypothesis = $p$; conclusion = $q$. Converse: $q \to p$. Inverse: $\sim p \to \sim q$. Contrapositive: $\sim q \to \sim p$. Crucial fact: a conditional and its contrapositive are logically equivalent; the converse and inverse are equivalent to each other, but neither follows from the original.
Take "If a figure is a square, then it is a rectangle." Identify hypothesis, conclusion. Write converse (false), inverse (false), contrapositive (true). Map out the truth values.
Problem Set 2.3, problems 1–4.
"If contrapositive is equivalent to original, why bother with two forms?" Sometimes one is easier to prove than the other (this anticipates indirect proof in §5.4).
Problem Set 2.3, problem 5.
"All humans are mortal. Socrates is a human. Therefore…" Students complete. Note the form.
Law of Detachment: if $p \to q$ is true and $p$ is true, then $q$ is true (modus ponens). Law of Syllogism: if $p \to q$ and $q \to r$ are both true, then $p \to r$ is true. Contrast inductive (probable) vs deductive (certain).
"If $x = 3$, then $x^2 = 9$. $x = 3$. Therefore?" ($x^2 = 9$.) Then a syllogism chain together.
Problem Set 2.4, problems 1–4.
"What's wrong with: 'If $x = 3$, then $x^2 = 9$. $x^2 = 9$. Therefore $x = 3$.'?" Affirming the consequent — a classic fallacy. ($x$ could be $-3$.)
Problem Set 2.4, problem 5.
"Through two points, how many lines? Through three non-collinear points, how many planes?" Establish that postulates are statements accepted without proof.
List key postulates (point-line, line-plane, intersection). Define theorem (a proved statement). Show structure of a paragraph proof: given, claim, reasoning written in prose. Walk through textbook Example 5.
Together, write a paragraph proof: "If two lines intersect, they intersect in exactly one point."
Problem Set 2.5, problems 1–4.
"Why do we need postulates at all? Why not prove everything?" Logical regress — somewhere we must start. Mention Euclid's five postulates.
Problem Set 2.5, problem 5.
Solve $2x + 5 = 11$ on the board, slowly, naming each step (subtraction property of equality; division property of equality). Students rarely realise solving is itself a proof.
Properties of equality: reflexive, symmetric, transitive, substitution, addition, subtraction, multiplication, division, distributive. Format a two-column proof: statements left, reasons right.
Two-column proof of solving $3(x - 2) = 12$ for $x = 6$. Students reproduce in notebooks.
Problem Set 2.6, problems 1–4.
"Why bother formalising what we already do?" The discipline transfers: tomorrow we'll write geometric proofs that look exactly like this.
Problem Set 2.6, problem 5.
Given collinear $A, B, C, D$ with $AB = CD$. "Does $AC = BD$?" Have students guess; we'll prove it.
Segment congruence is an equivalence relation (reflexive, symmetric, transitive). Walk through a two-column proof using Segment Addition.
Prove together: if $M$ is midpoint of $\overline{AB}$, then $AM = MB$. Two-column format.
Problem Set 2.7, problems 1–4.
Highlight a common student error: writing "given" as a reason for something that isn't given.
Problem Set 2.7, problem 5.
"Two intersecting lines. The four angles. Why are vertical angles equal?" Take guesses.
Prove Vertical Angles Theorem in two columns: let $\angle 1$ and $\angle 3$ be vertical, with $\angle 2$ adjacent to both. $\angle 1 + \angle 2 = 180°$ and $\angle 2 + \angle 3 = 180°$, so $\angle 1 = \angle 3$. Right Angle Congruence Theorem: all right angles are congruent.
Prove: supplements of congruent angles are congruent. Two-column.
Problem Set 2.8, problems 1–4.
Chapter 2 wrap: inductive vs deductive, structure of proof, key theorems. Brief quiz tomorrow.
Problem Set 2.8, problem 5; study guide for Chapter 2 quiz.
Sketch two parallel lines cut by a transversal. Number all 8 angles. "Which pairs do you think are equal?" Take guesses.
Parallel: coplanar, never meet. Transversal: a line cutting two or more lines. Name the four angle pair types (corresponding, alt. interior, alt. exterior, consecutive/same-side interior). Diagram with labels.
Identify each angle pair in a fresh diagram. Have students pair-share their identifications.
Problem Set 3.1, problems 1–4.
"Tomorrow: which of these pairs are congruent?" Tease the Postulate.
Problem Set 3.1, problem 5.
Two parallel lines cut by transversal. One angle marked $70°$. Find the other seven.
State the Corresponding Angles Postulate. Derive alternate interior angles congruent, alternate exterior congruent, consecutive interior supplementary. Emphasise: parallel is the hypothesis. Without it, none of this follows.
Solve textbook Example 2 together.
Problem Set 3.2, problems 1–4.
"What if I don't know the lines are parallel? Can I still apply these?" Sets up §3.5.
Problem Set 3.2, problem 5.
"Steep stairs vs gentle ramp. Which has greater slope?" Brief discussion of slope as rate.
Slope $m = \dfrac{y_2 - y_1}{x_2 - x_1}$. Positive (up to right), negative (down), zero (horizontal), undefined (vertical). Parallel: equal slopes. Perpendicular: slopes multiply to $-1$ (i.e., negative reciprocals).
Find slope through $(2, 3)$ and $(5, 9)$. ($m = 2$.) Then find slope of line perpendicular ($-\tfrac{1}{2}$).
Problem Set 3.3, problems 1–4.
"Why is vertical slope undefined?" Division by zero. Quick conceptual reminder.
Problem Set 3.3, problem 5.
Write the equation of a line with slope $3$ and $y$-intercept $-2$. Most will recall $y = mx + b$.
Slope-intercept $y = mx + b$. Point-slope $y - y_1 = m(x - x_1)$. Standard form $Ax + By = C$. Walk through writing a line parallel and perpendicular to a given line through a point.
Line through $(2, 5)$ parallel to $y = 3x + 1$: $y = 3x - 1$. Perpendicular: $y = -\tfrac{1}{3}x + \tfrac{17}{3}$.
Problem Set 3.4, problems 1–4.
"Why three different forms?" Each is convenient in different contexts.
Problem Set 3.4, problem 5.
"Two lines cut by a transversal. The alternate interior angles are congruent. Are the lines parallel?" Discuss before stating the converse.
Converses: if corresponding angles are congruent, lines are parallel; similarly for alternate interior, alternate exterior, consecutive interior supplementary. These give us tools to prove parallelism, where before we only assumed it.
Example: given $\angle 1 \cong \angle 5$ (corresponding), conclude lines are parallel. Two-column proof.
Problem Set 3.5, problems 1–4.
"What's the difference between this lesson and §3.2?" Direction of inference: there we assumed parallel; here we conclude it.
Problem Set 3.5, problem 5.
"How would you measure the distance from a tree to a fence?" Elicit: shortest distance = perpendicular.
Distance from point to line = length of perpendicular segment. Distance between parallel lines = constant. Demonstrate compass-and-straightedge construction of perpendicular from external point.
Students each construct a perpendicular from an external point to a given line.
Problem Set 3.6, problems 1–4.
Chapter 3 wrap: parallel + perpendicular relationships, slope criterion, equations of lines. Quiz next class.
Problem Set 3.6, problem 5.
Sketch four triangles of various proportions. "Group these. Why did you group them that way?" Surface intuitive classifications.
By angles: acute (all $< 90°$), right (one $= 90°$), obtuse (one $> 90°$), equiangular (all $= 60°$). By sides: scalene, isosceles, equilateral. Note the equilateral-equiangular equivalence is special to triangles.
Classify five sample triangles by both schemes. (E.g., a $3$-$4$-$5$ triangle is scalene and right.)
Problem Set 4.1, problems 1–4.
"Can a triangle be both right and isosceles?" Yes — legs equal, $45$-$45$-$90$.
Problem Set 4.1, problem 5.
Tear a paper triangle's three corners off, arrange them adjacent. They form a straight line. "What does this tell us?"
Triangle Angle-Sum: $180°$. Brief proof via parallel through a vertex (uses §3.2). Exterior Angle: equal to sum of two remote interior angles. Discuss why this is plausible.
Two interior angles are $50°$ and $70°$. Find third (60°). Find exterior at third vertex ($120°$). Verify Exterior Angle Theorem.
Problem Set 4.2, problems 1–4.
"On a sphere, do triangle angles sum to $180°$?" Foreshadow §12.7 — they sum to more.
Problem Set 4.2, problem 5.
"What makes two triangles 'the same'?" Discuss. Aim for: same size, same shape — i.e., all corresponding parts equal.
Definition: $\triangle ABC \cong \triangle DEF$ iff all six pairs of corresponding parts (three sides, three angles) are congruent. Order of letters matters — it encodes which vertices correspond. CPCTC: Corresponding Parts of Congruent Triangles are Congruent.
Given $\triangle ABC \cong \triangle XYZ$, name three pairs of congruent sides and three pairs of congruent angles.
Problem Set 4.3, problems 1–4.
"Do we really need all six congruences?" Sets up SSS, SAS, ASA, AAS.
Problem Set 4.3, problem 5.
"Build a triangle with sides $3, 4, 5$. Then another with same sides. Are they congruent?" Yes — SSS.
SSS Postulate: three pairs of congruent sides $\Rightarrow$ triangles congruent. SAS: two sides and the included angle. Included — the angle between the two sides. Demonstrate with sketches.
Walk through textbook Example 1 (SAS proof) on the board.
Problem Set 4.4, problems 1–4.
"Why doesn't SSA work?" Sketch a counterexample (the ambiguous case). Students must remember: the angle has to be included.
Problem Set 4.4, problem 5.
Two triangles with $\angle A = 30°$, $\angle B = 100°$, $\overline{AB} = 5$. Are they congruent? Discuss.
ASA: two angles and the included side. AAS: two angles and a non-included side. AAS works because two angles determine the third; we then have all three angles and a side, equivalent to ASA. Note: AAA alone is NOT congruence — only similarity.
Two-column ASA proof from textbook Example 2.
Problem Set 4.5, problems 1–4.
"List all triangle congruence shortcuts." SSS, SAS, ASA, AAS, HL (right triangles). Five total.
Problem Set 4.5, problem 5.
Sketch an isosceles triangle. "If the two legs are equal, what about the base angles?" Conjecture.
Isosceles Triangle Theorem: base angles congruent. Converse: equal base angles $\Rightarrow$ equal legs. Equilateral triangle: all sides equal $\Rightarrow$ all angles $60°$.
Apex angle $40°$. Find base angles. (Each $70°$.)
Problem Set 4.6, problems 1–4.
"In an equilateral triangle, is every altitude also a median? A perpendicular bisector? An angle bisector?" Yes — a remarkable concurrence.
Problem Set 4.6, problem 5.
"Slide a triangle. Flip it. Spin it. Is it still congruent to itself?" Yes — these are rigid motions.
Rigid motion: preserves distance (and hence angles). Reflection, translation, rotation. Any congruence between two triangles can be realised as a sequence of rigid motions.
Map $\triangle ABC$ at $(0,0), (3,0), (0,4)$ to $\triangle A'B'C'$ at $(2, 1), (5, 1), (2, 5)$. Identify the rigid motion (translation by $\langle 2, 1 \rangle$).
Problem Set 4.7, problems 1–4.
"Is a dilation a rigid motion?" No — it changes size. Sets up Chapter 7 (similarity) and §9.6.
Problem Set 4.7, problem 5.
"If I tell you triangle has vertices $(0,0), (a, 0), (b, c)$, how general is that?" Quite — any triangle can be positioned this way.
Coordinate proof: place a figure conveniently (often a vertex at origin, a side along an axis), then use distance and midpoint formulas to verify properties. Walk through textbook Example 8.
Coordinate proof: the midsegment of a triangle is half the length of the third side. (Place vertices conveniently; compute midpoints; use distance formula.)
Problem Set 4.8, problems 1–4.
Chapter 4 wrap. Quiz next class on triangle congruence and isosceles properties.
Problem Set 4.8, problem 5.
On a folded sheet, ask students to fold so two marked points coincide. The crease is the perpendicular bisector — geometry by laundry.
State the iff form of both bisector theorems. Construct the three perpendicular bisectors of a scalene triangle; they meet at the circumcenter, equidistant from vertices — draw the circumscribed circle. Repeat with angle bisectors; meeting point is incenter, equidistant from sides.
Example 1. Then: locate circumcenter of a right triangle (midpoint of hypotenuse — a useful diagnostic for the lazy).
Problem Set 5.1, problems 1–4.
Quick contrast: circumcenter is about vertices; incenter is about sides. Don't muddle them.
Problem Set 5.1, problem 5.
Balance a cardboard triangle on a pencil tip — eventually you find the centroid. The physical world cooperates beautifully.
Define median (vertex to midpoint of opposite side) and altitude (perpendicular from vertex to opposite line). The three medians meet at the centroid, which trisects each median in a $2:1$ ratio. The three altitudes meet at the orthocenter — which may be outside the triangle if it's obtuse. Mention the Euler line as cocktail-party material.
Example 2. Then sketch the orthocenter of an obtuse, right, and acute triangle in turn — note where it sits in each case.
Problem Set 5.2, problems 1–4.
Four triangle centers so far. In an equilateral triangle all coincide. In every other triangle they pull apart and the geometry gets more interesting.
Problem Set 5.2, problem 5.
Sketch a very tall thin triangle. Which side is longest? Which angle is largest? The intuition is reliable here.
State the Side–Angle Inequality both directions. Justify intuitively: a small angle "opens" onto a short side. Work Example 3 — label the angles, then the opposite sides — and order accordingly.
Given sides $4, 7, 9$, identify which angle (by vertex opposite each side) is largest.
Problem Set 5.3, problems 1–4.
Mnemonic: largest angle faces the largest side. They look across at one another.
Problem Set 5.3, problem 5.
Tell students: "Assume the moon is made of cheese." Now derive a contradiction with everyday astronomy. The move is the same in mathematics.
Three steps: (1) assume the opposite of what you want to prove; (2) derive a contradiction from your premises; (3) conclude the original. Walk Example 4. Note that indirect proof is the workhorse of irrationality proofs and many classical impossibility results.
Indirectly prove: in a triangle, you cannot have two obtuse angles. (Assume two; sum already exceeds $180°$; contradiction.)
Problem Set 5.4, problems 1–4.
The point of indirect proof is leverage: sometimes the negation is easier to manipulate than the positive claim.
Problem Set 5.4, problem 5.
Hand each pair three stirrers of lengths $2, 3, 6$. Try to make a triangle. Curious silence ensues.
State the Triangle Inequality. Geometric intuition: the straight path between two points is shortest — so the third side cannot equal or exceed the sum of the other two. Range of third side: $|a - b| < x < a + b$. Walk Examples 5 and 6.
Two sides are $6$ and $11$. Possible range for the third? ($5 < x < 17$.)
Problem Set 5.5, problems 1–4.
Equality ($a + b = c$) gives a degenerate "triangle" — three collinear points. Geometry, like grammar, distinguishes the well-formed from the merely close.
Problem Set 5.5, problem 5.
Open the classroom door slightly, then widely. Which case puts the doorknob farther from the jamb? The Hinge Theorem in three seconds.
State the Hinge Theorem (SAS Inequality) and its converse. Two sides held fixed; the included angle's size controls the opposite side's length. Walk Example 7.
Two triangles share sides of $5$ and $7$. Included angles $30°$ and $90°$. Compare opposite sides.
Problem Set 5.6, problems 1–4.
End-of-chapter recap: bisectors, medians, altitudes, inequalities. Quiz next class.
Problem Set 5.6, problem 5.
Draw a convex pentagon. From one vertex, draw all diagonals — count the triangles. That's the formula in pictures.
Triangulation argument: any convex $n$-gon yields $n - 2$ triangles, hence $(n-2)\cdot 180°$ in interior angles. Exterior angle sum is always $360°$ (one full turn as you walk the boundary). Work Example 1.
Sum of interior angles of a heptagon? Each angle of a regular heptagon? (Sum $900°$; each $\approx 128.57°$.)
Problem Set 6.1, problems 1–4.
The exterior-angle fact is the more surprising one: same total for every convex polygon, regardless of size or sides.
Problem Set 6.1, problem 5.
Trace a parallelogram; fold to overlap opposite sides — they match. Same with opposite angles. Diagonals: fold to find their intersection.
Enumerate the four properties: opposite sides $\cong$, opposite angles $\cong$, consecutive angles supplementary, diagonals bisect each other. Brief proof sketch using alternate interior angles and ASA. Walk Example 2.
Given $m\angle A = 73°$ in parallelogram $ABCD$, find the other three.
Problem Set 6.2, problems 1–4.
The parallelogram is the workhorse of the quadrilateral family. Rectangles, rhombi, and squares all inherit these properties.
Problem Set 6.2, problem 5.
Sketch a quadrilateral with one pair of parallel sides of equal length. Is it a parallelogram? (Yes — that single condition suffices.)
List the five sufficient conditions. Emphasize: the previous lesson gave consequences; this one gives diagnoses. Walk Example 3 using coordinates and slope/length.
Verify parallelogram by slope and length for the vertices $(1,1), (5,1), (6,4), (2,4)$.
Problem Set 6.3, problems 1–4.
One condition, not all five, is enough. Mathematicians like minimal sufficient conditions — less to check.
Problem Set 6.3, problem 5.
Two pieces of paper — one rectangular, one a parallelogram only. Fold each along a diagonal. Notice which fold matches the other diagonal in length.
Rectangle inherits all parallelogram properties, plus four right angles, plus congruent diagonals. The converse: a parallelogram with congruent diagonals is a rectangle. Walk Example 4 (Pythagoras for the diagonal).
Rectangle $9 \times 12$ — diagonal? ($15$.)
Problem Set 6.4, problems 1–4.
Pythagoras peeks around the corner here. We'll meet him formally in Chapter 8.
Problem Set 6.4, problem 5.
Hierarchy chart on the board: parallelogram → rectangle, rhombus → square (rectangle $\cap$ rhombus). The square has it all.
Rhombus = parallelogram with four congruent sides. Diagonals are perpendicular and bisect opposite angles. Square = rectangle $\cap$ rhombus: all properties together. Walk Example 5.
If a rhombus has diagonals $10$ and $24$, find its side. ($\sqrt{5^2 + 12^2} = 13$.)
Problem Set 6.5, problems 1–4.
Square: the overachiever of the quadrilateral family — congruent sides, right angles, perpendicular and congruent bisecting diagonals.
Problem Set 6.5, problem 5.
Show a kite shape (the toy variety). What distinguishes it from a rhombus? Consecutive vs opposite congruent sides.
Trapezoid: exactly one pair of parallel sides (Glencoe convention). Isosceles trapezoid: congruent legs ⇒ congruent base angles, congruent diagonals. Kite: two pairs of consecutive congruent sides, perpendicular diagonals, one diagonal bisects the other. Midsegment formula: average of the bases. Walk Example 6.
If midsegment is $12$ and one base is $7$, find the other. ($17$.)
Problem Set 6.6, problems 1–4.
Chapter 6 wrap. Quadrilaterals form a small but neatly nested family. Quiz next class.
Problem Set 6.6, problem 5.
"If 4 apples cost $\$3$, what do 10 cost?" Proportional reasoning in the wild.
Ratio and proportion definitions; cross-multiplication; equivalent forms ($\tfrac{a}{b} = \tfrac{c}{d} \Leftrightarrow \tfrac{a+b}{b} = \tfrac{c+d}{d}$, etc.). Work Example 1.
Solve $\tfrac{2x+1}{5} = \tfrac{3x-1}{8}$. ($16x + 8 = 15x - 5 \Rightarrow x = -13$.)
Problem Set 7.1, problems 1–4.
Proportional reasoning is the through-line of the chapter — and of much of the science curriculum besides.
Problem Set 7.1, problem 5.
Show two photos of the same object at different sizes. Same shape, different size: similar.
Two requirements for similarity: corresponding angles congruent AND corresponding sides proportional. Both are needed (a non-square rectangle and a square have all $90°$ but aren't similar). Walk Example 2.
$\triangle PQR \sim \triangle STU$ with scale $5:2$. If $PQ = 15$, find $ST$. ($6$.)
Problem Set 7.2, problems 1–4.
Similarity is congruence's looser sibling: shape without insistence on size.
Problem Set 7.2, problem 5.
Why is AAA enough for similarity but not congruence? Two angles fix the shape; the third is forced. But size remains free.
State AA, SAS, SSS for similarity. Walk Example 3. Emphasize the asymmetry with congruence postulates: AA needs only two angles; SSS uses proportional sides instead of congruent.
Triangles with sides $4, 6, 8$ and $6, 9, 12$: similar? (Ratios all $2:3$ — yes, SSS.)
Problem Set 7.3, problems 1–4.
AA is the standard workhorse — small data, big conclusion.
Problem Set 7.3, problem 5.
Draw a triangle. Pick a midpoint on each of two sides; join them. What do you notice? (Midsegment is parallel to the third side and half its length.)
State the Triangle Proportionality Theorem. Walk Example 4. Note the midsegment as a special case (proportion $1:1$).
$\triangle ABC$: $DE \parallel BC$, $AD = 4, DB = 8, AC = 18$. Find $AE$ and $EC$. ($AE = 6, EC = 12$.)
Problem Set 7.4, problems 1–4.
Parallel lines slice transversals in proportion — a fact that quietly underwrites surveying and perspective drawing.
Problem Set 7.4, problem 5.
If you double a square's side, what happens to perimeter? Area? Galileo noticed this — and used it to argue why elephants can't have spider legs.
Scale factor $k$ governs: perimeters by $k$, areas by $k^2$, volumes by $k^3$. Why: linear measures scale linearly; area is a product of two; volume of three. Walk Example 5.
Similar triangles with scale $3:4$. If smaller has area $27$, larger has area? ($27 \cdot \tfrac{16}{9} = 48$.)
Problem Set 7.5, problems 1–4.
The square-cube law: a single principle responsible for why small things scuttle and large things lumber.
Problem Set 7.5, problem 5.
Zoom on a phone screen: that's a dilation about the center of the pinch. Geometry in your pocket.
Dilation: $(x, y) \mapsto (kx, ky)$ centered at origin. Enlargement if $|k| > 1$, reduction if $0 < |k| < 1$. Walk Example 6.
Dilate $\triangle ABC$ with $A(1, 2), B(3, 1), C(2, 4)$ by factor $3$. Sketch image.
Problem Set 7.6, problems 1–4.
Dilation: similarity made dynamic. We'll return to it in Chapter 9.
Problem Set 7.6, problem 5.
Look at a road atlas — find the scale bar. Two cities four inches apart on the page: how far in reality?
Scale as a ratio of corresponding lengths. Once known, every measurement is a single proportion. Walk Example 7.
Architect's scale $1\text{ in.}:2\text{ ft}$. Hallway is $11.5$ in. on plan — how long in reality? ($23$ ft.)
Problem Set 7.7, problems 1–4.
Chapter 7 wrap. Similarity is one of geometry's deepest ideas: it's what makes scale possible. Quiz next class.
Problem Set 7.7, problem 5.
Arithmetic mean of $4$ and $16$: $10$. Geometric mean: $\sqrt{64} = 8$. The geometric mean is always at most the arithmetic mean (AM–GM).
Define geometric mean: $\sqrt{ab}$. State the Right Triangle Altitude Theorem and its three useful corollaries (altitude is geometric mean of segments; each leg is geometric mean of hypotenuse and adjacent segment). Walk Example 1.
Geometric mean of $3$ and $27$. ($9$.)
Problem Set 8.1, problems 1–4.
Right triangles harbor an unreasonable number of similar sub-triangles. We exploit this all chapter.
Problem Set 8.1, problem 5.
Three squares on the three sides of a right triangle. The biggest equals the other two combined. Pythagoras, in pictures.
State the theorem and converse. Note: Loomis catalogued $370+$ proofs. We'll skip them but the theorem is equivalent to "Euclidean space is flat." Common Pythagorean triples: $(3,4,5), (5,12,13), (7,24,25), (8,15,17)$. Walk Examples 2 and 3.
Classify $(6, 8, 11)$: $36 + 64 = 100 < 121$ ⇒ obtuse. Classify $(6, 8, 9)$: $100 > 81$ ⇒ acute.
Problem Set 8.2, problems 1–4.
The theorem is older than its namesake — Babylonian and Indian mathematicians knew it centuries earlier. Pythagoras gets the marquee.
Problem Set 8.2, problem 5.
Bisect a square along a diagonal: $45$–$45$–$90$. Bisect an equilateral triangle along an altitude: $30$–$60$–$90$. Two ratios fall out by inspection.
$45$–$45$–$90$: legs $x$, hypotenuse $x\sqrt{2}$. $30$–$60$–$90$: short leg $x$, long leg $x\sqrt{3}$, hypotenuse $2x$. Derive each by Pythagoras. Walk Examples 4 and 5.
$30$–$60$–$90$: hypotenuse $20$, find both legs. (Short $10$, long $10\sqrt{3}$.)
Problem Set 8.3, problems 1–4.
These two triangles are the unit-conversion currency for SAT, ACT, and every standardized geometry problem. Worth memorizing.
Problem Set 8.3, problem 5.
Draw two right triangles with one $40°$ angle, different sizes. Measure the opposite/hypotenuse ratio in each. Identical — because similar.
Define the three primary ratios; SOH-CAH-TOA mnemonic. Calculator practice: $\sin 35°$, $\tan^{-1}(0.4)$. Stress degree mode. Walk Examples 6 and 7.
Right triangle: $\theta = 50°$, adjacent $9$. Find opposite. ($9 \tan 50° \approx 10.72$.)
Problem Set 8.4, problems 1–4.
The ratios depend only on the angle because all right triangles with that angle are similar. Geometry's gift to navigation, astronomy, and everything in between.
Problem Set 8.4, problem 5.
From eye level, looking up at the ceiling: that's elevation. Looking down at the floor from a balcony: depression. Both measured from horizontal.
Always draw a diagram. Identify horizontal, line of sight, and the right triangle. Elevation and depression for the same sight line are equal (alternate interior). Walk Example 8.
A lighthouse beam strikes the water at $12°$ depression from a height of $80$ ft. Find horizontal distance. ($80 / \tan 12° \approx 376.4$ ft.)
Problem Set 8.5, problems 1–4.
Surveying, astronomy, even artillery — all built on these basics.
Problem Set 8.5, problem 5.
Pythagoras only works for right triangles. What if your triangle isn't right? Enter the cosine correction term.
State both laws. Selection rule: Law of Sines when you have an angle and its opposite side; Law of Cosines when you have two sides and the included angle (or three sides). Walk Examples 9 and 10.
$a = 6, b = 8, c = 10$. Find largest angle (opposite $c$). By Law of Cosines: $\cos C = (36 + 64 - 100)/(96) = 0$ ⇒ $C = 90°$. (Right triangle, as expected from $(6,8,10)$.)
Problem Set 8.6, problems 1–4.
Law of Cosines reduces to Pythagoras when $C = 90°$ because $\cos 90° = 0$. A pleasing nested result.
Problem Set 8.6, problem 5.
Walk three steps east, four steps north. Where are you, as the crow flies? (Five units, direction $\tan^{-1}(4/3)$.)
Vector as directed segment; component form $\langle a, b \rangle$. Magnitude by Pythagoras; direction by arctangent. Addition componentwise; scalar multiplication scales each component. Walk Example 11.
$\vec{u} = \langle 4, -3 \rangle$. Find magnitude. ($5$.)
Problem Set 8.7, problems 1–4.
Chapter 8 wrap. Pythagoras, trig, and vectors — geometry that travels well into physics. Quiz next class.
Problem Set 8.7, problem 5.
Hold a mirror perpendicular to the paper along a line. The reflected image is the geometric reflection. Note your left hand becomes a right hand — orientation reverses.
Definition: reflection line is the perpendicular bisector of segment from point to image. Coordinate rules for $x$-axis, $y$-axis, $y = x$, origin. Walk Example 1. Mention Klein's Erlangen program: classifying geometries by their transformation groups.
Reflect $(2, -5)$ across $y = x$. ($(-5, 2)$.)
Problem Set 9.1, problems 1–4.
Reflections are the building blocks: every rigid motion can be decomposed into at most three.
Problem Set 9.1, problem 5.
"Slide everything two units right and three down." That's a translation. Everyone moves the same way.
Translation rule: $(x, y) \to (x + a, y + b)$. Preserves shape, size, and orientation. Walk Example 2.
Translate $\triangle$ with vertices $(1,1), (3,1), (2,4)$ by $\langle -2, 3 \rangle$. ($(-1, 4), (1, 4), (0, 7)$.)
Problem Set 9.2, problems 1–4.
The mildest transformation: no rotation, no flipping, no scaling — just relocation.
Problem Set 9.2, problem 5.
Pin patty paper at a point on graph paper; mark a point and rotate. Trace and read off the image. Geometry by sleight of hand.
Rotation: center, angle, direction. Counter-clockwise positive. State the three coordinate rules about the origin. Walk Example 3. Note rotations are isometries and preserve orientation.
Rotate $\triangle(1,0), (3,0), (3,2)$ by $90°$ CCW about origin. ($(0,1), (0,3), (-2,3)$.)
Problem Set 9.3, problems 1–4.
$180°$ rotation about origin is the same as reflection through origin. Two ways to land on one place.
Problem Set 9.3, problem 5.
Take a step left, then turn right. Now: turn right, then take a step left. Different ending. Order matters.
Composition notation; right-to-left application. Two reflections across parallel lines = translation; across intersecting lines = rotation. This decomposition underlies the classification of plane isometries. Walk Example 4.
Reflect $(3, 5)$ across $x$-axis then across $y$-axis. What single transformation matches? ($180°$ rotation about origin.)
Problem Set 9.4, problems 1–4.
Composition is the operation in the group of rigid motions. Not commutative — geometry's modest reminder that order matters in the real world.
Problem Set 9.4, problem 5.
How many ways can a square map to itself? Eight — four rotations and four reflections. This is the dihedral group $D_4$.
Define line symmetry and rotational symmetry of order $n$. Regular $n$-gons have both, each of order $n$. Walk Example 5.
Identify symmetries of: rectangle (non-square), equilateral triangle, letter S.
Problem Set 9.5, problems 1–4.
Symmetry is geometry's deepest organizing principle — central to crystallography, particle physics, and your bathroom tile.
Problem Set 9.5, problem 5.
A photocopier set to $200\%$: dilation by factor $2$. The toner is doing geometry.
Dilation centered at origin: $(x, y) \to (kx, ky)$. Negative $k$: a half-turn plus a scaling. Dilations preserve angles and similarity but not distance. Walk Example 6.
Dilate $\triangle(2, 4), (4, 4), (4, 0)$ by factor $\tfrac{1}{2}$. ($(1,2), (2,2), (2,0)$.)
Problem Set 9.6, problems 1–4.
Chapter 9 wrap. Four transformations, four sets of rules — and from them, all of plane motion. Quiz next class.
Problem Set 9.6, problem 5.
Each pair: measure circumference and diameter of a tin or jar with string. Divide. Everyone gets roughly $3.14$. The fact that this works for every circle is what makes $\pi$ a constant — and a faintly unreasonable one.
Define centre, radius, diameter, chord; introduce circumference formula $C = 2\pi r = \pi d$. Note that proving $\pi$ is the same for every circle rigorously took two millennia (Archimedes' inscribed/circumscribed polygons, Cauchy's limit definitions, et cetera). Walk Example 1.
Find $C$ for $r = 4.5$ ($9\pi$). Find $d$ for $C = 100$ ($d = 100/\pi \approx 31.83$).
Problem Set 10.1, problems 1–4.
$\pi$ is transcendental — it isn't the root of any polynomial with rational coefficients. Lindemann proved this in 1882, settling forever the impossibility of squaring the circle. A satisfying loose end of antiquity, finally tied.
Problem Set 10.1, problem 5.
A clock face. From 12 to 3 is what fraction of the circle? What central angle? (One quarter; $90°$.)
Central angle = arc measure; minor arc $\leq 180°$, major arc $\geq 180°$; major + minor $= 360°$. Arc length formula. Walk Example 2. Distinguish arc measure (degrees) from arc length (linear units).
Arc length for $150°$ angle in $r = 6$ circle. ($150/360 \cdot 12\pi = 5\pi$.)
Problem Set 10.2, problems 1–4.
Two arcs share endpoints; together they exhaust the circle. Mind the distinction between an arc's degrees and its length in inches — they answer different questions.
Problem Set 10.2, problem 5.
Fold a paper circle so a chord lies along a crease. The fold passes through the centre — because the perpendicular bisector of a chord always does.
State the three chord–arc theorems. Key tool: dropping a perpendicular from centre to chord gives a right triangle. Walk Example 3 — note the Pythagorean step.
Chord of length $24$ in circle of radius $13$ — distance from centre? ($\sqrt{169 - 144} = 5$.)
Problem Set 10.3, problems 1–4.
Right triangles, hiding inside circles. They show up everywhere — a recurring theme.
Problem Set 10.3, problem 5.
Draw a circle, mark a diameter, then any point on the circle and connect. Measure the angle at that point. It will be $90°$ — every time. Thales smiles from antiquity.
Inscribed angle = half the intercepted arc. Three consequences: angles on the same arc are congruent; angle in a semicircle is right; opposite angles of a cyclic quadrilateral are supplementary. Mention Thales' anecdote (well, eclipse). Walk Example 4.
Inscribed angle of $25°$ intercepts arc of? ($50°$.) Two inscribed angles on the same arc — relationship? (Congruent.)
Problem Set 10.4, problems 1–4.
The Inscribed Angle Theorem is the single most generative result in circle geometry. Most of what follows is corollary or extension.
Problem Set 10.4, problem 5.
Roll a coin along a ruler. The point of contact at any instant: a single point. The ruler is tangent to the coin at that point.
Tangent meets circle at exactly one point; it's perpendicular to the radius at that point. From an external point, two tangent segments — congruent. Walk Example 5 (Pythagoras in the right triangle $OTP$).
From external point $P$, $OP = 17$, $r = 8$. Tangent length? ($\sqrt{289 - 64} = 15$.)
Problem Set 10.5, problems 1–4.
Tangents are the limiting case of secants — and the right-angle relation is what makes them computationally tractable.
Problem Set 10.5, problem 5.
Three cases to keep straight: vertex on the circle (half the arc), vertex inside (half the sum), vertex outside (half the difference). Memorize the locations and the operations follow.
State both theorems. The two-secant-outside case yields half the difference; the two-chord-inside case yields half the sum. Geometric intuition: as the vertex moves outward, the angle shrinks toward zero. Walk Example 6.
Two chords cross inside; arcs $80°$ and $40°$. Find the angle. ($60°$.)
Problem Set 10.6, problems 1–4.
Three positions of the vertex, three formulas — but the same idea: angles relate to arcs by halves.
Problem Set 10.6, problem 5.
Two chords crossing: $a \cdot b = c \cdot d$. Cute. The "power of a point" is a single invariant that absorbs three apparent rules.
State the three forms: two chords ($ab = cd$); two secants ($a_1 b_1 = a_2 b_2$, whole times external for each); tangent–secant ($t^2 = a \cdot b$). All express the same quantity — the "power" — relative to a fixed point. Walk Example 7.
Tangent $t = 10$ from external $P$; secant external part $5$ — whole secant? ($100 = 5b \Rightarrow b = 20$.)
Problem Set 10.7, problems 1–4.
The unified statement: for any line through point $P$ meeting the circle at $X, Y$, the product $\overline{PX} \cdot \overline{PY}$ (with signed lengths) is the same — the power of $P$. Three theorems collapse to one.
Problem Set 10.7, problem 5.
Distance formula: $(x - h)^2 + (y - k)^2 = r^2$ just says "every point on the circle is distance $r$ from the centre $(h, k)$." Geometry made algebraic.
Standard form derivation. Completing the square — review with a numerical example before doing the circle case. Walk Example 8.
Convert $x^2 + y^2 - 10x + 6y + 9 = 0$ to standard form. ($(x - 5)^2 + (y + 3)^2 = 25$; centre $(5, -3)$, radius $5$.)
Problem Set 10.8, problems 1–4.
Chapter 10 wrap. The circle is geometry's tidiest object — and one of its most prolific. Quiz next class.
Problem Set 10.8, problem 5.
Cut a parallelogram along a perpendicular and slide one piece to form a rectangle. Area unchanged. The base-times-height formula falls out of the demonstration.
State both formulas. Emphasize: height is the perpendicular distance, not the slant. The triangle is half a parallelogram on the same base and height. Walk Example 1.
Triangle with sides $8, 6$ and included angle $30°$. Area $= \tfrac{1}{2}(8)(6)\sin 30° = 12$. (Foreshadowing trigonometric area.)
Problem Set 11.1, problems 1–4.
Two parallelograms on the same base, between the same parallels, have the same area — Euclid's Proposition 35, and the seed of integral calculus.
Problem Set 11.1, problem 5.
Trapezoid area: average of the two bases, times the height. A trapezoid is morally just a rectangle whose width is the average of two slightly differing widths.
State trapezoid formula; derive by either dissection (cutting and recombining) or duplication (two trapezoids form a parallelogram). For rhombi and kites: $A = \tfrac{1}{2} d_1 d_2$. Walk Example 2.
Trapezoid: $b_1 = 9, b_2 = 15, h = 6$. ($A = 72$.)
Problem Set 11.2, problems 1–4.
Two formulas, half a dozen quadrilateral types. Mathematics often rewards a willingness to look for the same idea wearing different clothes.
Problem Set 11.2, problem 5.
Slice a circle into thin sectors and rearrange them alternating point-up and point-down. You get something that looks like a parallelogram with base $\pi r$ and height $r$ — area $\pi r^2$. Archimedes' argument, basically.
$A = \pi r^2$. Sector area: proportion of the circle's area equal to the fraction of $360°$ swept. Walk Example 3.
Sector area for $135°$, $r = 8$. ($\tfrac{135}{360} \cdot 64\pi = 24\pi$.)
Problem Set 11.3, problems 1–4.
$\pi$ does double duty: ratio of circumference to diameter, and (with $r^2$) the area constant. Same $\pi$, two appearances — calculus eventually explains why.
Problem Set 11.3, problem 5.
A regular hexagon = six congruent equilateral triangles. The apothem is the height of one triangle. $A = \tfrac{1}{2} a P$ is "six little triangle bases times their common height, halved."
$A = \tfrac{1}{2} a P$. Composite figures: decompose into known shapes; add or subtract. Walk Example 4.
Regular hexagon side $4$. Apothem $= 2\sqrt{3}$; perimeter $= 24$. Area $= \tfrac{1}{2}(2\sqrt{3})(24) = 24\sqrt{3}$.
Problem Set 11.4, problems 1–4.
As $n$ grows, the regular $n$-gon approaches its circumscribing circle. Apothem $\to r$, perimeter $\to 2\pi r$, area $\to \pi r^2$. Archimedes lurking again.
Problem Set 11.4, problem 5.
Galileo's argument: a kitten cannot grow into a tiger by scaling up. Volume scales as the cube of length; bone cross-section only as the square. The legs would crumble.
If linear scale is $k$, areas scale as $k^2$ — a direct consequence of area being a product of two linear measures. Mention the parallel volume rule ($k^3$) coming in Chapter 12. Walk Example 5.
Two similar rectangles, sides in ratio $1:4$. Area ratio? ($1:16$.) If areas are $4:9$, sides? ($2:3$.)
Problem Set 11.5, problems 1–4.
Chapter 11 wrap. The square-cube law is one of those quietly profound facts: dimensional scaling does much of the work in physics, biology, and engineering. Quiz next class.
Problem Set 11.5, problem 5.
Hand round a cube. Count vertices, edges, faces. Plug into $V - E + F$. The answer is always $2$ for any convex polyhedron — try a tetrahedron next.
Polyhedron, prism, pyramid, cylinder, cone, sphere — vocabulary. Euler's formula and its topological content (the $2$ is the Euler characteristic of the sphere; a torus would give $0$). Walk Example 1.
Pentagonal prism: $V = 10$, $F = 7$. Find $E$. ($E = 15$.)
Problem Set 12.1, problems 1–4.
Euler's formula opens the door to topology — geometry that ignores stretching but minds the holes. A doughnut and a coffee cup are topologically the same; a cube and a sphere also share their character.
Problem Set 12.1, problem 5.
Unwrap a soup-can label. It's a rectangle: width = circumference, height = can height. Cylinder lateral area in one move.
Prism $SA = 2B + Ph$. Cylinder $SA = 2\pi r^2 + 2\pi r h$. Both follow from unrolling. Walk Example 2.
Cube of side $5$ — surface area? ($6 \cdot 25 = 150$.)
Problem Set 12.2, problems 1–4.
Surface area is the dimension of paint, packaging, and heat loss. Volume is what's inside. Different units, different formulas, different intuitions.
Problem Set 12.2, problem 5.
Unroll a paper cone — it becomes a sector. Radius of sector = slant height of cone. Arc length of sector = circumference of cone's base. Area: $\tfrac{1}{2}(\text{arc})(\text{radius}) = \pi r \ell$. There's your lateral formula.
Pyramid $SA = B + \tfrac{1}{2} P\ell$. Cone $SA = \pi r^2 + \pi r \ell$. Distinguish slant height $\ell$ from vertical height $h$: related by Pythagoras, $\ell^2 = h^2 + r^2$ (for a cone). Walk Example 3.
Cone $r = 5, h = 12$. Find $\ell$ ($13$), then $SA$ ($25\pi + 65\pi = 90\pi$).
Problem Set 12.3, problems 1–4.
Mind the difference between $h$ and $\ell$. Confusing the two is the most common error on every quiz I've ever graded.
Problem Set 12.3, problem 5.
Stack of paper, perfectly aligned vs. skewed sideways: same volume. Cavalieri's principle in one demonstration — the cross-sectional area at every height is the same.
Volume = base area × height. Same formula for oblique and right prisms (Cavalieri). Cylinder: $\pi r^2 h$. Walk Example 4.
Cylinder $r = 4, h = 7$. $V = 112\pi$.
Problem Set 12.4, problems 1–4.
Cavalieri was a Jesuit priest and a forerunner of calculus. The "method of indivisibles" was a stepping stone from antiquity's geometric proofs to Newton's and Leibniz's integrals.
Problem Set 12.4, problem 5.
Fill the cone with rice, pour into the cylinder. Repeat. Three cones fill one cylinder. Hence the $\tfrac{1}{3}$.
Pyramid: $V = \tfrac{1}{3} Bh$. Cone: $V = \tfrac{1}{3}\pi r^2 h$. The $\tfrac{1}{3}$ comes from Cavalieri or, equivalently, from integration ($\int_0^h \pi (r \tfrac{x}{h})^2 dx = \tfrac{1}{3}\pi r^2 h$). Walk Example 5.
Pyramid: base $6 \times 6$, height $10$. ($V = \tfrac{1}{3}(36)(10) = 120$.)
Problem Set 12.5, problems 1–4.
A pyramid is one-third of its bounding prism. Same height, same base: factor of three. A satisfyingly clean ratio.
Problem Set 12.5, problem 5.
Archimedes wanted a sphere inscribed in a cylinder carved on his tombstone. The ratio of their volumes is exactly $2:3$ — and he was so taken with this that he counted it among his greatest discoveries. Worth more than a $5$ on the AP, in his book.
$SA = 4\pi r^2$ — exactly four great-circle areas. $V = \tfrac{4}{3}\pi r^3$. Both are within reach of Cavalieri (sphere $\leftrightarrow$ cylinder-minus-two-cones). Walk Example 6.
Sphere of radius $4$. $SA = 64\pi$; $V = \tfrac{256}{3}\pi$.
Problem Set 12.6, problems 1–4.
Among all solids of a given surface area, the sphere encloses the most volume. The isoperimetric principle — nature's preference for round things, from soap bubbles to droplets.
Problem Set 12.6, problem 5.
On the globe, stretch a string from London to Tokyo. It hugs the Arctic — not the Mercator-projection "straight line." That's the great circle. Geometry without flat paper.
Great circle as geodesic. Spherical triangle: angle sum $> 180°$; the excess is proportional to the triangle's area. Euclid's fifth postulate fails — there are no parallels on a sphere, since any two great circles intersect. This was the door through which non-Euclidean geometry walked in (Gauss, Lobachevsky, Riemann).
Walk students through the polar-equatorial triangle: three right angles. Angle sum $270°$, exceeding $180°$ by $90°$ — and the triangle's area is exactly one-eighth of the sphere's surface.
Problem Set 12.7, problems 1–4.
General relativity uses curved spacetime, where geometry differs from Euclid's in much the same way. Light bends; triangles' angles don't add to $180°$; great-circle thinking gets you to the truth.
Problem Set 12.7, problem 5.
Two cubes of side $2$ and $3$: side ratio $2:3$. Surface area ratio? Volume ratio? ($4:9$ and $8:27$.) Squares and cubes of the scale factor.
Similar solids: linear $k$, surface area $k^2$, volume $k^3$. This is the same square-cube law that limits the size of land animals, governs heat dissipation, and explains why ice cubes melt faster than ice cubes shaped like spheres of the same mass. Walk Example 8.
Two similar pyramids: edge ratio $3:5$. Volume ratio $27:125$.
Problem Set 12.8, problems 1–4.
Chapter 12 wrap. Three dimensions, three scaling laws. Cumulative quiz next class on surface area and volume.
Problem Set 12.8, problem 5.
Flip a coin twice. Without lifting your pen, list every possible outcome. ($HH, HT, TH, TT$.) Order matters — $HT \neq TH$ in a sample space.
Sample space $S$, event $E \subseteq S$, equally-likely probability $P(E) = |E|/|S|$. Distinguish from non-equally-likely cases (a biased coin). Walk Example 1.
Roll two dice. $|S| = 36$. $P(\text{sum} = 7)$? Six favourable: $(1,6), (2,5), \dots, (6,1)$. $P = 6/36 = 1/6$.
Problem Set 13.1, problems 1–4.
Probability lives between $0$ and $1$. Anything outside that range is a sign of arithmetic gone astray.
Problem Set 13.1, problem 5.
From $\{A, B, C\}$, list all ordered pairs (no repetition): $AB, AC, BA, BC, CA, CB$. Six. Now list all unordered pairs: $\{A,B\}, \{A,C\}, \{B,C\}$. Three. The factor of $2$ between them is $2!$ — the number of ways to order each pair.
Fundamental Counting Principle. Permutations $P(n, r) = n!/(n-r)!$. Combinations $C(n, r) = n!/(r!(n-r)!)$. Distinction: order matters for permutations, not for combinations. Walk Example 2.
How many $5$-card poker hands from a $52$-card deck? $\binom{52}{5} = 2{,}598{,}960$.
Problem Set 13.2, problems 1–4.
"Combination locks" are misnomers — order does matter, so they're really permutation locks. Mathematics is petty about its vocabulary.
Problem Set 13.2, problem 5.
A bus arrives every $20$ minutes. If you arrive at the stop at a random time, what's the probability of waiting under $5$ minutes? Geometric probability on a line: $5/20 = 1/4$.
For continuous sample spaces (lines, regions, volumes), probability is a ratio of measures. Walk Example 3 (square with bullseye).
Probability a random point in a $4 \times 4$ square lies within $1$ unit of the centre. ($\pi/16 \approx 0.196$.)
Problem Set 13.3, problems 1–4.
Geometric probability bridges chapters 11–12 to chapter 13: areas and volumes become probabilities. The connection is no coincidence — measure theory underlies both.
Problem Set 13.3, problem 5.
"What's the probability a family of four children has exactly two girls?" You could enumerate ($\binom{4}{2}/16 = 3/8$), or you could flip four coins, many times, and tally. Both are legitimate. The latter is a simulation.
Simulation: a randomized procedure standing in for the experiment. The Law of Large Numbers says relative frequency converges to true probability as trials $\to \infty$. Walk Example 4.
Design a die-based simulation of a basketball player who makes $1/3$ of free throws. ($1$ or $2$ = miss; $3$–$6$ = miss too actually; better: rolls of $1$ or $2$ count as a make, $3$–$6$ as a miss.)
Problem Set 13.4, problems 1–4.
Monte Carlo methods, invented at Los Alamos in the 1940s, are simulations gone industrial. Today they price options, model climates, and design particle accelerators.
Problem Set 13.4, problem 5.
Draw two cards from a deck. With replacement vs without: independent vs dependent. The second draw's odds change in the latter — fewer cards left.
Independent: $P(A \cap B) = P(A)P(B)$. Dependent: $P(A \cap B) = P(A) P(B \mid A)$. Conditional probability shrinks the sample space to events compatible with $A$. Walk Example 5.
Bag of $5$ red, $3$ blue. Draw $2$ without replacement. $P(\text{both red}) = \tfrac{5}{8} \cdot \tfrac{4}{7} = \tfrac{20}{56} = \tfrac{5}{14}$.
Problem Set 13.5, problems 1–4.
Conditional probability is the engine of Bayes' theorem and modern statistical inference. We've planted the seed; calculus and statistics will tend it.
Problem Set 13.5, problem 5.
From a deck, draw one card. $P(\text{king}) = 4/52$. $P(\text{heart}) = 13/52$. $P(\text{king or heart})$? Add them and subtract the overlap — the king of hearts, which we'd otherwise double-count.
$P(A \cup B) = P(A) + P(B) - P(A \cap B)$. Mutually exclusive: $P(A \cap B) = 0$, so the third term vanishes. Inclusion–exclusion generalises to many sets. Walk Example 6.
Roll a die. $P(\text{multiple of } 2 \text{ or } 3)$? Multiples of $2$: $\{2,4,6\}$, of $3$: $\{3,6\}$, intersection: $\{6\}$. $P = 3/6 + 2/6 - 1/6 = 4/6 = 2/3$.
Problem Set 13.6, problems 1–4.
End of course. Geometry — from Euclid's postulates through Riemann's curved spaces, with a side trip into Bayesian uncertainty. Cumulative final next class. You've done splendidly.
Problem Set 13.6, problem 5.
— End of lesson plans —