Geometry

Level · High-school Euclidean geometry Chapters · 11 Explorers · 3 interactive Feeds into · SAT · ACT · IGCSE

Foundations & Reasoning

Geometry begins with undefined terms — point, line, plane — and builds outward through definitions and logical argument. Most early results come from a handful of angle relationships.

● Definition 1.1 — Angle pairs

Two angles are complementary if they sum to $90^{\circ}$ and supplementary if they sum to $180^{\circ}$ . A linear pair (adjacent angles on a straight line) is always supplementary, and vertical angles (opposite at an intersection) are always equal.

Vertical & Linear Pair Angles

Vertical angles (opposite each other) are equal: 140° = 140° and 40° = 40°. Adjacent angles along a line form a linear pair and sum to 180°.

Midpoint

M=\Big(\tfrac{x_1+x_2}{2},\tfrac{y_1+y_2}{2}\Big)

Complementary

\alpha+\beta=90^{\circ}

Supplementary

\alpha+\beta=180^{\circ}

Vertical angles

equal in measure

Worked example 1.A

Find the complement of $35^{\circ}$ .

90^{\circ}-35^{\circ}=55^{\circ}.

Worked example 1.B

Angles $\angle 1$ and $\angle 2$ form a linear pair with $\angle 1=125^{\circ}$ . Find $\angle 2$ .

180^{\circ}-125^{\circ}=55^{\circ}.

Common Pitfall — Assuming Angles from the Drawing

Never assume two angles are complementary, supplementary, or equal simply because they look that way in a drawing. Unless an angle has an explicit right-angle box ( $90^\circ$ ), parallel markings, or the problem states that points lie on a straight line, figures on standardized tests (SAT, ACT, EST) are not drawn to scale. Always verify relationships with a theorem or postulate.

Two angles

\angle P

and

\angle Q

form a linear pair. If

m\angle P = (4x + 12)^{\circ}

and

m\angle Q = (2x + 18)^{\circ}

, find the measure of the smaller angle.

Solution: Since they form a linear pair, their measures sum to $180^{\circ}$ :

(4x + 12) + (2x + 18) = 180 \implies 6x + 30 = 180 \implies 6x = 150 \implies x = 25.

Now substitute $x = 25$ back into each expression:

m\angle P = 4(25) + 12 = 112^{\circ}, \quad m\angle Q = 2(25) + 18 = 68^{\circ}.

The smaller angle has measure $68^{\circ}$ .

Two angles

\angle 1

and

\angle 2

are vertical angles. If

m\angle 1 = (5x - 23)^{\circ}

and

m\angle 2 = (3x + 17)^{\circ}

, find the measure of

m\angle 1

and the measure of its supplement.

Step 1: Set up the equation using vertical angles: Since vertical angles are equal in measure:

5x - 23 = 3x + 17 \implies 2x = 40 \implies x = 20.

Step 2: Find the measure of $m\angle 1$ :

m\angle 1 = 5(20) - 23 = 100 - 23 = 77^{\circ}.

Step 3: Find its supplement:

180^{\circ} - 77^{\circ} = 103^{\circ}.

📝 Chapter Quiz

Questions Level

Lines & Angles

When a transversal cuts two parallel lines, eight angles appear — but only two distinct measures. Knowing which pairs are equal and which are supplementary unlocks most angle-chase problems.

▲ Reference — Parallel lines & a transversal

Corresponding angles are equal; alternate (interior or exterior) angles are equal; co-interior (same-side interior) angles are supplementary, summing to $180^{\circ}$ .

Transversal Across Parallel Lines

Corresponding angles (such as ∠1 and ∠5) occupy the same relative position at each intersection and are congruent.

Tilt the transversal below. Every acute angle stays equal to every other acute angle (corresponding and alternate), and each is supplementary to the obtuse one.

⊙ Explorer · Transversal anglesinteractive

Tilt the transversal

acute angle—

obtuse angle—

The two marked angles are corresponding — always equal. Acute + obtuse $=180^{\circ}$ .

Corresponding

equal

Alternate

equal

Co-interior

sum to

180^{\circ}

Angles on a line

sum to

180^{\circ}

Worked example 2.A

Two parallel lines are cut by a transversal. One interior angle is $72^{\circ}$ . Find its alternate interior angle.

Alternate interior angles are equal:

72^{\circ}.

Worked example 2.B

A co-interior angle measures $110^{\circ}$ . Find the other co-interior angle.

180^{\circ}-110^{\circ}=70^{\circ}.

Common Pitfall — Conflating Alternate and Consecutive Angles

A frequent error on SAT and ACT geometry questions is equating consecutive (same-side) interior angles instead of setting their sum to $180^{\circ}$ . Remember the letter shapes: Z-angles (alternate interior) are equal; F-angles (corresponding) are equal; but C-angles (consecutive interior) are supplementary.

Two parallel lines are intersected by a transversal line. Two consecutive interior angles have measures

(3x - 10)^{\circ}

and

(2x + 25)^{\circ}

. Find the measure of the acute angle.

Solution: Consecutive interior angles are supplementary, so they sum to $180^{\circ}$ :

(3x - 10) + (2x + 25) = 180 \implies 5x + 15 = 180 \implies 5x = 165 \implies x = 33.

Substitute $x = 33$ into both angle expressions:

3(33) - 10 = 89^{\circ}, \quad 2(33) + 25 = 91^{\circ}.

The acute angle has measure $89^{\circ}$ .

Two parallel lines

L_1

and

L_2

are cut by a transversal

T

. A pair of alternate exterior angles are given by

(7x - 14)^{\circ}

and

(4x + 31)^{\circ}

. Find the value of

x

and the measure of each alternate exterior angle.

Step 1: Set up the equation using alternate exterior angles: Alternate exterior angles created by parallel lines are congruent:

7x - 14 = 4x + 31 \implies 3x = 45 \implies x = 15.

Step 2: Evaluate the angle measure:

7(15) - 14 = 105 - 14 = 91^{\circ}.

Check: $4(15) + 31 = 60 + 31 = 91^{\circ}$ . Both alternate exterior angles measure $91^{\circ}$ .

📝 Chapter Quiz

Questions Level

Triangles & Congruence

The angles of every triangle sum to $180^{\circ}$ — no matter its shape. Congruence criteria then let us prove two triangles identical from only three pieces of information.

▲ Reference — Triangle facts

Interior angles sum to $180^{\circ}$ . An exterior angle equals the sum of the two remote interior angles. Triangles are congruent by SSS, SAS, ASA, AAS, or HL (right triangles).

Triangle Angle Sum & Exterior Angles

The three interior angles sum to 180°: 60° + 45° + 75° = 180°. Any exterior angle equals the sum of the two remote interior angles.

Drag the apex. The three interior angles change, yet their sum holds at exactly $180^{\circ}$ .

⊙ Explorer · The angle suminteractive

Slide the apex left ↔ right

∠A—

∠B—

∠C—

sum—

$\angle A+\angle B+\angle C=180^{\circ}$ for every triangle you can make.

Angle sum

A+B+C=180^{\circ}

Exterior angle

=

sum of remote interior

Isosceles

base angles equal

Congruence

SSS · SAS · ASA · AAS · HL

Worked example 3.A

Two angles of a triangle are $40^{\circ}$ and $60^{\circ}$ . Find the third.

180^{\circ}-40^{\circ}-60^{\circ}=80^{\circ}.

Worked example 3.B

The angles of a triangle are in the ratio $1:2:3$ . Find them.

Six parts total $180^{\circ}$ , so one part is $30^{\circ}$ :

30^{\circ},\ 60^{\circ},\ 90^{\circ}.

Common Pitfall — The 'SSA' Ambiguity Trap

Do not attempt to prove triangles congruent using SSA (Side-Side-Angle) or AAA (Angle-Angle-Angle). SSA does not guarantee unique triangles (two non-congruent triangles can share two sides and a non-included angle, known as the ambiguous case), while AAA proves similarity, not congruence. The only valid congruence criteria are SSS, SAS, ASA, AAS, and HL (hypotenuse-leg in right triangles).

\triangle ABC

, the exterior angle at vertex

C

measures

134^{\circ}

. The two remote interior angles have measures

m\angle A = (5x - 4)^{\circ}

and

m\angle B = (3x + 10)^{\circ}

. Find the measure of

\angle A

Solution: By the Exterior Angle Theorem, an exterior angle equals the sum of the two remote interior angles:

(5x - 4) + (3x + 10) = 134 \implies 8x + 6 = 134 \implies 8x = 128 \implies x = 16.

Substitute $x = 16$ into $m\angle A$ :

m\angle A = 5(16) - 4 = 80 - 4 = 76^{\circ}.

In isosceles triangle

\triangle PQR

PQ = PR

. The vertex angle

\angle P

measures

40^{\circ}

. An angle bisector is drawn from base angle

\angle Q

to point

D

on side

PR

. What is the measure of

\angle QDR

Step 1: Find the base angles of $\triangle PQR$ : Since $PQ = PR$ , the base angles $\angle Q$ and $\angle R$ are equal:

m\angle Q = m\angle R = \frac{180^{\circ} - 40^{\circ}}{2} = \frac{140^{\circ}}{2} = 70^{\circ}.

Step 2: Find the bisected angle measure: $QD$ bisects $\angle PQR$ , so:

m\angle DQR = \frac{70^{\circ}}{2} = 35^{\circ}.

Step 3: Apply the triangle angle sum to $\triangle QDR$ :

m\angle QDR = 180^{\circ} - (m\angle DQR + m\angle R) = 180^{\circ} - (35^{\circ} + 70^{\circ}) = 180^{\circ} - 105^{\circ} = 75^{\circ}.

📝 Chapter Quiz

Questions Level

Similarity

Similar figures have the same shape but not necessarily the same size: equal angles and sides in a constant ratio. The ratio — the scale factor — governs lengths, areas and volumes differently.

● Definition 4.1 — Similar polygons

Two polygons are similar when corresponding angles are equal and corresponding sides are in proportion. Triangles need only two equal angles (AA) to be similar.

Similar Triangles & Scale Factor

When linear side lengths scale by k = 0.6, all corresponding angles remain identical, perimeters scale by 0.6, and areas scale by k² = 0.36.

Side ratio

\dfrac{a'}{a}=k

Perimeter ratio

Area ratio

k^{2}

Triangle test

AA similarity

Worked example 4.A

A $6$ -ft person casts a $4$ -ft shadow while a tree casts a $20$ -ft shadow. Find the tree's height.

Similar triangles give $\dfrac{6}{4}=\dfrac{h}{20}$ :

h=\frac{6\times 20}{4}=30\ \text{ft}.

Worked example 4.B

Two similar figures have side ratio $1:3$ . What is the ratio of their areas?

1^{2}:3^{2}=1:9.

Common Pitfall — Scaling Areas vs. Lengths

When figures are similar with linear scale factor $k$ , lengths and perimeters scale by $k$ , but areas scale by $k^2$ and volumes scale by $k^3$ . If an exam question states: 'If a triangle's side lengths are tripled, by what factor does its area increase?', the answer is $3^2 = 9$ , NOT $3$ .

Two similar triangles

\triangle T_1

and

\triangle T_2

have areas of

45\ \text{cm}^2

and

80\ \text{cm}^2

respectively. If the perimeter of

\triangle T_1

27\ \text{cm}

, find the perimeter of

\triangle T_2

Solution: The ratio of areas is the square of the linear scale factor:

\frac{\text{Area}(T_2)}{\text{Area}(T_1)} = \frac{80}{45} = \frac{16}{9} = k^2.

Taking the positive square root gives the linear ratio $k$ :

k = \sqrt{\frac{16}{9}} = \frac{4}{3}.

Perimeters scale linearly by $k$ :

\text{Perimeter}(T_2) = \frac{4}{3} \times 27 = 36\ \text{cm}.

A 6-foot-tall surveyor casts an 8-foot shadow on level ground. At the exact same time, a nearby telecommunications tower casts a 60-foot shadow. How tall is the telecommunications tower?

Step 1: Set up the proportion of similar triangles: The sun's rays strike level ground at the same angle, creating similar right triangles between each object's height and shadow length:

\frac{\text{Height of Tower}}{\text{Shadow of Tower}} = \frac{\text{Height of Surveyor}}{\text{Shadow of Surveyor}} \implies \frac{H}{60} = \frac{6}{8}.

Step 2: Solve for the tower's height $H$ :

H = 60 \times \frac{6}{8} = 60 \times \frac{3}{4} = 45\text{ feet}.

The telecommunications tower is $45$ feet tall.

📝 Chapter Quiz

Questions Level

Right Triangles & Trigonometry

The right angle is geometry's workhorse. Pythagoras links the three sides; the special $45\text{–}45\text{–}90$ and $30\text{–}60\text{–}90$ triangles give exact ratios; and SOH-CAH-TOA handles the rest.

▲ Reference — Right-triangle toolkit

Pythagoras: $a^{2}+b^{2}=c^{2}$ . A $45\text{–}45\text{–}90$ triangle has sides $1:1:\sqrt2$ ; a $30\text{–}60\text{–}90$ triangle has sides $1:\sqrt3:2$ . Ratios: $\sin=\tfrac{\text{opp}}{\text{hyp}}$ , $\cos=\tfrac{\text{adj}}{\text{hyp}}$ , $\tan=\tfrac{\text{opp}}{\text{adj}}$ .

Right Triangle Trigonometry

For acute angle θ: sin(θ) = 5/13, cos(θ) = 12/13, and tan(θ) = 5/12. Pythagoras confirms: 5² + 12² = 25 + 144 = 169 = 13².

Pythagoras is really a statement about areas: the square on the hypotenuse equals the sum of the squares on the legs. Resize the legs and watch the three areas balance.

⊙ Explorer · Squares on the sidesinteractive

Leg a = 3

Leg b = 4

a² + b²9 + 16

c² (hypotenuse)25

The amber squares (on the legs) always add up to the indigo square (on the hypotenuse).

Pythagoras

a^{2}+b^{2}=c^{2}

45–45–90

1:1:\sqrt2

30–60–90

1:\sqrt3:2

Ratios

SOH-CAH-TOA

Worked example 5.A

A right triangle has legs $5$ and $12$ . Find the hypotenuse.

\sqrt{5^{2}+12^{2}}=\sqrt{169}=13.

Worked example 5.B

In a $30\text{–}60\text{–}90$ triangle the short leg is $1$ . Find the hypotenuse.

The hypotenuse is twice the short leg:

2.

Common Pitfall — Trigonometry on Non-Right Triangles

SOH-CAH-TOA definitions apply only to right-angled triangles. If a triangle does not contain a $90^\circ$ angle, you must either drop an altitude to create right triangles or use the Law of Sines / Law of Cosines. Also watch out for 45-45-90 vs 30-60-90 ratios: in a 30-60-90 triangle, the hypotenuse is $2x$ , while the side opposite $60^\circ$ is $x\sqrt{3}$ , NOT $x\sqrt{2}$ .

An equilateral triangle has side lengths of

12\ \text{cm}

. An altitude is drawn from one vertex to the opposite side. Find the exact height and area of the triangle.

Solution: The altitude bisects the vertex angle into two $30^{\circ}$ angles and bisects the base into two equal segments of length $6\ \text{cm}$ , forming two congruent $30\text{–}60\text{–}90$ right triangles.

The shorter leg opposite $30^{\circ}$ is $x = 6$ . The altitude (height) is the longer leg opposite $60^{\circ}$ :

h = x\sqrt{3} = 6\sqrt{3}\ \text{cm}.

Now compute the area using $\text{Area} = \tfrac{1}{2} b h$ :

\text{Area} = \frac{1}{2} (12) (6\sqrt{3}) = 36\sqrt{3}\ \text{cm}^2.

In a

45^{\circ}\text{–}45^{\circ}\text{–}90^{\circ}

right triangle, the hypotenuse has length

10\sqrt{2}\ \text{cm}

. What is the length of each leg, and what is the area of the triangle?

Step 1: Find leg length $s$ : In a $45^{\circ}\text{–}45^{\circ}\text{–}90^{\circ}$ special right triangle, the hypotenuse is $s\sqrt{2}$ :

s\sqrt{2} = 10\sqrt{2} \implies s = 10\text{ cm}.

Step 2: Calculate the area: Since both legs form the base and height:

\text{Area} = \frac{1}{2} s^2 = \frac{1}{2}(10^2) = \frac{1}{2}(100) = 50\text{ cm}^2.

📝 Chapter Quiz

Questions Level

Polygons & Quadrilaterals

Beyond the triangle lie quadrilaterals and regular polygons, each with predictable angle sums and — for the special quadrilaterals — distinctive side, angle and diagonal properties.

▲ Reference — Polygon angles

Interior angles of an $n$ -gon sum to $(n-2)\times 180^{\circ}$ ; exterior angles always sum to $360^{\circ}$ . In a parallelogram, opposite sides and opposite angles are equal and the diagonals bisect each other.

Parallelogram Properties

In any parallelogram, opposite sides are equal (AB = CD and BC = DA), opposite angles are equal (∠A = ∠C, ∠B = ∠D), and consecutive angles are supplementary.

Interior sum

(n-2)\,180^{\circ}

Regular interior

\dfrac{(n-2)180^{\circ}}{n}

Exterior angle

\dfrac{360^{\circ}}{n}

Parallelogram

opposite angles equal

Worked example 6.A

Find the sum of the interior angles of an octagon.

(8-2)\times 180^{\circ}=1080^{\circ}.

Worked example 6.B

Find each interior angle of a regular hexagon.

\frac{(6-2)\times 180^{\circ}}{6}=120^{\circ}.

Common Pitfall — Exterior vs Interior Angle Sums

A universal rule often forgotten: the sum of the exterior angles of any convex polygon is always exactly $360^\circ$ , regardless of whether it has 4 sides, 10 sides, or 100 sides! Only the interior sum depends on the number of sides $n$ via $(n-2) \cdot 180^\circ$ . Calculating the exterior angle first ( $360^\circ / n$ ) is usually the fastest shortcut for finding the interior angle ( $180^\circ - \text{exterior}$ ).

Each interior angle of a regular polygon measures

144^{\circ}

. How many sides does the polygon have?

Solution: The fastest approach is to find the exterior angle first. Since interior and exterior angles form a linear pair:

\text{Exterior angle} = 180^{\circ} - 144^{\circ} = 36^{\circ}.

Since the exterior angles of any regular $n$ -gon sum to $360^{\circ}$ :

n = \frac{360^{\circ}}{36^{\circ}} = 10.

The polygon is a regular decagon with $10$ sides.

A rhombus has diagonals of lengths

16\ \text{cm}

and

12\ \text{cm}

. Find the perimeter and area of the rhombus.

Step 1: Use diagonal properties of a rhombus: The diagonals of a rhombus are perpendicular bisectors of each other. They divide the rhombus into four congruent right triangles with legs of lengths $\frac{16}{2} = 8\text{ cm}$ and $\frac{12}{2} = 6\text{ cm}$ .

Step 2: Find the side length $s$ : Using the Pythagorean theorem:

s = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10\text{ cm}.

Step 3: Calculate the perimeter:

\text{Perimeter} = 4s = 4(10) = 40\text{ cm}.

Step 4: Calculate the area:

\text{Area} = \frac{1}{2} d_1 d_2 = \frac{1}{2}(16)(12) = 96\text{ cm}^2.

📝 Chapter Quiz

Questions Level

Circles

Circles bring their own angle and length rules: the relationship between central and inscribed angles, the right angle in a semicircle, and arcs and sectors as fractions of the whole.

▲ Reference — Circle facts

Circumference $C=2\pi r$ ; area $A=\pi r^{2}$ . The inscribed angle is half the central angle on the same arc, so the angle in a semicircle is $90^{\circ}$ . An arc and sector scale by $\dfrac{\theta}{360}$ .

Circle Sector & Central Angle

A 90° central sector represents exactly 1/4 of the total circle area: Area = (90/360) · π(4)² = 4π ≈ 12.57.

Circumference

C=2\pi r

Area

A=\pi r^{2}

Arc length

\dfrac{\theta}{360}\,2\pi r

Sector area

\dfrac{\theta}{360}\,\pi r^{2}

⊙ Explorer · Arc length & sector areainteractive

radius r: 4

angle θ: 90

arc length6.283

sector area12.566

Default r = 4, θ = 90° matches Worked example 7.B — that example rounds π to 3.14 and gets 12.56; this explorer uses exact π (≈12.566). Both a sector's arc length and its area scale by θ/360, exactly as the Reference box states.

Worked example 7.A

A triangle is inscribed in a circle with one side as the diameter. Find the angle opposite that side.

The angle in a semicircle is a right angle:

90^{\circ}.

Worked example 7.B

Find the area of a $90^{\circ}$ sector of a circle with $r=4$ . Take $\pi=3.14$ .

\frac{90}{360}\,\pi r^{2}=\tfrac14\times 3.14\times 16=12.56.

Common Pitfall — Inscribed Angle vs. Central Angle

A central angle has its vertex at the center of the circle, while an inscribed angle has its vertex on the circle's circumference. The inscribed angle is half the measure of the central angle subtending the same arc. Confusing the two is one of the most common pitfalls on circle questions. Additionally, any inscribed angle that intercepts a diameter (semicircle) is guaranteed to be $90^\circ$ (Thales' theorem).

Points

A, B, C

lie on a circle with center

O

. Inscribed angle

\angle ABC

intercepts minor arc

AC

and has measure

38^{\circ}

. What is the measure of the central angle

\angle AOC

intercepting the same arc?

Solution: The measure of a central angle is twice the measure of an inscribed angle that intercepts the same arc:

m\angle AOC = 2 \times m\angle ABC = 2 \times 38^{\circ} = 76^{\circ}.

The central angle $\angle AOC$ has measure $76^{\circ}$ , and the measure of minor arc $AC$ is also $76^{\circ}$ .

From an external point

P

, two tangents

PA

and

PB

are drawn to a circle with radius

r = 5\ \text{cm}

. The distance from

P

to the center of the circle

O

13\ \text{cm}

. Find the length of the tangents

PA

and

PB

, and find the perimeter of quadrilateral

PAOB

Step 1: Recognize tangent-radius perpendicularity: Tangents are perpendicular to the radius at the point of tangency, so $\angle PAO = \angle PBO = 90^{\circ}$ . $\triangle PAO$ is a right triangle with hypotenuse $PO = 13$ and leg $OA = 5$ .

Step 2: Solve for tangent length using the Pythagorean theorem:

PA = \sqrt{PO^2 - OA^2} = \sqrt{13^2 - 5^2} = \sqrt{169 - 25} = \sqrt{144} = 12\text{ cm}.

By equal tangent lengths from an external point, $PB = PA = 12\text{ cm}$ .

Step 3: Compute the perimeter of $PAOB$ :

\text{Perimeter} = PA + PB + OA + OB = 12 + 12 + 5 + 5 = 34\text{ cm}.

📝 Chapter Quiz

Questions Level

Area, Surface Area & Volume

The measurement strand: areas of plane figures and the surface areas and volumes of the common solids — prisms, cylinders, cones, pyramids and spheres.

▲ Reference — Key formulas

Rectangle area $=bh$ ; triangle $=\tfrac12 bh$ . Prism volume $=A_{\text{base}}\times h$ ; cylinder $=\pi r^{2}h$ ; sphere $=\tfrac43\pi r^{3}$ ; cone $=\tfrac13\pi r^{2}h$ .

Cylinder Dimensions & Volume

For a cylinder with base radius r = 3 and height h = 8, the base area is B = π(3)² = 9π and volume is V = Bh = 9π(8) = 72π ≈ 226.2.

Rectangle

A=bh

Triangle

A=\tfrac12 bh

Prism / cylinder

V=A_{\text{base}}h

Sphere / cone

\tfrac43\pi r^{3},\ \tfrac13\pi r^{2}h

Worked example 8.A

Find the volume of a cube of side $4$ .

4^{3}=64.

Worked example 8.B

Find the volume of a cylinder with $r=2$ , $h=5$ . Take $\pi=3.14$ .

\pi r^{2}h=3.14\times 4\times 5=62.8.

Common Pitfall — Cones and Pyramids Have a 1/3 Factor

A notorious test trap is forgetting the $\tfrac{1}{3}$ factor for solids that taper to an apex (cones and pyramids): $V = \tfrac{1}{3} B h$ . A cone with the same radius and height as a cylinder holds exactly one-third the volume. Also take care with surface area vs volume units: volume is in cubic units ( $\text{cm}^3$ ), surface area is in square units ( $\text{cm}^2$ ).

A cone and a cylinder have identical base radii of

6\ \text{cm}

. If both solids have the same volume of

180\pi\ \text{cm}^3

, find the height of the cylinder and the height of the cone.

Solution: The base area for both solids is $B = \pi r^2 = \pi (6^2) = 36\pi\ \text{cm}^2$ .

For the cylinder:

V_{\text{cyl}} = B \cdot h_{\text{cyl}} \implies 180\pi = 36\pi \cdot h_{\text{cyl}} \implies h_{\text{cyl}} = \frac{180\pi}{36\pi} = 5\ \text{cm}.

For the cone:

V_{\text{cone}} = \frac{1}{3} B \cdot h_{\text{cone}} \implies 180\pi = \frac{1}{3} (36\pi) \cdot h_{\text{cone}} = 12\pi \cdot h_{\text{cone}} \implies h_{\text{cone}} = \frac{180\pi}{12\pi} = 15\ \text{cm}.

The cylinder has height $5\ \text{cm}$ and the cone has height $15\ \text{cm}$ (three times as tall to achieve equal volume).

A solid metal sphere of radius

3\ \text{cm}

is melted down and recast into a cylinder of radius

2\ \text{cm}

. What is the height of the recast cylinder in cm?

Step 1: Calculate the volume of the sphere:

V_{\text{sphere}} = \frac{4}{3}\pi r^3 = \frac{4}{3}\pi (3^3) = \frac{4}{3}\pi (27) = 36\pi\text{ cm}^3.

Step 2: Equate with the volume of the cylinder:

V_{\text{cyl}} = \pi R^2 h = \pi (2^2) h = 4\pi h.

Step 3: Solve for height $h$ :

4\pi h = 36\pi \implies h = \frac{36\pi}{4\pi} = 9\text{ cm}.

The height of the cylinder is $9$ cm.

📝 Chapter Quiz

Questions Level

Transformations & Symmetry

Rigid motions — translations, reflections and rotations — preserve size and shape; dilations scale a figure.

▲ Reference — Transformations

Translations, reflections and rotations are rigid motions that preserve distance. A dilation by factor $k$ scales lengths by $k$ and areas by $k^{2}$ .

Reflection Across the y-Axis

Reflecting across the y-axis maps (x, y) to (-x, y), preserving all side lengths, angles, and area (an isometry / rigid motion).

Translation

slide

Reflection

flip

Rotation

turn

Dilation

scale by

k

⊙ Explorer · Dilationinteractive

scale factor k: 3

image side length12.00

image area54.000

The faint triangle has a side of length 4, dilated from the origin. Default k = 3 matches Worked example 9.A exactly (4 → 12). Length scales by k, but area scales by k² — drag k and watch the area readout grow much faster than the length one.

Worked example 9.A

A dilation with scale factor $3$ maps a side of length $4$ . Find the image length.

3\times 4=12.

Worked example 9.B

Point $P(3, -5)$ is rotated $90^{\circ}$ counterclockwise about the origin. Find the coordinates of its image $P'$ .

A counterclockwise rotation of $90^{\circ}$ follows the coordinate rule $(x, y) \to (-y, x)$ :

P(3, -5) \longrightarrow P'(-(-5), 3) = P'(5, 3).

Common Pitfall — Rotations vs Reflections & Dilation Centers

Watch the orientation of transformed figures: reflections invert orientation (a clockwise vertex cycle becomes counterclockwise), whereas rotations and translations preserve orientation. For dilations, remember that coordinates scale as $(kx, ky)$ only when the center of dilation is the origin $(0, 0)$ . If a dilation has center $(a, b)$ , the rule is $(x, y) \to (a + k(x - a), b + k(y - b))$ .

Point

A(-2, 4)

is reflected across the line

y = x

, and the resulting point is then translated by the vector

\langle 3, -1 \rangle

. What are the final coordinates?

Solution: Apply the transformations step by step:

Step 1: Reflection across $y = x$ swaps the coordinates: $(x, y) \to (y, x)$ :

A(-2, 4) \longrightarrow A'(4, -2).

Step 2: Translation by $\langle 3, -1 \rangle$ adds 3 to the x-coordinate and subtracts 1 from the y-coordinate:

A'(4, -2) \longrightarrow A''(4 + 3, -2 - 1) = A''(7, -3).

The final coordinates are $(7, -3)$ .

Triangle

\triangle ABC

has vertices

A(1, 2)

B(3, 2)

, and

C(1, 5)

. The triangle is dilated by a scale factor of

k = 3

centered at the origin, and then translated by vector

\langle -2, 4 \rangle

. What are the coordinates of vertex

C''

Step 1: Dilation centered at the origin: Multiply each coordinate of $C(1, 5)$ by the scale factor $k = 3$ :

C(1, 5) \longrightarrow C'(3 \times 1, 3 \times 5) = C'(3, 15).

Step 2: Translation by vector $\langle -2, 4 \rangle$ : Add $-2$ to the x-coordinate and $4$ to the y-coordinate:

C'(3, 15) \longrightarrow C''(3 - 2, 15 + 4) = C''(1, 19).

The final coordinates of vertex $C''$ are $(1, 19)$ .

📝 Chapter Quiz

Questions Level

Coordinate Geometry

Placing figures on the plane lets us use distance, midpoint and slope to prove geometric facts.

▲ Reference — On the plane

Distance $\sqrt{(x_2-x_1)^{2}+(y_2-y_1)^{2}}$ ; midpoint $\left(\tfrac{x_1+x_2}{2},\tfrac{y_1+y_2}{2}\right)$ . Parallel lines share a slope; perpendicular slopes multiply to $-1$ .

Perpendicular Lines on the Coordinate Plane

Line L₁ with slope m₁ = 2 and line L₂ with slope m₂ = -1/2 intersect at (2, 3) at a right angle (m₁ · m₂ = -1).

Distance

\sqrt{\Delta x^{2}+\Delta y^{2}}

Midpoint

average the ends

Parallel

equal slopes

Perpendicular

m_1m_2=-1

⊙ Explorer · Distance, midpoint & slopeinteractive

A_x: 0

A_y: 0

B_x: 6

B_y: 8

distance AB10.000

midpoint(3.00, 4.00)

Default A = (0, 0), B = (6, 8) matches Worked example 10.A exactly (distance 10). Drag either point and watch the distance and midpoint formulas from the Reference box update live.

Worked example 10.A

Find the distance from $(0,0)$ to $(6,8)$ .

\sqrt{36+64}=10.

Worked example 10.B

Line $L_1$ passes through $(1, 2)$ and $(4, 8)$ . Find the slope of a line $L_2$ perpendicular to $L_1$ .

First compute the slope of $L_1$ :

m_1 = \frac{8 - 2}{4 - 1} = \frac{6}{3} = 2.

Perpendicular slopes are negative reciprocals: $m_1 \cdot m_2 = -1$ :

m_2 = -\frac{1}{m_1} = -\frac{1}{2}.

Common Pitfall — Opposite vs Negative Reciprocal

A common mistake is taking only the negative of the slope without inverting it (e.g. thinking the perpendicular slope to $3$ is $-3$ ). Perpendicular lines have negative reciprocals: both flip the sign and invert the fraction ( $3 \to -\tfrac{1}{3}$ ). Special case: horizontal lines ( $m = 0$ ) are perpendicular to vertical lines (slope undefined).

Segment

AB

has endpoints

A(2, -1)

and

B(6, 7)

. Find the equation of the perpendicular bisector of

AB

in slope-intercept form.

Solution: The perpendicular bisector passes through the midpoint and has a perpendicular slope.

Step 1 (Midpoint):

M = \left(\frac{2 + 6}{2},\ \frac{-1 + 7}{2}\right) = (4, 3).

Step 2 (Slope of AB):

m_{AB} = \frac{7 - (-1)}{6 - 2} = \frac{8}{4} = 2.

Step 3 (Perpendicular slope):

m_{\perp} = -\frac{1}{2}.

Step 4 (Point-slope equation through M(4, 3)):

y - 3 = -\frac{1}{2}(x - 4) \implies y - 3 = -\frac{1}{2}x + 2 \implies y = -\frac{1}{2}x + 5.

Find the coordinates of the point

P

on directed line segment from

A(-3, 2)

B(7, 12)

that partitions the segment in the ratio

2 : 3

Step 1: Determine the fractional distance: A ratio of $2 : 3$ means $P$ is $\frac{2}{2 + 3} = \frac{2}{5}$ of the way from $A$ to $B$ .

Step 2: Apply the partition formula:

x_P = x_1 + \frac{2}{5}(x_2 - x_1) = -3 + \frac{2}{5}(7 - (-3)) = -3 + \frac{2}{5}(10) = -3 + 4 = 1.

y_P = y_1 + \frac{2}{5}(y_2 - y_1) = 2 + \frac{2}{5}(12 - 2) = 2 + \frac{2}{5}(10) = 2 + 4 = 6.

The partitioning point is $P(1, 6)$ .

📝 Chapter Quiz

Questions Level

Probability

Probability measures the chance of an event, including area-based (geometric) models.

▲ Reference — Probability

$P=\tfrac{\text{favourable}}{\text{total}}$ . Geometric probability compares lengths or areas. Independent events multiply, and $P(\text{not }A)=1-P(A)$ .

Geometric Probability: Target Area

In geometric probability, P(Event) = (Area of Target) / (Total Area). Scaling linear radius by 1/3 reduces target area to (1/3)² = 1/9.

Probability

\tfrac{\text{fav}}{\text{total}}

Geometric

ratio of areas

Independent

P(A)P(B)

Complement

1-P

Worked example 11.A

A point lands in a $10\times10$ square. Find the chance it lands in a $2\times2$ corner.

\frac{4}{100}=0.04.

Worked example 11.B

A circular dartboard with radius $r = 6$ has a bullseye of radius $2$ . If a dart lands randomly on the board, find the probability it lands in the bullseye.

Compute the ratio of target area to total area:

P(\text{bullseye}) = \frac{\pi(2)^2}{\pi(6)^2} = \frac{4\pi}{36\pi} = \frac{1}{9} \approx 0.111.

Common Pitfall — Ring Area in Concentric Targets

When computing the probability of landing in a circular ring (annulus) between radius $r_1$ and $r_2$ (where $r_2 > r_1$ ), the favorable area is $\pi r_2^2 - \pi r_1^2$ . A classic trap is calculating $\pi(r_2 - r_1)^2$ , which severely underestimates the ring's area because $(r_2 - r_1)^2 \neq r_2^2 - r_1^2$ .

A circle is inscribed inside a square of side length

10\ \text{cm}

. A point is chosen at random inside the square. What is the exact probability that the point lands outside the circle (in the four shaded corners)?

Solution: The diameter of the inscribed circle equals the side of the square ( $10\ \text{cm}$ ), so the radius is $r = 5\ \text{cm}$ .

Step 1 (Total area):

\text{Area}(\text{square}) = 10^2 = 100\ \text{cm}^2.

Step 2 (Circle area):

\text{Area}(\text{circle}) = \pi(5^2) = 25\pi\ \text{cm}^2.

Step 3 (Corner area):

\text{Area}(\text{outside}) = 100 - 25\pi.

Step 4 (Probability):

P(\text{outside}) = \frac{100 - 25\pi}{100} = 1 - \frac{\pi}{4} \approx 1 - 0.7854 = 0.2146\ (21.5\%).

A circular archery target has three concentric rings of radii

10\ \text{cm}

20\ \text{cm}

, and

30\ \text{cm}

. If an arrow hits the target completely at random, what is the exact probability that it lands in the middle ring (between radii

10\ \text{cm}

and

20\ \text{cm}

Step 1: Calculate the total area of the target: The outer circle has radius $R = 30\text{ cm}$ :

\text{Area}_{\text{total}} = \pi (30^2) = 900\pi\text{ cm}^2.

Step 2: Calculate the area of the middle ring (annulus): The middle ring is bounded between $r_1 = 10\text{ cm}$ and $r_2 = 20\text{ cm}$ :

\text{Area}_{\text{ring}} = \pi (20^2) - \pi (10^2) = 400\pi - 100\pi = 300\pi\text{ cm}^2.

Step 3: Compute the geometric probability:

P(\text{middle ring}) = \frac{\text{Area}_{\text{ring}}}{\text{Area}_{\text{total}}} = \frac{300\pi}{900\pi} = \frac{1}{3} \approx 33.3\%.

📝 Chapter Quiz

Questions Level

Practice set

Fifty problems spanning all eight chapters — foundations through area and volume — tagged by difficulty. Click any problem to reveal a full worked solution.

$M$ is the midpoint of $\overline{AB}$ with $AB=12$ . Find $AM$ .

Basic

\tfrac12\times 12=6.

Find the complement of $35^{\circ}$ .

Basic

90^{\circ}-35^{\circ}=55^{\circ}.

Find the supplement of $110^{\circ}$ .

Basic

180^{\circ}-110^{\circ}=70^{\circ}.

Two angles are vertical and one measures $70^{\circ}$ . Find the other.

Intermediate

Vertical angles are equal:

70^{\circ}.

An angle bisector splits an $80^{\circ}$ angle. Find each part.

Intermediate

\tfrac12\times 80^{\circ}=40^{\circ}.

$\angle 1$ and $\angle 2$ form a linear pair with $\angle 1=125^{\circ}$ . Find $\angle 2$ .

Intermediate

180^{\circ}-125^{\circ}=55^{\circ}.

Parallel lines are cut by a transversal. Find the corresponding angle to $65^{\circ}$ .

Basic

Corresponding angles are equal:

65^{\circ}.

Find the alternate interior angle to $72^{\circ}$ .

Intermediate

72^{\circ}.

One co-interior angle is $110^{\circ}$ . Find the other.

Intermediate

180^{\circ}-110^{\circ}=70^{\circ}.

Two corresponding angles measure $3x$ and $75^{\circ}$ . Find $x$ .

Intermediate

Equal, so $3x=75$ :

x=25.

Alternate angles measure $2x$ and $x+40$ . Find $x$ .

Intermediate

$2x=x+40$ :

x=40.

Angles on a straight line are $120^{\circ}$ and $x$ . Find $x$ .

Basic

180^{\circ}-120^{\circ}=60^{\circ}.

Two angles of a triangle are $40^{\circ}$ and $60^{\circ}$ . Find the third.

Basic

180^{\circ}-40^{\circ}-60^{\circ}=80^{\circ}.

An isosceles triangle has an apex angle of $40^{\circ}$ . Find each base angle.

Intermediate

Base angles are equal: $\tfrac{180^{\circ}-40^{\circ}}{2}$ :

70^{\circ}.

State the measure of each angle in an equilateral triangle.

Intermediate

\frac{180^{\circ}}{3}=60^{\circ}.

An exterior angle of a triangle has remote interior angles $50^{\circ}$ and $60^{\circ}$ . Find it.

Intermediate

Exterior angle = sum of remote interior angles:

50^{\circ}+60^{\circ}=110^{\circ}.

A right triangle has one acute angle of $35^{\circ}$ . Find the other acute angle.

Intermediate

90^{\circ}-35^{\circ}=55^{\circ}.

A triangle has angles $x$ , $x$ and $80^{\circ}$ . Find $x$ .

Intermediate

$2x+80=180$ :

x=50.

The angles of a triangle are in the ratio $1:2:3$ . Find the largest.

Advanced

Six parts $=180^{\circ}$ , one part $=30^{\circ}$ ; largest is 3 parts:

90^{\circ}.

State the sum of the exterior angles of any triangle.

Intermediate

360^{\circ}.

Two similar triangles have scale factor 2. A side of the small one is 5. Find the matching side of the large one.

Basic

5\times 2=10.

Similar triangles satisfy $\dfrac{3}{6}=\dfrac{x}{8}$ . Find $x$ .

Intermediate

x=\frac{3\times 8}{6}=4.

Two similar figures have side ratio $1:3$ . Find the ratio of their areas.

Intermediate

1^{2}:3^{2}=1:9.

A $6$ -ft person casts a $4$ -ft shadow; a tree casts a $20$ -ft shadow. Find the tree's height.

Intermediate

\frac{6}{4}=\frac{h}{20}\Rightarrow h=30\ \text{ft}.

In similar triangles, sides 4 and 10 correspond. A second side of the small triangle is 6. Find its match.

Intermediate

Scale factor $\tfrac{10}{4}=2.5$ :

6\times 2.5=15.

Two similar solids have scale factor 3. By what factor does the area scale?

Advanced

3^{2}=9.

A right triangle has legs 3 and 4. Find the hypotenuse.

Basic

\sqrt{3^{2}+4^{2}}=5.

A right triangle has legs 5 and 12. Find the hypotenuse.

Intermediate

\sqrt{5^{2}+12^{2}}=\sqrt{169}=13.

In a $45\text{–}45\text{–}90$ triangle each leg is 1. Find the hypotenuse.

Intermediate

\sqrt{1^{2}+1^{2}}=\sqrt2.

In a $30\text{–}60\text{–}90$ triangle the short leg is 1. Find the hypotenuse.

Intermediate

2.

A right triangle has hypotenuse 10 and one leg 6. Find the other leg.

Intermediate

\sqrt{10^{2}-6^{2}}=\sqrt{64}=8.

Evaluate $\sin 30^{\circ}$ .

Intermediate

\tfrac12.

Evaluate $\cos 60^{\circ}$ .

Advanced

\tfrac12.

Find the sum of the interior angles of a quadrilateral.

Basic

(4-2)\times 180^{\circ}=360^{\circ}.

Find each interior angle of a regular hexagon.

Intermediate

\frac{(6-2)\times 180^{\circ}}{6}=120^{\circ}.

Find the sum of the interior angles of an octagon.

Intermediate

(8-2)\times 180^{\circ}=1080^{\circ}.

In a parallelogram one angle is $70^{\circ}$ . Find its opposite angle.

Intermediate

Opposite angles of a parallelogram are equal:

70^{\circ}.

A rectangle has one diagonal of length 10. Find the length of the other diagonal.

Intermediate

A rectangle's diagonals are equal:

10.

Find the exterior angle of a regular pentagon.

Intermediate

\frac{360^{\circ}}{5}=72^{\circ}.

A polygon has interior angles summing to $1440^{\circ}$ . How many sides has it?

Advanced

$(n-2)180=1440\Rightarrow n-2=8$ :

n=10.

Find the circumference of a circle of radius 7. Take $\pi=\tfrac{22}{7}$ .

Basic

2\pi r=2\times\tfrac{22}{7}\times 7=44.

Find the area of a circle of radius 5. Take $\pi=3.14$ .

Intermediate

\pi r^{2}=3.14\times 25=78.5.

A triangle is inscribed in a circle with one side a diameter. Find the angle opposite that side.

Intermediate

Angle in a semicircle:

90^{\circ}.

A central angle is $100^{\circ}$ . Find the inscribed angle on the same arc.

Intermediate

\tfrac12\times 100^{\circ}=50^{\circ}.

Find the arc length of a quarter circle of radius 8. Take $\pi=3.14$ .

Intermediate

\tfrac14\times 2\pi r=\tfrac14\times 2\times 3.14\times 8=12.56.

Find the area of a $90^{\circ}$ sector of a circle with $r=4$ . Take $\pi=3.14$ .

Advanced

\tfrac14\,\pi r^{2}=\tfrac14\times 3.14\times 16=12.56.

Find the area of a rectangle measuring $8\times 5$ .

Basic

8\times 5=40.

Find the volume of a cube of side 4.

Intermediate

4^{3}=64.

Find the volume of a cylinder with $r=2$ , $h=5$ . Take $\pi=3.14$ .

Intermediate

\pi r^{2}h=3.14\times 4\times 5=62.8.

A triangle has base 12 and height 5. Find its area.

Advanced

\tfrac12\times 12\times 5=30.

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