Course Review

Study Sheet

Course Review

Every key definition, theorem, and formula from all ten units

Points, Lines, Planes & Segments

Tip

How to use this sheet. Each section is one unit. Concept boxes give the definitions and rules, example boxes show a worked calculation, and the red tip box at the end of each unit warns you about the most common mistake. Keep answers exact (leave π\pi and radicals) unless a decimal is requested.

Concept
The undefined terms and basic figures
  • [leftmargin=6mm,itemsep=2pt]
  • Point: a location, no size (named AA). Line: extends forever in two directions (AB\overleftrightarrow{AB}). Plane: a flat surface extending forever.
  • Collinear points lie on one line; coplanar points lie in one plane.
  • Segment AB\overline{AB}: two endpoints and all points between (has length ABAB). Ray AB\overrightarrow{AB}: one endpoint, goes forever one way.
  • Congruent segments have equal length: ABCD\overline{AB}\cong\overline{CD} means AB=CDAB = CD.
Concept
Segment addition, midpoint, and distance
  • [leftmargin=6mm,itemsep=2pt]
  • Segment Addition Postulate: if BB is between AA and CC, then AB+BC=ACAB + BC = AC.
  • Midpoint MM of AB\overline{AB}: the point with AM=MBAM = MB (it bisects the segment).

On a coordinate plane, for A(x1,y1)A(x_1,y_1) and B(x2,y2)B(x_2,y_2):

Midpoint=(x1+x22, y1+y22)Distance AB=(x2x1)2+(y2y1)2\begin{aligned} \text{Midpoint} &= \left(\frac{x_1+x_2}{2},\ \frac{y_1+y_2}{2}\right)\\[2pt] \text{Distance } AB &= \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2} \end{aligned}
Example
Midpoint and distance

For A(1,2)A(1,2) and B(7,10)B(7,10):

M=(1+72,2+102)=(4,6),AB=(71)2+(102)2=36+64=100=10.M = \left(\tfrac{1+7}{2},\tfrac{2+10}{2}\right) = (4,6),\qquad AB = \sqrt{(7-1)^2+(10-2)^2} = \sqrt{36+64} = \sqrt{100} = 10.
Tip

Midpoint adds, distance subtracts. The midpoint formula averages (add the coordinates, divide by 2); the distance formula subtracts coordinates first, then squares. Squaring makes signs irrelevant, so order of subtraction does not matter.

Reminder — Distance and midpoint:d=(x2x1)2+(y2y1)2,M=(x1+x22, y1+y22)d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}, \qquad M=\left(\frac{x_1+x_2}{2},\ \frac{y_1+y_2}{2}\right)

Angles & Angle Relationships

Concept
Types of angles

Measured in degrees (^\circ): acute <90<90^\circ, right =90=90^\circ, obtuse between 9090^\circ and 180180^\circ, straight =180=180^\circ. An angle bisector splits an angle into two equal halves.

Concept
Angle pairs
  • [leftmargin=6mm,itemsep=2pt]
  • Complementary: two angles that add to 9090^\circ.
  • Supplementary: two angles that add to 180180^\circ.
  • Linear pair: adjacent angles on a straight line --- they are supplementary (=180=180^\circ).
  • Vertical angles: the opposite angles formed by two crossing lines --- they are congruent (equal).
  • Angle Addition Postulate: if DD is inside ABC\angle ABC, then ABD+DBC=ABC\angle ABD + \angle DBC = \angle ABC.
Example
Vertical and linear pair

Two lines cross. One angle is 7070^\circ. Its vertical angle is also 7070^\circ. Each angle next to it (linear pair) is 18070=110180^\circ - 70^\circ = 110^\circ.

Tip

Vertical vs. linear. Vertical angles are equal; a linear pair is supplementary (adds to 180180^\circ). Do not confuse “complementary” (9090^\circ) with “supplementary” (180180^\circ) --- c comes before s, and 9090 before 180180.

Parallel & Perpendicular Lines

Concept
Angles formed by a transversal

When a transversal cuts two parallel lines, these pairs are formed:

  • [leftmargin=6mm,itemsep=2pt]
  • Corresponding angles: congruent (equal).
  • Alternate interior angles: congruent.
  • Alternate exterior angles: congruent.
  • Co-interior (same-side interior) angles: supplementary (=180=180^\circ).

The converse also works: if any of these relationships holds, the lines are parallel.

Concept
Slopes of parallel and perpendicular lines

For slope mm (recall m=y2y1x2x1m = \dfrac{y_2-y_1}{x_2-x_1}, “rise over run”):

  • [leftmargin=6mm,itemsep=2pt]
  • Parallel lines have equal slopes: m1=m2m_1 = m_2.
  • Perpendicular lines have opposite-reciprocal slopes: m1m2=1m_1 \cdot m_2 = -1, i.e. m2=1m1m_2 = -\dfrac{1}{m_1}.

Horizontal (m=0m=0) and vertical (undefined slope) lines are perpendicular to each other.

Example
Parallel and perpendicular slopes

A line has slope m=23m = \tfrac{2}{3}. A line parallel to it also has slope 23\tfrac{2}{3}. A line perpendicular to it has slope 32-\tfrac{3}{2} (flip and negate).

Tip

Flip and negate. Perpendicular slope == reciprocal with the sign changed. Just flipping (2332\tfrac{2}{3}\to\tfrac{3}{2}) or just negating (2323\tfrac{2}{3}\to-\tfrac{2}{3}) is not enough --- you must do both.

Triangles & Congruence

Concept
Angle facts in triangles
  • [leftmargin=6mm,itemsep=2pt]
  • Triangle Angle Sum: the three interior angles add to 180180^\circ.
  • Exterior Angle Theorem: an exterior angle equals the sum of the two remote (non-adjacent) interior angles.
  • Classify by sides: scalene (none equal), isosceles (two equal), equilateral (all equal). By angles: acute, right, obtuse.
Concept
Proving triangles congruent

Two triangles are congruent if any one of these matches: multicols2

  • [leftmargin=5mm,itemsep=1pt]
  • SSS --- three sides
  • SAS --- two sides + included angle
  • ASA --- two angles + included side
  • AAS --- two angles + a non-included side
  • HL --- (right triangles) hypotenuse + a leg

multicols SSA and AAA do not prove congruence. After proving congruence, CPCTC lets you conclude the matching parts are equal.

Concept
Isosceles triangle & the triangle inequality
  • [leftmargin=6mm,itemsep=2pt]
  • Isosceles Triangle Theorem: the angles opposite the two equal sides (base angles) are equal --- and the converse holds.
  • Triangle Inequality: the sum of any two sides is greater than the third: a+b>ca+b>c. The third side lies between ab|a-b| and a+ba+b.
Example
Missing angle and side range

A triangle has angles 5050^\circ and 6060^\circ; the third is 1805060=70180^\circ - 50^\circ - 60^\circ = 70^\circ. If two sides are 77 and 1010, the third side xx satisfies 107<x<10+710-7 < x < 10+7, i.e. 3<x<173 < x < 17.

Tip

“Included” is everything. SAS needs the angle between the two sides; ASA needs the side between the two angles. If the parts are not in that position, the shortcut does not apply (that is why SSA fails).

Similarity

Concept
Similar triangles

Similar figures (\sim) have equal corresponding angles and proportional corresponding sides. Prove triangles similar by:

  • [leftmargin=6mm,itemsep=2pt]
  • AA --- two pairs of equal angles.
  • SSS\sim --- all three side ratios equal.
  • SAS\sim --- two side ratios equal with equal included angle.

The common ratio of the sides is the scale factor kk.

Concept
Proportions, midsegment, and scaling
  • [leftmargin=6mm,itemsep=2pt]
  • Solve proportions by cross-multiplying: ab=cdad=bc\dfrac{a}{b} = \dfrac{c}{d} \Rightarrow ad = bc.
  • Triangle Midsegment: the segment joining midpoints of two sides is parallel to the third side and half its length.
  • Side-Splitter: a line parallel to one side cuts the other two sides proportionally.

For scale factor kk:   ratio of perimeters =k= k,   ratio of areas =k2= k^2,   ratio of volumes =k3= k^3.

Example
Scale factor and area

Two similar triangles have sides in ratio k=32k = \tfrac{3}{2}. Their perimeters are in ratio 32\tfrac{3}{2}, and their areas are in ratio (32)2=94\left(\tfrac{3}{2}\right)^2 = \tfrac{9}{4}.

Tip

Square the scale factor for area. If lengths scale by kk, area scales by k2k^2 (and volume by k3k^3). Doubling every side does not double the area --- it makes it 44 times bigger.

Right Triangles & Trigonometry

Concept
Pythagorean Theorem and its converse

In a right triangle with legs a,ba,b and hypotenuse cc (the side opposite the right angle):

a2+b2=c2.a^2 + b^2 = c^2 .

Converse: if a2+b2=c2a^2+b^2 = c^2, the triangle is right. If a2+b2>c2a^2+b^2 > c^2 it is acute; if a2+b2<c2a^2+b^2 < c^2 it is obtuse. Common triples: 3-4-53\text{-}4\text{-}5, 5-12-135\text{-}12\text{-}13, 8-15-178\text{-}15\text{-}17, 7-24-257\text{-}24\text{-}25 (and their multiples).

Concept
Special right triangles
45-45-90:legs equal x, hypotenuse =x230-60-90:short leg x, long leg =x3, hyp =2x\begin{aligned} 45^\circ\text{-}45^\circ\text{-}90^\circ:&\quad \text{legs equal } x,\ \text{hypotenuse } = x\sqrt{2}\\ 30^\circ\text{-}60^\circ\text{-}90^\circ:&\quad \text{short leg } x,\ \text{long leg } = x\sqrt{3},\ \text{hyp } = 2x \end{aligned}

(In 3030-6060-9090 the short leg is opposite 3030^\circ; the long leg is opposite 6060^\circ.)

Concept
Trigonometric ratios (SOH-CAH-TOA)

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

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

To find a missing angle, use inverse trig: θ=sin1 ⁣(opphyp)\theta = \sin^{-1}\!\left(\tfrac{\text{opp}}{\text{hyp}}\right), and likewise cos1\cos^{-1}, tan1\tan^{-1}.

Example
Finding a side and an angle

A right triangle has an angle θ\theta with opposite side 33 and hypotenuse 55. Then sinθ=35\sin\theta = \tfrac{3}{5}, so θ=sin1(0.6)36.9\theta = \sin^{-1}(0.6) \approx 36.9^\circ. The adjacent leg is 5232=4\sqrt{5^2-3^2} = 4.

Tip

Ratio for sides, inverse for angles. If you know the angle and want a side, use sin/cos/tan\sin/\cos/\tan. If you know two sides and want the angle, use sin1/cos1/tan1\sin^{-1}/\cos^{-1}/\tan^{-1}. Set your calculator to degrees.

Quadrilaterals & Polygons

Concept
Angle sums in any polygon

For a polygon with nn sides:

sum of interior angles=(n2)180,sum of exterior angles=360.\text{sum of interior angles} = (n-2)\cdot 180^\circ,\qquad \text{sum of exterior angles} = 360^\circ .

For a regular polygon: each interior angle =(n2)180n= \dfrac{(n-2)180^\circ}{n}, each exterior angle =360n= \dfrac{360^\circ}{n}.

Concept
Parallelogram properties

In every parallelogram: opposite sides are parallel and congruent; opposite angles are congruent; consecutive angles are supplementary; and the diagonals bisect each other.

Concept
The special quadrilaterals
  • [leftmargin=6mm,itemsep=2pt]
  • Rectangle: parallelogram with 44 right angles; diagonals are congruent.
  • Rhombus: parallelogram with 44 equal sides; diagonals are perpendicular and bisect the angles.
  • Square: both a rectangle and a rhombus (all properties of each).
  • Trapezoid: exactly one pair of parallel sides; isosceles trapezoid has congruent legs, congruent base angles, and congruent diagonals.
  • Kite: two pairs of adjacent congruent sides; diagonals are perpendicular.
Example
Interior angle of a regular polygon

A regular hexagon has n=6n=6: interior-angle sum =(62)180=720= (6-2)180^\circ = 720^\circ, so each angle =7206=120= \tfrac{720^\circ}{6} = 120^\circ. Each exterior angle =3606=60= \tfrac{360^\circ}{6} = 60^\circ.

Tip

Exterior angles always total 360360^\circ. No matter how many sides, the exterior angles sum to 360360^\circ. That is the fast way to find a regular polygon's angle: exterior =360n= \tfrac{360^\circ}{n}, then interior =180= 180^\circ - exterior.

Circles

Concept
Circumference, area, and arcs

With radius rr and diameter d=2rd = 2r:

C=2πr=πd,A=πr2.C = 2\pi r = \pi d,\qquad A = \pi r^2 .

For a central angle of nn^\circ:   arc length =n3602πr= \dfrac{n}{360}\cdot 2\pi r,   sector area =n360πr2= \dfrac{n}{360}\cdot \pi r^2.

Concept
Central, inscribed, and other angles
  • [leftmargin=6mm,itemsep=2pt]
  • Central angle (vertex at center) == its intercepted arc.
  • Inscribed angle (vertex on circle) =12= \tfrac{1}{2} its intercepted arc. An angle inscribed in a semicircle is 9090^\circ.
  • Tangent \perp radius at the point of tangency; two tangents from one outside point are congruent.
Reminder — Inscribed angle theorem:inscribed angle=12central angle on the same arc\text{inscribed angle}=\tfrac12\,\text{central angle on the same arc}
Concept
Power of a Point

From a point, the products of the two chord/secant/tangent pieces are equal:

Two chords:ab=cdTwo secants:(whole1)(outside1)=(whole2)(outside2)Tangent & secant:t2=(whole)(outside)\begin{aligned} \text{Two chords:} &\quad a\cdot b = c\cdot d\\ \text{Two secants:} &\quad (\text{whole}_1)(\text{outside}_1) = (\text{whole}_2)(\text{outside}_2)\\ \text{Tangent \& secant:} &\quad t^2 = (\text{whole})(\text{outside}) \end{aligned}
Concept
Equation of a circle

A circle with center (h,k)(h,k) and radius rr:

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

The center uses the opposite sign of what appears; the right side is r2r^2, so take a square root for rr.

Example
Inscribed angle and a circle equation

An inscribed angle intercepting a 100100^\circ arc measures 12(100)=50\tfrac{1}{2}(100^\circ) = 50^\circ. The equation (x2)2+(y+3)2=25(x-2)^2+(y+3)^2 = 25 has center (2,3)(2,-3) and radius 25=5\sqrt{25} = 5.

Tip

Central equals, inscribed halves. A central angle equals its arc; an inscribed angle is half its arc. And length answers (circumference, arc) use plain units, while area answers (circle, sector) use square units.

Perimeter, Area & Volume

Concept
Area of 2D figures

multicols2

  • [leftmargin=5mm,itemsep=1pt]
  • Rectangle: A=bhA = bh
  • Parallelogram: A=bhA = bh
  • Triangle: A=12bhA = \tfrac{1}{2}bh
  • Trapezoid: A=12(b1+b2)hA = \tfrac{1}{2}(b_1+b_2)h
  • Rhombus / kite: A=12d1d2A = \tfrac{1}{2}d_1 d_2
  • Circle: A=πr2A = \pi r^2
  • Regular polygon: A=12(apothem)(perimeter)A = \tfrac{1}{2}(\text{apothem})(\text{perimeter})

multicols Perimeter is the distance around; the circumference C=2πrC=2\pi r is a circle's perimeter.

Concept
Surface area and volume of solids

Let BB = area of the base, PP = base perimeter, hh = height, \ell = slant height.

Prism:V=Bh,SA=2B+PhCylinder:V=πr2h,SA=2πr2+2πrhPyramid:V=13Bh,SA=B+12PCone:V=13πr2h,SA=πr2+πrSphere:V=43πr3,SA=4πr2\begin{aligned} \text{Prism:} &\quad V = Bh, &\quad SA &= 2B + Ph\\ \text{Cylinder:} &\quad V = \pi r^2 h, &\quad SA &= 2\pi r^2 + 2\pi r h\\ \text{Pyramid:} &\quad V = \tfrac{1}{3}Bh, &\quad SA &= B + \tfrac{1}{2}P\ell\\ \text{Cone:} &\quad V = \tfrac{1}{3}\pi r^2 h, &\quad SA &= \pi r^2 + \pi r \ell\\ \text{Sphere:} &\quad V = \tfrac{4}{3}\pi r^3, &\quad SA &= 4\pi r^2 \end{aligned}
Example
Volume of a cylinder and a cone

A cylinder with r=3r=3, h=10h=10: V=π(3)2(10)=90πV = \pi(3)^2(10) = 90\pi. A cone with the same base and height holds one third as much: V=13π(3)2(10)=30πV = \tfrac{1}{3}\pi(3)^2(10) = 30\pi.

Tip

The one-third family. Pyramids and cones each get a 13\tfrac{1}{3} compared to the prism or cylinder with the same base and height. Area uses square units; volume uses cubic units --- check your units to catch errors.

Transformations & Symmetry

Concept
Rigid motions (isometries)

These preserve size and shape (image \cong pre-image):

Translation by (a,b):(x,y)(x+a, y+b)Reflection over x-axis:(x,y)(x, y)Reflection over y-axis:(x,y)(x, y)Reflection over y=x:(x,y)(y, x)Reflection over y=x:(x,y)(y, x)\begin{aligned} \text{Translation by }(a,b):&\quad (x,y)\to(x+a,\ y+b)\\ \text{Reflection over }x\text{-axis:}&\quad (x,y)\to(x,\ -y)\\ \text{Reflection over }y\text{-axis:}&\quad (x,y)\to(-x,\ y)\\ \text{Reflection over }y=x:&\quad (x,y)\to(y,\ x)\\ \text{Reflection over }y=-x:&\quad (x,y)\to(-y,\ -x) \end{aligned}
Concept
Rotations and dilations

Rotations about the origin (counterclockwise):

90:(x,y)(y, x)180:(x,y)(x, y)270:(x,y)(y, x)\begin{aligned} 90^\circ:&\quad (x,y)\to(-y,\ x)\\ 180^\circ:&\quad (x,y)\to(-x,\ -y)\\ 270^\circ:&\quad (x,y)\to(y,\ -x) \end{aligned}

Dilation (center origin, scale factor kk): (x,y)(kx, ky)(x,y)\to(kx,\ ky). This is not rigid --- it changes size (lengths ×k\times k, area ×k2\times k^2) but keeps shape, so figures are similar.

Concept
Symmetry
  • [leftmargin=6mm,itemsep=2pt]
  • Line (reflection) symmetry: a fold line maps the figure onto itself.
  • Rotational symmetry: a turn of less than 360360^\circ about a center maps the figure onto itself (a regular nn-gon has order nn, smallest angle 360n\tfrac{360^\circ}{n}).
Example
Applying a rule

Reflect P(3,5)P(3,5) over the xx-axis: (x,y)(x,y)(x,y)\to(x,-y) gives P(3,5)P'(3,-5). Then rotate PP' by 9090^\circ: (x,y)(y,x)(x,y)\to(-y,x) gives (5,3)(5,3).

Tip

Rigid motions keep congruence; dilations make similarity. Translations, reflections, and rotations preserve length and angle (image \cong pre-image). Only a dilation changes size --- and it scales lengths by kk, area by k2k^2.