Geometry SOL“Things to Know:”

• G.1Hypothesis and Conclusions (if … then …) Examples

p q (original) If today if Friday, then tomorrow is Saturday.

q p (converse) If tomorrow is Saturday, then today is Friday.

~p ~q (inverse) If today is not Friday, then tomorrow is not Saturday.

~q ~p (contrapostive) If tomorrow is not Saturday, then today is not Friday.

~ means “not”

Law of Syllogism: If p q is true and q r is true, then p r is true.

Example: (pq) If fossil fuels are burned, then acid rain is produced.

(qr) If acid rain is produced, then wildlife suffers.

(pr) if fossil fuels are burned, then wildlife suffers.

• G.2

Distance (d) = √(x2-x1)2 + (y2-y1)2 Midpoint (midpt): ½ sum (an average)

Examples: Find the distance between (2,5) and 3,10) Find the midpt of AB

d = √(3-2)2 + (10-5)2

d = √12 + 52

d = √1 + 25 = √26 = 5.099 midpt = ½ (-10+5) = ½ (-5) = -2.5

Ex: Find the length of AB Ex: Find midpt of AB

d = √(x2-x1)2 + (y2-y1)2

= √(3-(-5))2 + (4-(-2))2

= √(3+5)2 + (4+2)2

= √(8)2 + (6)2

= √64 + 36

= √100 = 10

BCDA

50-3-10

y

x

A (-5,-2)

B (3,4)

C (4,-3)

D (7,1)

x2+x1 y2+y1Midpt = -------, ------- 2 2

3 + -5 4 + -2-------, ------- 2 2

-2 2-----, ----- 2 2

-1 , 1

rise y2 - y1Slope = ------- or m = -------- run x2 - x1

rise

run

y2 - y1 1 - (-3)Ex: Find the slope of CD m = -------- = ---------- x2 - x1 7 – 4

4 m = ----- 3

y2 - y1 7 – 1 6Ex: Find the slope of the line going through (1,1) and (6,7) Slope: m = -------- = ------ = --- x2 - x1 6 – 1 5

• G3Complimentary angles: 2’s whose sum = 90°Supplementary angles: 2’s whose sum = 180°Corresponding ’s, alternate interior ’s, and alternate exterior ’s are when parallel

lines are cut by a transversalConsecutive ’s are supplementary when parallel lines are cut by a transversal

lines m // n, line t is a transversalCorresponding ’s: 1&5, 3&7, 2&6, 4&8Alternate Int ’s: 3&6, 4&5Alternate Ext ’s: 1&8, 2&7Consecutive ’s: 3&5, 4&6Vertical ’s: 1&4, 2&3, 5&8, 6&7

Sum of interior angles in a polygon: S = (# of sides – 2)*180°Sum of exterior angles in a polygon: Always 360°Each exterior angle in the polygon: 360 ÷ n (# of sides or # of angles)Each interior angle in the polygon: 180 – exterior (linear pairs are supplementary)

Examples: Find mABD, mCBE, mABC Find m1, m2, m3, m4, m5, m6

5x+12 + 3x+8 = 180° (linear pair) 5x – 2 = 3x + 8 (corresponding ’s )

8x + 20 = 180 -3x -3x -20 -20 2x – 2 = 8 8x = 160 + 2 + 2 x = 20° 2x = 10 x = 5mABD = 5(20) + 12 = 112°mCBE = 112° (vertical with ABD) 5(5) – 2 = 23° mABC = 3(20) + 8 = 68° m1 = 157 (vertical to 2)

m2 = 157° (alt int with 5)m3 = 23° (corresponding to

4)m4 = 23° (vertical with 5x-2)m5 = 157° (corresponding to

1)m6 = 157° (linear pair)

Vertical angles are congruent ()

5 6

87

1 2

3 4

t

n

m

A

B

C

DE

5x + 12

3x + 8

2 3x + 8

13

6 5x – 2

4 5

t

n

m

• G3 examples contFor a regular (all sides and angles ) octagon find:1. Sum of interior angles: S = (n-2)*180 = (8-2)*180 = 6*180 = 1080°2. Sum of exterior angles: 360°3. each exterior angle: 360 ÷ 8 = 45°4. each interior angle: 180 – 45 = 135°

• G4 example

• G5Ways to prove ∆’s (congruent – copies, all corresponding ’s & sides )1.SSS2.SAS3.ASA4.AAS

5.(additional ways for right ∆’s only)

6.HL – hypotenuse-leg (also SSS after use of Pythagorean Thrm)7.HA – hypotenuse-angle (also AAS)8.LA – leg-angle (also AAS or ASA depending on which leg )9.LL – leg-leg (also SAS with 90° angle)

Ways to prove ∆’s ~ (similar – angles , all sides proportional)1.AA (third angle must be because all add to 180°)2.SSS3.SAS

150°

150°

t

b

a Is a // b? YesWhy? Corresponding ’s are when two parallellines are cut by a transversal!

40°4

53

A

BC

40°12

159

LK

J

∆ABC ~ ∆JKLAB * scaling factor = JK3 * 3 = 9AC * scaling factor = JL5 * 3 = 15

A

B

C

JL

K

If AC JL (side), A J (included angle)and AB JK (side)then ∆ABC ∆JKL (SAS)

A

B

C

SSS

XZ

Y

23

4

812

16

A

B

C

SAS

XZ

Y3

4

12

16

• G6 examplesThe sum of two sides of a triangle must be greater than the third side.Is this triangle possible? 1, 2, 3 Why not? Because 1+2 is not greater than 3

Are the following triangles possible?5, 6, 7 Yes 5+6 > 73, 4, 5 Yes 3+4 > 5 (also a Pythagorean triple)7, 10, 11 Yes 7+10 > 112, 6, 3 No 2+3 < 6

The longest side is opposite the largest ; smallest side is opposite the smallest

What is the shortest side? BCWhat is the longest side? ABWhat if the middle side? AC

• G7 Pythagorean Theorem (c2 = a2 + b2)

Solve for x 52 + 22 = x2 x2 + 122 = 132

25 + 4 = x2 x2 + 144 = 169 29 = x2 - 144 = -144 √29 = x x2 = 25 x = 5

Is 3, 4, 5 a right triangle? Is 11, 17, 23 a right triangle?

Yes, because 32 + 42 = 52 No, because 112 + 172 ≠ 232

9 + 16 = 25 121 + 289 ≠ 529 25 = 25 650 ≠ 529

Special Case Right Triangles

30-60-90 45-45-90

Side opposite 30°: ½ hypotenuse Side opposite 45°: ½ hypotenuse √2

Side opposite 60°: ½ hypotenuse √3hyp = leg * √2

short leg = long leg ÷ √3 leg = hyp ÷ √2long leg = short leg * √3short leg = hyp ÷ 2hyp = short leg * 2

120°

40°20°

B

A

C

2

5

x

12

13x

5

30°

y60°

x

y5

x

45°

45°

10y

x

x = 5y = 5√2

x = 10÷√2 10 * √2 10√2x = ----- ----- = -------- √2 * √2 2x = 5√2y = 5√2

y = 2*5 = 10x = 5√3

x

30°

1560°

y

x = ½*15 = 7.5y = 7.5√3

x

30°

y60°

7

x = 7÷√3 7 * √3 7√3x = ----- ----- = -------- √3 * √3 3

7√3 2 14√3y = ----- * --- = -------- 3 1 3

• G7 continuedTrigonometrySome old hippy

caught another hippytripping on acid.

orSOH CAH TOA

• G8 Quadrilaterals

Properties of a parallelogram1. Opposite sides 2. Opposite ’s 3. Consecutive ’s are supplementary (=180°)4.Diagonals bisect each other5.Opposite sides parallel

Properties of a Rectangle1-5 same as a parallelogram6. 4 right ’s7. Diagonals

Properties of a Rhombus1-5 same as a parallelogram6. 4 sides7. Diagonals bisect opposite ’s8. Diagonals perpendicular

Properties of a SquareAll of the above properties of parallelogram, rectangle, and rhombus

• G.9 TessellationsThe sum of all angles around a point is 360°A regular polygon will tessellate a plane if it’s interior angle will ÷ into

360° evenly.

x

5

30°

oppositesin x = ------------------- hypotenuse

adjacentcos x = ------------------ hypotenuse

oppositetan x = ------------------ adjacent

oppositetan x = ------------------ adjacent

trig function

angle

sides of ∆

oppositetan-1 --------------- = x adjacent

Always press 2nd then the function when finding angles

Inversetrig function

angle

sides of ∆

3

4x°

5

5 tan 30° = ----- xx tan 30° = 5

5x = ---------- = 8.66 tan 30°

3 4 3sin x = ---- cos x = ---- tan x = ----- 5 5 4

sin-1 cos-1 tan-1

2nd sin (3/5) 2nd cos (4/5) 2nd tan (3/4)

x≈ 37° x ≈ 37° x ≈ 37°

Will a regular hexagon tessellate a plane?ext = 360°÷6 = 60° int = 180° - 60° = 120°Will 120 go into 360 with no remainder? yes

Will a regular octagon tessellate a plane?ext = 360°÷8 = 45° int = 180° - 45° = 135°Will 135 go into 360 with no remainder? no!!

50x+5

20x+65A

B C

D

9 9

9 9

• G.10 Circles

Chords, Secants and Tangents

Area of a sector (% of total area)

θA = -------- * πr2 where θ is central angle 360°

Central x = 60°

to get “x”5*x = 2*10 5x = 20 x = 4

Inscribed x = ½(60°)x = (30°)

Exterior x = 50°-20° = 30° = 15° 2 2

Inscribed x = 100°

Exterior x = 240°-120° = 120° = 60° 2 2

Interior x = 100°+120° = 220° = 110° 2 2

Exterior x = 100°-30° = 70° = 35° 2 2

“outside part times whole thing ”3(3 + x) = 2*(2 + 5) 9 + 3x = 14 3x = 5 x = 5/3

“outside part times whole thing = tangent squared”3(3 + x) = 102

9 + 3x = 100 3x = 91 x = 30.33

x

60°

x

200°

x

50°

20°

x

x = 12

12

x

240°

120°

x 100°30°x

60°

x100° 120°

x

5

2

10x

5

3

2

x

10

3

60°5

θA = ------ * πr2

360°

60° 25πA = ------ * π52 = ------ ≈ 13.083 360° 6

“tangents = from same point”x2 = 122

x = 12

G.10 Circles continued

Length of an Arc (% of Circumference)

θArc Length = -------- * 2πr where θ is central angle 360°

• G.11 Constructions (see pages in your book for figures)

1. Angle Bisector (pg 32 - 33)A. Place compass point at the vertex B. With the center at B, draw an arc that

intersects the sides of the angle. Label the point of intersection X and Y. B. Place the compass point at X and draw an arc in the interior of ABC. Place

the compass point at Y. Using the same radius, draw an arc that intersects the previous arc.

C. Label the point of intersection Z. Draw BZ, which is the angle bisector of ABC. ABZ CBZ.

2. Congruent Angle to a Given Angle (pg 31)A. Top copy ABC, place the compass point at the vertex, B. With the center at B,

draw an arc that intersects the sides of the angle. Label the points of intersection D and E.

B. Draw a new line m and mark a point P on the line. Place your compass point at P. Use the same radius that you used in step 1, and draw an arc with its center at P that intersects line m. Label the intersection point Q.

C. On the original angle, measure arc ED with your compass (see figure 3). Then, on line m, place your compass point at Q. Use a radius equal to arc ED, draw an arc with center at Q. Lavel the intersection point of the two arcs R.

D. Draw PR. The measure of PQR is the same as the measure of ABC.

3. Perpendicular Bisector (pg 24, 44, 236)A. Place compass point on point A. Use a radius that is more than half the length

of AB. Draw an arc that intersects AB. B. Using the same radius, place the compass point at B, and draw an arc that

intersects the previous arc BOTH above and below AB. Label the intersection points of the arcs P and Q.

C. Draw line PQ, which is the perpendicular bisector of AB. The diagram shows PW as a line which intersects AB. You can use this construction to find the midpoint of any line segment.

θ 60° 1 10πA = ------ * 2πr = ------- * 2π5 = ---- * 10 π = ------ ≈ 5.236 360° 360° 6 6

60°5

arc length

4. Perpendicular Thru a Point on a Line (pg 44)A. To draw a perpendicular to line m at point P, first place the compass point at

P. Draw 2 arcs of the same radius, one on either side of P, that intersect line m. Label the points of intersection A and B.

B. Place the compass point at A. Use a radius greater than the previous one, and draw an arc above with center at A. Place the compass point at B. Using the same radius, draw an arc with the center at B that intersects the previous arc.

C. Label the point of intersection of the two arcs Q. Draw PQ, which is perpendicular to m.

5. Perpendicular to a Line From a Point Off the Line (pg 44)A. To draw a perpendicular to line m through point P, first place the compass at

point P. Draw an arc that intersects line m at 2 points. B. Label the intersection points S and T. Place the compass point at T. Using

any radius longer than half the distance between S and T, draw and arc with center at T on the opposite side of the line from P.

C. Repeat this same step from point S, using the same radius. Label the intersection point of the 2 arcs Q. Draw PQ, which is perpendicular to line m.

• G.12 Making a model of a 3-D figure from a 2-D drawing

• G.13 Surface Area & Volume * remember Area is measured in square units

Volume is measured in cubic units

• G.14 Proportional reasoning

Remember: tree to shadow tree to shadow

30 ft x

shadow shadow 15 ft 7 ft

30 x --- = ---- 15 7

15x = 210

x = 14

2

5

x

7

2 x--- = ---- 5 7

5x = 14

x = 14/5 = 2.8

Area of a RhombusA = ½ diag * diag