Transcript Geometry SOL “Things to Know:”

Geometry SOL
“Things to Know:”
• G.1
Hypothesis 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: (pq) If fossil fuels are burned, then acid rain is produced.
(qr) If acid rain is produced, then wildlife suffers.
(pr) 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
-10
-3
0
5
d = √(3-2)2 + (10-5)2
d = √12 + 52
A
D
C
B
d = √1 + 25 = √26 = 5.099
midpt = ½ (-10+5) = ½ (-5) = -2.5
y
Ex: Find the length of AB
x2+x1 y2+y1
Midpt = -------, ------2
2
B (3,4)
√(x2-x1)2 + (y2-y1)2
√(3-(-5))2 + (4-(-2))2
d=
=
= √(3+5)2 + (4+2)2
= √(8)2 + (6)2
= √64 + 36
= √100 = 10
Ex: Find midpt of AB
D (7,1)
x
3 + -5 4 + -2
-------, ------2
2
-2
2
-----, ----2
2
A (-5,-2)
C (4,-3)
-1 , 1
rise
y2 - y1
Slope = ------- or m = -------run
x 2 - x1
Ex: Find the slope of CD
rise
y2 - y1
1 - (-3)
m = -------- = ---------x2 - x1
7–4
4
m = ----3
run
Ex: Find the slope of the line going through (1,1) and (6,7)
y2 - y1 7 – 1
6
Slope: m = -------- = ------ = --x2 - x1 6 – 1
5
• G3
Complimentary angles: 2’s whose sum = 90°
Vertical angles are congruent ()
Supplementary angles: 2’s whose sum = 180°
Corresponding ’s, alternate interior ’s, and alternate exterior ’s are  when parallel
lines are cut by a transversal
Consecutive ’s are supplementary when parallel lines are cut by a transversal
t
1
m
3
5
n
7
lines m // n, line t is a transversal
Corresponding ’s: 1&5, 3&7, 2&6, 4&8
Alternate Int ’s: 3&6, 4&5
Alternate Ext ’s: 1&8, 2&7
Consecutive ’s: 3&5, 4&6
Vertical ’s: 1&4, 2&3, 5&8, 6&7
2
4
6
8
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 mABD, mCBE, mABC
Find m1, m2, m3, m4, m5, m6
t
C
A
5x – 2
6
m
3x + 8
B
4
5
5x + 12
D
n
E
5x+12 + 3x+8 = 180° (linear pair)
8x + 20 = 180
-20 -20
8x
= 160
x = 20°
mABD = 5(20) + 12 = 112°
mCBE = 112° (vertical  with ABD)
mABC = 3(20) + 8 = 68°
2
3
3x + 8
1
5x – 2 = 3x + 8 (corresponding ’s )
-3x
-3x
2x – 2 =
8
+2
+2
2x
=
10
x=5
5(5) – 2 = 23°
m1 = 157 (vertical  to 2)
m2 = 157° (alt int  with 5)
m3 = 23° (corresponding to 4)
m4 = 23° (vertical  with 5x-2)
m5 = 157° (corresponding to 1)
m6 = 157° (linear pair)
• G3 examples cont
For 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
t
150°
a
Is a // b? Yes
Why?
Corresponding ’s are  when two parallel
lines are cut by a transversal!
150°
b
• G5
Ways to prove ∆’s  (congruent – copies, all corresponding ’s & sides )
1. SSS
K
C
2. SAS
3. ASA
If AC  JL (side),
A  J (included angle)
4. AAS
A
and AB  JK (side)
then ∆ABC  ∆JKL (SAS)
L
B
J
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. SSS
A
3. SAS
C
5
3
12
40°
8
B
A
SSS
3
X
16
Y
4
∆ABC ~ ∆JKL
AB * scaling factor = JK
3*3=9
AC * scaling factor = JL
5 * 3 = 15
4
C
J
B
15
C
12
2
Z
9
A
K
SAS
3
X
16
Y
4
B
Z
40°
12
L
• G6 examples
The 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 > 7
3, 4, 5
Yes
3+4 > 5
7, 10, 11
Yes
7+10 > 11
2, 6, 3
No
2+3 < 6
(also a Pythagorean triple)
B
The longest side is opposite the largest ;
smallest side is opposite the smallest 
120°
40°
What is the shortest side? BC
What is the longest side? AB
What if the middle side?
AC
C
20°
A
• G7 Pythagorean Theorem
(c2 = a2 + b2)
Solve for x
52 + 22 = x2
25 + 4 = x2
29 = x2
√29 = x
x2 + 122 = 132
x2 + 144 = 169
- 144 = -144
x2
= 25
x=5
x
2
5
Is 3, 4, 5 a right triangle?
13
x
12
Is 11, 17, 23 a right triangle?
No, because 112 + 172 ≠ 232
121 + 289 ≠ 529
650 ≠ 529
Yes, because 32 + 42 = 52
9 + 16 = 25
25 = 25
Special Case Right Triangles
30-60-90
45-45-90
Side opposite 45°: ½ hypotenuse √2
Side opposite 30°: ½ hypotenuse
Side opposite 60°: ½ hypotenuse √3
short leg = long leg ÷ √3
long leg = short leg * √3
short leg = hyp ÷ 2
hyp = short leg * 2
y = 2*5 = 10 5
x = 5√3
y
30°
x
15
x = ½*15 = 7.5 x 60°
y = 7.5√3
x=5
y = 5√2
30°
y
x = 7÷√3
7 * √3
7√3
x = ----- ----- = -------√3 * √3
3
60°
hyp = leg * √2
leg = hyp ÷ √2
7√3 2
14√3
y = ----- * --- = -------3
1
3
x
5
45°
y
45°
x
y
60°
30°
7
x = 10÷√2
10 * √2
10√2
x = ----- ----- = -------√2 * √2
2
y
x = 5√2
y = 5√2
10
x
• G7 continued
x
opposite
tan x = -----------------adjacent
adjacent
cos x = -----------------hypotenuse
angle
opposite
tan x = -----------------adjacent
Inverse
trig function
30°
5
5
tan 30° = ----x
x tan 30° = 5
5
x = ---------tan 30°
= 8.66
sides of ∆
trig function
Trigonometry
Some old hippy
caught another hippy
tripping on acid.
or
SOH CAH TOA
opposite
sin x = ------------------hypotenuse
sides of ∆
opposite
tan-1 --------------adjacent
=x
angle
x°
Always press 2nd then the
function when finding angles
3
4
sin x = ---cos x = ---5
5
• G8 Quadrilaterals
5
3
sin-1
2nd sin (3/5)
cos-1
2nd cos (4/5)
x≈ 37°
Properties of a parallelogram
1. Opposite sides 
2. Opposite ’s 
3. Consecutive ’s are supplementary (=180°)
4. Diagonals bisect each other
5. Opposite sides parallel
4
3
tan x = ----4
tan-1
2nd tan (3/4)
x ≈ 37°
B
x ≈ 37°
C
50x+5
20x+65
A
D
Properties of a Rectangle
1-5 same as a parallelogram
6. 4 right ’s
7. Diagonals 
Properties of a Rhombus
1-5 same as a parallelogram
6. 4  sides
7. Diagonals bisect opposite ’s
8. Diagonals perpendicular
9
9
9
9
Properties of a Square
All of the above properties of parallelogram, rectangle, and rhombus
• G.9 Tessellations
The 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.
Will a regular hexagon tessellate a plane?
Will a regular octagon tessellate a plane?
ext  = 360°÷6 = 60° int  = 180° - 60° = 120°
Will 120 go into 360 with no remainder? yes
ext  = 360°÷8 = 45° int  = 180° - 45° = 135°
Will 135 go into 360 with no remainder? no!!
•
G.10 Circles
60°
200°
240°
50°
x
x
x
20°
120°
x
Central 
x = 60°
Inscribed 
x = 100°
Exterior 
x = 50°-20° = 30° = 15°
2
2
Exterior 
x = 240°-120° = 120° = 60°
2
2
60°
x
100°
x
120°
Interior 
x = 100°+120° = 220° = 110°
2
2
Inscribed 
x = ½(60°)
x = (30°)
x
30°
100°
Exterior 
x = 100°-30° = 70° = 35°
2
2
Chords, Secants and Tangents
x
5
x
10
3
2
x
2
to get “x”
5*x = 2*10
5x = 20
x=4
3
5
“outside part times whole thing ”
3(3 + x) = 2*(2 + 5)
9 + 3x = 14
3x = 5
x = 5/3
10
“outside part times whole
thing = tangent squared”
3(3 + x) = 102
9 + 3x = 100
3x = 91
x = 30.33
Area of a sector (% of total area)
θ
A = -------- * πr2 where θ is central angle
360°
θ
A = ------ * πr2
360°
60°
5
60°
25π
A = ------ * π52 = ------ ≈ 13.083
360°
6
x
x = 12
12
“tangents = from same point”
x2 = 122
x = 12
G.10 Circles continued
Length of an Arc (% of Circumference)
θ
Arc Length = -------- * 2πr where θ is central angle
360°
θ
A = ------ * 2πr =
360°
arc length
60°
5
60°
1
10π
------- * 2π5 = ---- * 10 π = ------ ≈ 5.236
360°
6
6
•
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.
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
•
Area of a Rhombus
A = ½ diag * diag
G.14 Proportional reasoning
Remember: tree to shadow
tree to shadow
30 ft
x
shadow
15 ft
shadow
7 ft
30
x
--- = ---15
7
15x = 210
x = 14
2
2
x
--- = ---5
7
x
5x = 14
5
7
x = 14/5 = 2.8