1. Geometry Review Lines and Angles Triangles Polygons Circles
3-D Geometry
Similarity & Scales

2. Lines and Angles Intersecting Lines: Vertical angles: congruent
Linear angles: supplementary
Vertical angles: congruent

3. Lines and Angles Parallel Lines: Corresponding angles: congruent
Alternate angles: congruent

4. Example
Points A, B, C, D, E and F lie, in that order, on AF, dividing it into five segments, each of length 1. Point G is not on line . Point H lies on GD, and point J lies on GF. The line segments HC, JE and AG are parallel. Find HC/JE. G H J A B C D E F

5. Example EJ // AG, we get: FEJ = FAG, FJE = FGA Thus  FEJ   FAGB C D E F
EJ // AG, we get: FEJ = FAG, FJE = FGA
Thus  FEJ   FAG
CH // AG, we get: DCH = DAG, DHC = DGA
Thus  DCH   DAG

6. Example  FEJ   FAG, we get: EJ/AG = EF/AF = 1/5H J A B C D E F
 FEJ   FAG, we get: EJ/AG = EF/AF = 1/5
Hence EJ = 1/5 AG (1)
 DCH   DAG, we get: CH/AG = CD/AD = 1/3
Hence CH = 1/3 AG (2)
From (2), (1), we get: CH / EJ = 5/3

7. Triangles a + b > c a + c > b b + c > a a - b < ch C a B
Triangle Inequality:
a + b > c
a + c > b
b + c > a
a - b < c
a - c < b
b - c < a
Area of Triangle:
area(ABC) = ½ * h * c

8. Example
In a triangle with integer side lengths, one side is three times as long as a second side, and the length of the third side is 15. What is the greatest possible perimeter of the triangle? 15 X 3X
By triangle inequality, we have: 3 X – X < 15
Hence 2X < 15  X < 15/2 = 7.5
X must be an integer, so the largest X = 7
And the largest perimeter = 7 + 3* = 43

9. Example
In quadrilateral ABCD, AB=5, BC=17, CD=5, DA=9, and BD is an integer. What is BD?
By triangle Inequality: BD < AD + AB = = 14
By triangle Inequality: BD > BC – CD = 17 – 5 = 12
Hence we have: 12 < BD < 14, and BD is an integer
Answer: BD = 13

10. Angles of Triangle a + b + c = 180 c =  a + b
Sum of the interior angles: b a c
a + b + c = 180
Exterior angle = sum of the non-adjacent angles b a c
c =  a + b

11. Special Triangles h Isosceles triangle Equilateral triangleb a c A B C b c B A C h
Isosceles triangle
Equilateral triangle
|AB| = |BC| = |CA|
|AB | = |AC|
a + b + c = 180
b + c
a = b = c = 60
Height h = 3/2 R
Area S = 3/4 R2

12. Special Right TrianglesB C
45
30 B A C
60
30 -60 Right Triangle
45 - 45 Right Triangle
|AB | = ½ * |AC|
|AB| = |BC| = 2/2 *|AC|
|BC| = 3 * |AB| = 3/2 * |AC|
Area S = ½ |AB| * |BC| = ¼ |AC|2

13. Example
What’s the area of the stop sign whose sides are 1 foot long? A C B
STOP
Note that ABC is a right triangle
Area of ABC = ¼ |BC| = ¼ * 1 = 1/4
And |AB| = |AC| = 2/2 |BC| = 2/2 * 1 = 2/2
Area of the green square = (1 + 2/2 + 2/2)2 = 3 + 22
Area of the stop-sign = 3 + 22 – 4 ( ¼) = 2 + 22

14. Example
Regular Hexagon ABCDEF has sides measuring 6 cm. What’s the area of triangle ACE? A B F C P E D
Draw AD intersecting EC at P
By symmetry, we have AD  EC
EPD is a right triangle!
ED = 6cm  |EP| = 3/2 * |EC| = 3/2 * 6 = 33
ACE is equilateral. |CE| = 2 * |EP| = 2 * 33 = 63
Area of ACE = 3/4 |CE|2 = 3/4 *(63) 2 = 273

15. Bi-Sector Theorem |AB | |BD| --------- = ------------ |AC| |DC|
For any ABC. AD bisects BAC, we have:
|AB | |BD| =
|AC| |DC|

16. Example
Non-degenerate ABC has integer side lengths, BD is an angle bisector, AD = 3, and DC = 8. What is the smallest possible value of the perimeter? B A 3 D 8 C
From Bi-Sector theorem, we have AB / BC = 3/ 8
Hence AB = 3/8 * BC
The smallest BC to make AB an integer will be 8
This gives AB = 3/8 * 8 = 3, and AB + BC = 11
and we get: AB + BC = AC --- invalid triangle!

17. Example
Non-degenerate ABC has integer side lengths, BD is an angle bisector, AD = 3, and DC = 8. What is the smallest possible value of the perimeter? B A 3 D 8 C
So BC = 8 is not a valid answer!!!
The next smallest BC to make AB = 3/8 * BC an integer is 16.
This gives AB = 3/8 * BC = 3/8 * 16 = 6
And we get the smallest perimeter: = 33

18. Pythagorean Theorem |AB|2 + |BC|2 = |AC|2
For any right ABC, where ABC forms the right angle. We have:
|AB|2 + |BC|2 = |AC|2
Right triangles whose side lengths are integers:
3-4-5 triangles: right triangle with side length 3, 4, 5.
triangles: right triangle with side length 5, 12, 13.
triangles: right triangle with side length 7, 24, 25.
triangles: right triangle with side length 8, 15, 17.

19. Example
A rectangle with diagonal length 1, height is twice as long as it is wide. What is the area of the rectangle? A B 1 D C
Assume h and w are height & width of the rectangle
We have: h = 2 * w
By Pythagorean theorem, we have: w2 + (2w)2 = 1  5 w2 = 1  w = 5/5
Thus the area of the rectangle: A = w * h = w * (2w) = 5/5 * 2 * 5/5 = 2/5

20. Example
In ABC, we have AB=AC=7 and BC=2. Suppose that D is a point on line BC such that C lies between B and D and AD=8. What is CD? B A C 7 2 8 E D
ABC is isosceles. Draw AE  BC, we have: BE = EC = 1
Also, ABE a right triangle, and by AE2 = AB2 – BE2: |AE| = (72 – 12 ) = (48)
Also, ADE is a right triangle, and by ED2 = AD2 – AE2: |ED| = (82 – 48 ) = ( ) = = (16) = 4
Hence CD = ED – EC = 4 – 1 = 3

21. Example
Let XOY be a right triangle with XOY=90. Let M and N be the midpoints of legs OX and OY, respectively. Given that XN=19 and YM=22, find XY. O X Y 19 ? M 22 N
Let OM=X, ON=Y. By Pythagorean Theorem on XON, MOY:
(2X)2 + Y2 = (1) X2 + (2Y)2 = (2)
Sum up (1) & (2), we get: 5X2 + 5Y2 = 845  X2 + Y2 = 169
By Pythagorean Theorem on OXY: XY2 = (2X)2 + (2Y)2 = 4*(X2 + Y2) = 4*169  XY = 2*13 = 26

22. Polygon Sum of exterior angles = 360
A N-polygon can be divided into N – 2 triangles.
Sum of interior angles of a N-polygon  * (N – 2)

23. Example
A regular polygon has interior angles measuring 179, and a side length of 2cm. What is the perimeter of the polygon?
Since the interior angle is 179, the exterior angle is 1
Thus there are 360 / 1 = 360 exterior angles.
And there are 360 sides, each of length 2 cm.
Hence the perimeter = 2 * 360 = 720

24. Example
Equiangular hexagon ABCDEF has side lengths AB=CD=EF=1 and BC=DE=FA=r. The area of  ACE is 70% of the area of the hexagon. What is the sum of all possible values of r ? A 1 B r r F C 1 1 r D E

25. Example Extend AB to form right BMC
Equiangular hexagon ABCDEF has side lengths AB=CD=EF=1 and BC=DE=FA=r. The area of  ACE is 70% of the area of the hexagon. What is the sum of all possible values of r ? A B C D E F 1 r M
Extend AB to form right BMC
We have CBM = 360 / 6 = 60, BCM=90  MC = 3/2 * r ; BM = ½ r
Area of ABC = ½ * AB * MC = ½ * 1 * 3/2 * r = ¾* r
Hence AC = (MC2 + (BM + 1)2) = (r2 + r + 1)
Hence AC = (MC2 + (BM + 1)2) = (r2 + r + 1) = CD = EA
Area ACE = ¾ AC2 = ¾(r2 + r + 1)
= 7/10 (area of ACE + 3 * area of ABC)

26. Example = 7/10 (area of ACE + 3 * area of ABC)
Equiangular hexagon ABCDEF has side lengths AB=CD=EF=1 and BC=DE=FA=r. The area of  ACE is 70% of the area of the hexagon. What is the sum of all possible values of r ? A B C D E F 1 r M
Area ACE = ¾ AC2 = ¾(r2 + r + 1)
= 7/10 (area of ACE + 3 * area of ABC)
Hence ¾(r2 + r + 1) = 7/10 (¾(r2 + r + 1) + 3 * ¾ * r)
10r r + 10 = 7r2 + 7r r
3r2 - 18r + 3 = 0  r2 - 6r + 1 = 0
From the sum of the roots theorem, we get: The sum of the possible values of r = 6

27. Special Polygons trapezoids – quadrilateral with two parallel bases
parallelogram – two pairs of parallel sides
rhombus – parallelograms that’s equilateral,
with diagonals perpendicular to each other
rectangle – parallelograms that’s equal angular, with diagonals of same length
square – regular rectangle

28. Example
The long diagonal of a rhombus is 3 times the length of the short diagonal. The perimeter of the rhombus is 40cm. What is its area? B C A O D
Let BO = OD = X, we have AO = OC = 3 * X
X2 + (3X)2 = (40/4)2 = 100
10 X2 =  X = 10
Hence OB = 10, AO = 310, area(ABO) = ½*10*310 = 15
Area of rhombus = 4 * 15 = 60 (cm2)

29. Example
Trapezoid ABCD has bases AB and CD and diagonals intersecting at K. Suppose that AB=9, DC=12, and the area of AKD is 24. What is the area of trapezoid ABCD? A B D C K
ABK  CDK  BK/DK = AB/CD = 9/12 = ¾  AK/CK = AB/CD = 9/12 = ¾
ABK & ADK share same height, hence: area(ABK) / area(ADK) = BK/DC = 3/4
Area(ADK) = 24  area(ADK) = ¾ * 24 = 18

30. Example
Trapezoid ABCD has bases AB and CD and diagonals intersecting at K. Suppose that AB=9, DC=12, and the area of AKD is 24. What is the area of trapezoid ABCD? A B D C K
Similarly, ADK & CDK share same height, hence: area(ADK) / area(CDK) = AK/DK = 3/4
Area(ADK) = 24  area(CDK) = 4/3 * 24 = 32
Note that area(ADC) = area(BDC), we get:
area(BCK) = area(BDC) - area(KDC) = area(ADC) - area(KDC) = area(ADK) = 24

31. Example area(ABCD) = area(ABK) + area(CDK) + area(ADK) + area(BCK)
Trapezoid ABCD has bases AB and CD and diagonals intersecting at K. Suppose that AB=9, DC=12, and the area of AKD is 24. What is the area of trapezoid ABCD? A B D C K
area(ABCD) = area(ABK) + area(CDK) area(ADK) + area(BCK)
area(BCK) =
= 98

32. Circle secant chord tangent radius Definitions:
Area of the circle = R2 * 
Perimeter of the circle = 2 * R * 

33. Circle Inscribed α = ½ x where x is the opposite arc.y 
Inscribed α = ½ x where x is the opposite arc.
scant angle  = ½ (y - x )

34. Example Find the angle measure X in the diagram below
80
100 y x
150
The measure of the circle = 360
Hence Y = 360 – 150 – 100 – 80 = 30
Hence X = ½ (100 – Y) = ½ (100 – 30) = ½ * 70 = 35

35. Example Find the measure of arc X in the diagram below
125 x
80
The measure of the circle = 360
Observe that X = 125 * * 2
We have: X = – 360 = 50

36. Example Let P be center of small circle. By AOB = 60,
Points A and B lie on a circle centered at O, AOB= 60. A second circle is internally tangent to the first and tangent to both OA and OB. What is the ratio of the area of the smaller circle to that of the larger circle? A C P B O M
Let P be center of small circle. By AOB = 60,
OMP is right triangle  OP = 2 PM
OC = OP + PC = OP + PM = 3 PM  PM/OC = 1/3
Area(P)/Area(O) = (PM2* )/(OC2* ) = 1/9

37. Inscribed Triangle O 
For any inscribed triangle that has one edge through the center, it must be a right triangle.

38. Inscribed Quadrilateral
180 - X x
For any inscribed quadrilateral, the sum of the two opposite angles = 180

39. Example
In the given circle, the diameter EB is parallel to DC, and AB is parallel to ED. The angles AEB and ABE are in the ratio 4:5. What is the degree measure of angle BCD?

40. Example Since BAE over the center  BAE = 90 Hence ABE +AEB = 90
In the given circle, the diameter EB is parallel to DC, and AB is parallel to ED. The angles AEB and ABE are in the ratio 4:5. What is the degree measure of angle BCD?
Since BAE over the center  BAE = 90
Hence ABE +AEB = 90
Since AEB : ABE = 4:5  AEB = 40, ABE = 50
Now AB//ED  BED = ABE = 50
BCDE is an inscribed quadrilateral  BED + BCD = 180
BCD = 180 - BED = 180 - 50 = 130

41. Power of a Pointer AE * EB = CE * ED AB * AC = AD * AE B C C  E B A E

42. Example
In the figure below, AB and CD are diameters of the circle with center O, ABCD, and chord DF intersects AB at E. If DE=6 and EF=2, what’s the area of the circle?

43. Example By Power of a Pointer, DE * EF = AE * EB
In the figure below, AB and CD are diameters of the circle with center O, ABCD, and chord DF intersects AB at E. If DE=6 and EF=2, what’s the area of the circle?
By Power of a Pointer, DE * EF = AE * EB
Let X = OE, and AO = OB = R we have: AE = X+R, EB= X-R
Hence: 6 * 2 = (R + X) (R – X)  R2 – X2 = (1)
DOE is right triangle, and OD = OE = R
R2 + X2 = DE2 = 62 = (2)
By (1) + (2), 2 R2 =  R2 = 24  Area = 24

44. Example
A 45 arc of circle A is equal in length to a 30 arc of circle B. What is the ratio of circle A's area and circle B's area?
Let R1 and R2 be the radii of the two circles A & B.
The arc length opposite to 45 in R1 = 45/360 * 2 R1  = 1/4 R1 
The arc length opposite to 30 in R2 = 30/360 * 2 R2  = 1/6 R2 
Hence we get: 1/4 R1  = 1/6 R2   R1 / R2 = 3/2
area(A) / area(B) = (R12  )/ (R22 ) = 9/4

45. 3-D Geometry Volume of prism = height * area-of-base
Volume of cone = ⅓ * H * R2 *  H
Surface are of cone = S * R *  + R2 *  S R h
Volume of prism = height * area-of-base
Volume of pyramid = ⅓ * height * area-of-base
Volume of cylinder = height * R2 * 

46. 3-D Geometry Volume of sphere V = 4/3 * R3 * 
Surface are of sphere A = 4 * R2 * 

47. Example
The height of a cone is 4m and its radius is 3m. What’s the surface area and the volume of the cone?
Volume = ⅓ * H * R2 *  = ⅓ * 4 * 9 *  = 12  (m3)
Surface Area = r * s *  + r2 *  = 3 * 5 *  + 9 * 
= 24  (m2)

48. Example
A solid cube has side length 3 inches. A 2-inch by 2-inch square hole is cut into the center of each face. The edges of each cut are parallel to the edges of the cube, and each hole goes all the way through the cube. What is the volume, in cubic inches, of the remaining solid?

49. Example Volume of whole cube = 3 * 3 * 3 = 27
A solid cube has side length 3 inches. A 2-inch by 2-inch square hole is cut into the center of each face. The edges of each cut are parallel to the edges of the cube, and each hole goes all the way through the cube. What is the volume, in cubic inches, of the remaining solid?
Volume of whole cube = 3 * 3 * 3 = 27
Volume of the prism = 3 * 2 * 2 = 12
There are 3 prisms, total volume = 3 * 12 = 36
Note that the intersection of 3 prisms is a 2x2x2 cube
And we need to add two of them back.
Hence the answer = 27 – * 8 = 7

50. Scale & Similarity Rules for similar triangles: SAS, AAA, SSS
The ratio between any corresponding edges is the same between any similar polygons
Correspond angles have the same measure between any similar polygons
When the side lengths of any polygon (or radius of circle) are increased by a factor S, then the area is increased by a factor S2
When the side lengths of any 3-D solid are increased by a scale factor of S, then the surface area is increased by a scale factor of S2, and the volume is increased by a scale factor of S3

51. Example
Rectangle ABCD has AB=4 and BC=3. Segment EF is constructed through B so that EF is perpendicular to DB. A and C lie on DE and EF, respectively. What is EF? D C F 3 4 A B E

52. Example By Pythagorean theorem: DB = 5C F 3
Rectangle ABCD has AB=4 and BC=3. Segment EF is constructed through B so that EF is perpendicular to DB. A and C lie on DE and EF, respectively. What is EF? 4 A B E
By Pythagorean theorem: DB = 5
Observe that ABE  ABD  EB/AB=DB/AD
EB = AB * DB / AD = 4 * 5 / 3 = 20/3
Observe that BCF  ABD FB/CB=DB/AB
FB = CB * DB / AB = 3 * 5 / 4 = 15/4
EF = EB + BF = 20/3 + 15/4 = 125/12

53. Example
Points E and F are located on unit square ABCD so that BEF is equilateral. What is the ratio of the area of DEF to that of ABE?

54. Example Observe that BEF is equilateral, and EA = FC
Points E and F are located on unit square ABCD so that BEF is equilateral. What is the ratio of the area of DEF to that of ABE? 
Observe that BEF is equilateral, and EA = FC
We have ABE  BCF  DE = DE & DEF = 45
Let DE = DE = X
We get: EF = EB = FB = 2X
From ABE, we get: (2X)2 = (1 –X)2 + 12
2X2 = 1 – 2X + X2 + 1  X2 + 2X -2 = 0  X = 3 - 1
area(DEF)/area(ABE) = (½ X2)/(½(1- X)) = X2/(1-X) =2