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Geometry Review

Geometry Review. Lines and Angles. Triangles. Polygons. Circles. 3-D Geometry. Similarity & Scales. Lines and Angles. Intersecting Lines:. Vertical angles: congruent. Linear angles: supplementary. Lines and Angles. Parallel Lines:. Corresponding angles: congruent.

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Geometry Review

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  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

  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 G H J A B 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 G H 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 h C B a 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*7 + 15 = 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 = 9 + 5 = 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 Sum of the interior angles: b a c a + b + c = 180 Exterior angle = sum of the non-adjacent angles b c a c =  a + b

  11. Special Triangles A A a b c h B C b c B C 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 Triangles A A 45 60 30 B C 45 B C 30-60 Right Triangle 45- 45 Right Triangle |AB| = |BC| = 2/2 *|AC| |AB | = ½ * |AC| Area S = ½ |AB| * |BC| = ¼ |AC|2 |BC| = 3 * |AB| = 3/2 * |AC|

  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 45-45 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? B A F C P E D Draw AD intersecting EC at P By symmetry, we have AD  EC EPD is a 30-60 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 A B C D 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: 6 + 16 + 11 = 33

  18. Pythagorean Theorem A B C 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. 5-12-13 triangles: right triangle with side length 5, 12, 13. 7-24-25 triangles: right triangle with side length 7, 24, 25. 8-15-17 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? A 7 7 8 C E D B 2 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 ) = (64 - 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. X O Y 19 ? M 22 N Let OM=X, ON=Y. By Pythagorean Theorem on XON, MOY: (2X)2 + Y2 = 192 ------- (1) X2 + (2Y)2 = 222 ------- (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 180 * (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 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 ? M A B Extend AB to form right BMC 1 We have CBM = 360 / 6 = 60, BCM=90.  MC = 3/2 * r ; BM = ½ r r r F Area of ABC = ½ * AB * MC = ½ * 1 * 3/2 * r = ¾* r C Hence AC = (MC2 + (BM + 1)2) = (r2+ r + 1) 1 1 Hence AC = (MC2 + (BM + 1)2) = (r2+ r + 1) = CD = EA E D r Area ACE = ¾ AC2 = ¾(r2+ r + 1) = 7/10 (area of ACE + 3 * area of ABC)

  26. 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 ? M A B 1 Area ACE = ¾ AC2 = ¾(r2+ r + 1) = 7/10 (area of ACE + 3 * area of ABC) r r Hence ¾(r2+ r + 1) = 7/10 (¾(r2+ r + 1) + 3 * ¾ * r) F C 10r2+ 10r + 10 = 7r2+ 7r + 7 + 21r 1 1 3r2- 18r + 3 = 0  r2- 6r + 1 = 0 E D From the sum of the roots theorem, we get: The sum of the possible values of r = 6 r

  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 = 100  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 K D C 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 K D C 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 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 K D C area(ABCD) = area(ABK) + area(CDK) + area(ADK) + area(BCK) area(BCK) = 18 + 32 + 24 + 24 = 98

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

  33. Circle  x y α x 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 + 360 = 125 * 2 + 80 * 2 We have: X = 250 + 160 – 360 = 50

  36. Example 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 30-60 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 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 B C C  E B A E D A D AB * AC = AD * AE AE * EB = CE * ED

  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 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 = 12 ----- (1) DOE is right triangle, and OD = OE = R R2 + X2 = DE2 = 62 = 36 -------- (2) By (1) + (2), 2 R2 = 12 + 36  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 S h H R Volume of prism = height * area-of-base Volume of pyramid = ⅓ * height * area-of-base Volume of cylinder = height * R2 *  Volume of cone = ⅓ * H * R2 *  Surface are of cone = S * R *  + 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 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 – 36 + 2 * 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

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