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

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Presentation on theme: "Geometry Review Lines and Angles Triangles Polygons 1 Circles 3-D Geometry Similarity & Scales."— Presentation transcript:

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

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

3 Lines and Angles 3 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. 4 ABCDEF G J H

5 5 ABCDEF G J H EJ // AG, we get:  FEJ =  FAG,  FJE =  FGA Thus  FEJ   FAG CH // AG, we get:  DCH =  DAG,  DHC =  DGA Thus  DCH   DAG Example

6 6 ABCDEF G J H  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 Example

7 Triangles 7 a + b > c a + c > b b + c > a b c a a - b < c a - c < b b - c < a Triangle Inequality: Area of Triangle: area(ABC) = ½ * h * c A BC h

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? 8 By triangle inequality, we have: 3 X – X < 15 Hence 2X < 15  X < 15/2 = 7.5 X 15 3X 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? 9 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 Sum of the interior angles: 10 a b c  a +  b +  c = 180  Exterior angle = sum of the non-adjacent angles a b c  c =  a +  b

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

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

13 What’s the area of the stop sign whose sides are 1 foot long? 13 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 =  2 STOP A B C Area of the stop-sign =  2 – 4 ( ¼) =  2 Example

14 Regular Hexagon ABCDEF has sides measuring 6 cm. What’s the area of triangle ACE? 14 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 A B C D F E P Area of  ACE =  3/4 |CE| 2 =  3/4 *(6  3) 2 = 27  3 Example

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

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? 16 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 A B C D 3 8 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? 17 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 A B C D 3 8

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

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

20 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? 20  ABC is isosceles. Draw AE  BC, we have: BE = EC = 1 Also,  ABE a right triangle, and by AE 2 = AB 2 – BE 2 : |AE| =  (7 2 – 1 2 ) =  (48) B A C 77 2 D 8 E Also,  ADE is a right triangle, and by ED 2 = AD 2 – AE 2 : |ED| =  (8 2 – 48 ) =  ( ) = =  (16) = 4 Hence CD = ED – EC = 4 – 1 = 3 Example

21 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. 21 Let OM=X, ON=Y. By Pythagorean Theorem on  XON,  MOY: (2X) 2 + Y 2 = (1) X 2 + (2Y) 2 = (2) Sum up (1) & (2), we get: 5X 2 + 5Y 2 = 845  X 2 + Y 2 = 169 O X Y M N ? By Pythagorean Theorem on  OXY: XY 2 = (2X) 2 + (2Y) 2 = 4*(X 2 + Y 2 ) = 4*169  XY = 2*13 = 26 Example

22 Polygon 22 Sum of interior angles of a N-polygon 180  * (N – 2) A N-polygon can be divided into N – 2 triangles. Sum of exterior angles = 360 

23 A regular polygon has interior angles measuring 179 , and a side length of 2cm. What is the perimeter of the polygon? 23 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 Example

24 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 ? 24 A B C D E F rr r Example

25 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 ? 25 AB C DE F rr r Extend AB to form right  BMC M 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 =  (MC 2 + (BM + 1) 2 ) =  (r 2 + r + 1) Hence AC =  (MC 2 + (BM + 1) 2 ) =  (r 2 + r + 1) = CD = EA Area  ACE =  ¾ AC 2 =  ¾(r 2 + r + 1) = 7/10 (area of  ACE + 3 * area of  ABC) Example

26 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 ? 26 AB C DE F rr r M Area  ACE =  ¾ AC 2 =  ¾(r 2 + r + 1) = 7/10 (area of  ACE + 3 * area of  ABC) Hence  ¾(r 2 + r + 1) = 7/10 (  ¾(r 2 + r + 1) + 3 *  ¾ * r) 10r r + 10 = 7r 2 + 7r r 3r r + 3 = 0  r 2 - 6r + 1 = 0 From the sum of the roots theorem, we get: The sum of the possible values of r = 6 Example

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

28 28 Let BO = OD = X, we have AO = OC = 3 * X 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? A B D C O X 2 + (3X) 2 = (40/4) 2 = X 2 = 100  X =  10 Hence OB =  10, AO = 3  10, area(  ABO) = ½*  10*3  10 = 15 Area of rhombus = 4 * 15 = 60 (cm 2 ) Example

29 29 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?  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 AB DC K Example

30 30 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? Similarly,  ADK &  CDK share same height, hence: area(  ADK) / area(  CDK) = AK/DK = 3/4 Area(  ADK) = 24  area(  CDK) = 4/3 * 24 = 32 AB DC K Note that area(  ADC) = area(  BDC), we get: area(  BCK) = area(  BDC) - area(  KDC) = area(  ADC) - area(  KDC) = area(  ADK) = 24 Example

31 31 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? area(ABCD) = area(  ABK) + area(  CDK) + area(  ADK) + area(  BCK) AB DC K area(  BCK) = = 98 Example

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

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

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

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

36 36 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?  OMP is right triangle  OP = 2 PM O A B P M C OC = OP + PC = OP + PM = 3 PM  PM/OC = 1/3 Area(P)/Area(O) = (PM 2 *  )/(OC 2 *  ) = 1/9 Example

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

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

39 39 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? Example

40 40 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? Hence  ABE +  AEB = 90  Since  AEB :  ABE = 4:5   AEB = 40 ,  ABE = 50  Now AB//ED   BED =  ABE = 50  Since  BAE over the center   BAE = 90  BCDE is an inscribed quadrilateral   BED +  BCD = 180   BCD = 180  -  BED = 180  - 50  = 130  Example

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

42 42 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? Example

43 43 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)  R 2 – X 2 = (1) DOE is right triangle, and OD = OE = R R 2 + X 2 = DE 2 = 6 2 = (2) By (1) + (2), 2 R 2 =  R 2 = 24  Area = 24  Example

44 44 Let R 1 and R 2 be the radii of the two circles A & B. 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? The arc length opposite to 45 in R 1 = 45/360 * 2 R 1  = 1/4 R 1  The arc length opposite to 30 in R 2 = 30/360 * 2 R 2  = 1/6 R 2  Hence we get: 1/4 R 1  = 1/6 R 2   R 1 / R 2 = 3/2 area(A) / area(B) = (R 1 2  )/ (R 2 2  ) = 9/4 Example

45 3-D Geometry 45 Volume of prism = height * area-of-base Volume of pyramid = ⅓ * height * area-of-base Volume of cylinder = height * R 2 *  Volume of cone = ⅓ * H * R 2 *  H h Surface are of cone = S * R *  + R 2 *  S R

46 3-D Geometry 46 Volume of sphere V = 4/3 * R 3 *  Surface are of sphere A = 4 * R 2 * 

47 47 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 * R 2 *  = ⅓ * 4 * 9 *  = 12  (m 3 ) Surface Area = r * s *  + r 2 *  = 3 * 5 *  + 9 *  = 24  (m 2 ) Example

48 48 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? Example

49 49 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 Example

50 Scale & Similarity 50 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 S 2 The ratio between any corresponding edges is the same between any similar polygons Rules for similar triangles: SAS, AAA, SSS 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 S 2, and the volume is increased by a scale factor of S 3 Correspond angles have the same measure between any similar polygons

51 51 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? DC A B 3 4 E F Example

52 52 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? DC A B 3 4 E F 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 Example

53 53 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? Example

54 54 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  We get: EF = EB = FB =  2X Let DE = DE = X  From  ABE, we get: (  2X) 2 = (1 –X) X 2 = 1 – 2X + X  X 2 + 2X -2 = 0  X =  area(  DEF)/area(  ABE) = (½ X 2 )/(½(1- X)) = X 2 /(1-X) =2 Example

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