4 ExamplePoints 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.GHJABCDEF
5 Example EJ // AG, we get: FEJ = FAG, FJE = FGA Thus FEJ FAG BCDEFEJ // AG, we get: FEJ = FAG, FJE = FGAThus FEJ FAGCH // AG, we get: DCH = DAG, DHC = DGAThus DCH DAG
6 Example FEJ FAG, we get: EJ/AG = EF/AF = 1/5 HJABCDEF FEJ FAG, we get: EJ/AG = EF/AF = 1/5Hence EJ = 1/5 AG (1) DCH DAG, we get: CH/AG = CD/AD = 1/3Hence 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 < c hCaBTriangle Inequality:a + b > ca + c > bb + c > aa - b < ca - c < bb - c < aArea of Triangle:area(ABC) = ½ * h * c
8 ExampleIn 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?15X3XBy triangle inequality, we have: 3 X – X < 15Hence 2X < 15 X < 15/2 = 7.5X must be an integer, so the largest X = 7And the largest perimeter = 7 + 3* = 43
9 ExampleIn quadrilateral ABCD, AB=5, BC=17, CD=5, DA=9, and BD is an integer. What is BD?By triangle Inequality: BD < AD + AB = = 14By triangle Inequality: BD > BC – CD = 17 – 5 = 12Hence we have: 12 < BD < 14, and BD is an integerAnswer: BD = 13
10 Angles of Triangle a + b + c = 180 c = a + b Sum of the interior angles:baca + b + c = 180Exterior angle = sum of the non-adjacent anglesbacc = a + b
11 Special Triangles h Isosceles triangle Equilateral triangle bacABCbcBAChIsosceles triangleEquilateral triangle|AB| = |BC| = |CA||AB | = |AC|a + b + c = 180b + ca = b = c = 60Height h = 3/2 RArea S = 3/4 R2
12 Special Right Triangles BC4530BAC6030 -60 Right Triangle45 - 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 ExampleWhat’s the area of the stop sign whose sides are 1 foot long?ACBSTOPNote that ABC is a right triangleArea of ABC = ¼ |BC| = ¼ * 1 = 1/4And |AB| = |AC| = 2/2 |BC| = 2/2 * 1 = 2/2Area of the green square = (1 + 2/2 + 2/2)2 = 3 + 22Area of the stop-sign = 3 + 22 – 4 ( ¼) = 2 + 22
14 ExampleRegular Hexagon ABCDEF has sides measuring 6 cm. What’s the area of triangle ACE?ABFCPEDDraw AD intersecting EC at PBy symmetry, we have AD ECEPD is a right triangle!ED = 6cm |EP| = 3/2 * |EC| = 3/2 * 6 = 33ACE is equilateral. |CE| = 2 * |EP| = 2 * 33 = 63Area of ACE = 3/4 |CE|2 = 3/4 *(63) 2 = 273
15 Bi-Sector Theorem |AB | |BD| --------- = ------------ |AC| |DC| For any ABC. AD bisects BAC, we have:|AB | |BD|=|AC| |DC|
16 ExampleNon-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?BA3D8CFrom Bi-Sector theorem, we have AB / BC = 3/ 8Hence AB = 3/8 * BCThe smallest BC to make AB an integer will be 8This gives AB = 3/8 * 8 = 3, and AB + BC = 11and we get: AB + BC = AC --- invalid triangle!
17 ExampleNon-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?BA3D8CSo 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 = 6And 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|2Right 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 ExampleA rectangle with diagonal length 1, height is twice as long as it is wide. What is the area of the rectangle?AB1DCAssume h and w are height & width of the rectangleWe have: h = 2 * wBy Pythagorean theorem, we have: w2 + (2w)2 = 1 5 w2 = 1 w = 5/5Thus the area of the rectangle: A = w * h = w * (2w) = 5/5 * 2 * 5/5 = 2/5
20 ExampleIn 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?BAC728EDABC is isosceles. Draw AE BC, we have: BE = EC = 1Also, 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) = 4Hence CD = ED – EC = 4 – 1 = 3
21 ExampleLet 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.OXY19?M22NLet 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 = 169By 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 ExampleA 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 1Thus there are 360 / 1 = 360 exterior angles.And there are 360 sides, each of length 2 cm.Hence the perimeter = 2 * 360 = 720
24 ExampleEquiangular 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 ?A1BrrFC11rDE
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 ?ABCDEF1rMExtend AB to form right BMCWe have CBM = 360 / 6 = 60, BCM=90 MC = 3/2 * r ; BM = ½ rArea of ABC = ½ * AB * MC = ½ * 1 * 3/2 * r = ¾* rHence AC = (MC2 + (BM + 1)2) = (r2 + r + 1)Hence AC = (MC2 + (BM + 1)2) = (r2 + r + 1) = CD = EAArea 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 ?ABCDEF1rMArea 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 r3r2 - 18r + 3 = 0 r2 - 6r + 1 = 0From 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 sidesrhombus – parallelograms that’s equilateral,with diagonals perpendicular to each otherrectangle – parallelograms that’s equal angular, with diagonals of same lengthsquare – regular rectangle
28 ExampleThe 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?BCAODLet BO = OD = X, we have AO = OC = 3 * XX2 + (3X)2 = (40/4)2 = 10010 X2 = X = 10Hence OB = 10, AO = 310, area(ABO) = ½*10*310 = 15Area of rhombus = 4 * 15 = 60 (cm2)
29 ExampleTrapezoid 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?ABDCKABK CDK BK/DK = AB/CD = 9/12 = ¾ AK/CK = AB/CD = 9/12 = ¾ABK & ADK share same height, hence: area(ABK) / area(ADK) = BK/DC = 3/4Area(ADK) = 24 area(ADK) = ¾ * 24 = 18
30 ExampleTrapezoid 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?ABDCKSimilarly, ADK & CDK share same height, hence: area(ADK) / area(CDK) = AK/DK = 3/4Area(ADK) = 24 area(CDK) = 4/3 * 24 = 32Note 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?ABDCKarea(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. yInscribed 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 80100yx150The measure of the circle = 360Hence Y = 360 – 150 – 100 – 80 = 30Hence X = ½ (100 – Y) = ½ (100 – 30) = ½ * 70 = 35
35 Example Find the measure of arc X in the diagram below 125x80The measure of the circle = 360Observe that X = 125 * * 2We 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?ACPBOMLet P be center of small circle. By AOB = 60,OMP is right triangle OP = 2 PMOC = OP + PC = OP + PM = 3 PM PM/OC = 1/3Area(P)/Area(O) = (PM2* )/(OC2* ) = 1/9
37 Inscribed TriangleOFor any inscribed triangle that has one edge through the center, it must be a right triangle.
38 Inscribed Quadrilateral 180 - XxFor any inscribed quadrilateral, the sum of the two opposite angles = 180
39 ExampleIn 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 = 90Hence ABE +AEB = 90Since AEB : ABE = 4:5 AEB = 40, ABE = 50Now AB//ED BED = ABE = 50BCDE is an inscribed quadrilateral BED + BCD = 180BCD = 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 ExampleIn the figure below, AB and CD are diameters of the circle with center O, ABCD, 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, ABCD, 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 * EBLet X = OE, and AO = OB = R we have: AE = X+R, EB= X-RHence: 6 * 2 = (R + X) (R – X) R2 – X2 = (1)DOE is right triangle, and OD = OE = RR2 + X2 = DE2 = 62 = (2)By (1) + (2), 2 R2 = R2 = 24 Area = 24
44 ExampleA 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 = 3/2area(A) / area(B) = (R12 )/ (R22 ) = 9/4
45 3-D Geometry Volume of prism = height * area-of-base Volume of cone = ⅓ * H * R2 * HSurface are of cone = S * R * + R2 * SRhVolume of prism = height * area-of-baseVolume of pyramid = ⅓ * height * area-of-baseVolume of cylinder = height * R2 *
46 3-D Geometry Volume of sphere V = 4/3 * R3 * Surface are of sphere A = 4 * R2 *
47 ExampleThe 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 ExampleA 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 = 27Volume of the prism = 3 * 2 * 2 = 12There are 3 prisms, total volume = 3 * 12 = 36Note that the intersection of 3 prisms is a 2x2x2 cubeAnd 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 polygonsCorrespond angles have the same measure between any similar polygonsWhen the side lengths of any polygon (or radius of circle) are increased by a factor S, then the area is increased by a factor S2When 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 ExampleRectangle 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?DCF34ABE
52 Example By Pythagorean theorem: DB = 5 CF3Rectangle 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?4ABEBy Pythagorean theorem: DB = 5Observe that ABE ABD EB/AB=DB/ADEB = AB * DB / AD = 4 * 5 / 3 = 20/3Observe that BCF ABD FB/CB=DB/ABFB = CB * DB / AB = 3 * 5 / 4 = 15/4EF = EB + BF = 20/3 + 15/4 = 125/12
53 ExamplePoints 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 = FCWe have ABE BCF DE = DE & DEF = 45Let DE = DE = XWe get: EF = EB = FB = 2XFrom ABE, we get: (2X)2 = (1 –X)2 + 122X2 = 1 – 2X + X2 + 1 X2 + 2X -2 = 0 X = 3 - 1area(DEF)/area(ABE) = (½ X2)/(½(1- X)) = X2/(1-X) =2