Vert. opp. angles equal. Base angle isos equal and
|
|
|
- Harry Craig
- 5 years ago
- Views:
Transcription
1 AS 1.6 Geometric Reasoning Answers Page 5 (other reasons are possible) 1. A = 50 Angle sum = 180. B = 62 Angle sum = C = 58 Angle sum = 180. D = 122 Adj. angles str. line 180. = 58 Page 6 (other reasons are possible) Vert. opp. angles equal. 3. F = 80 Adj. angles str. line 180. G = 70 Adj. angles str. line = 180 Angle sum = 180. = 12 y = 324 Angle sum of a point = H = 72 Angle sum of a point = 360. I = 72 terior angles poly = 360. J = 54 Base angle isos equal. 6. = 60 quilateral. N = 120 Adj. angles str. line 180. P = 30 Base angle isos equal. 7. Q = 297 Angle sum heagon = 720. R = 63 Angle sum of a point = S = 108 Angle sum pentagon = 540. T = 36 Base angles isos. equal. U = 72 T + U = 108. Int. angle regular pentagon. 9. V = 28 Adj. angles str. line 180 and Angle sum = 180. W = 288 Angle sum of a point = 360 and symmetrical angles. 10. Y = 87 Adj. angles str. line 180 Z = 46.5 Page 7 (other reasons are possible) Base angle isos equal. 11. A = 32 Base angle isos equal. B = 64 C = 52 Angle sum = 180 and Adj. angles str. line 180. Base angle isos equal and Angle sum = y+ 90 = 180 Adj. angles str. line 180. y = R = 117 Adj. angles str. line 180. S = 27 Adj. angles str. line 180. Q = 63 Vert. opp. angles equal. 14. T = 40 terior angles poly = 360. U = 120 Angle sum of a point 360. V = 140 Interior angles 9 sided poly. W = 220 terior angle Page 7 cont = 180 Angle sum = 180. = Y = 42 Angle sum = 180. Z = 318 Angle sum of a point A = 135 Interior angle of a octagon. B = 120 Interior angle of a heagon. C = 105 Angle sum of a point D = 211 Angle of a quad. sum 360. = 149 Angle sum of a point 360. Page 9 (other reasons are possible) 19. A = 41 Alt. angles // lines =. B = 139 Co-int. angles // = P = 46 Co-int. angles // = 180. Q = 134 Corr. angles // lines =. Page 10 (other reasons are possible) 21. B = 34 Corr. angles // lines =. A = 34 Vert. opp. angles equal. 22. C = 23 Alt. angles // lines =. D = 71 Alt. angles // lines =. = 86 Angle sum = 180. F = 94 Adj. angles str. line G = 44 Alt. angles // lines =. H = 99 Angle sum = J = 5 Angle sum = 180. K = 124 Alt. angles // lines =. 25. L = 38 Alt. angles // lines =. = 105 Adj. angles str. line 180. N = 37 Alt. angles // lines =. 26. D = 32 Corr. angles // lines =. XD = 32 Base angle isos equal. DX = 116 Angle sum = R = 112 Vert. opp. angles equal. S = 68 Co-int. angles // = 180. T = 65 Angle sum = U = 109 Corr. angles // lines =. V = 71 Adj. angles str. line 180. W = 61 Angle sum = 180. Page 11 (other reasons are possible) = 180 Co-int. angles // = 180. = 12 7y = 112 Co-int. angles // = 180. y = Z = Co-int. angles // = 180. Z = 293 AS 1.6 Year 11 athematics and Statistics Published by NuLake Ltd New Zealand Robert Lakeland & Carl Nugent
2 66 AS 1.6 Geometric Reasoning Page 11 cont A = 5 Co-int. angles // = 180. B = 18 Alt. angles // lines =. C = 157 Angle sum = D = 127 Alt. angles // lines =. = 74 F = 53 Base angle isos equal. and Angle sum = 180. Vert. opp. angles equal. 33. G = 101 Corr. angles // lines =. and Adj. angles str. line H = 123 Alt. angles // lines =. and Adj. angles str. line 180. and Page 14 (other reasons are possible) t. angle of a = 2 opp. int. L s. 35. X = 98 Angle sum of a point 360. Y = 82 Tangent at 90 to radius. and Angle of a quad. sum 360. Page 15 (other reasons are possible) 36. A = 42 B = 21 Angles same arc =. 37. C = 107 D = 73 and angle sum of a point is = 90 Angle in semi circle 90. F = G = 31 Base angles isos are =. H = 59 Angle in semi circle 90. I = 59 Base angles isos are =. 40. J = 18 Angles same arc =. K = 18 Alt. angles // lines =. = 144 Angle sum of a = N = 226 P = 134 Angle sum of a point is Q = 41 Angles same arc =. R = 131 Angle in semi circle 90. and Adj. angles str. line 180. S = 24.5 Base angles isos are =. 43. T = 67 Tangent at Rt. angle to radius. U = 33.5 Page 16 (other reasons are possible) 44. Q = 22 Angle between a tangent and a radius is a Rt. angle. R = 136 S = 68 Base angles isos are = and Angle sum of a = 180. T = 44 Angle sum of a quad. = 360. Page 16 cont U = 90 Angle in semi circle 90. V = 33 Tangent at 90 to radius. W = 33 Angle sum of a = = 32 Angles same arc =. y = 35 Angle sum of a = = 41 Alt. angles // lines =. y = 98 Angles same arc = and Angle sum of a = y = 12 Angles same arc =. = y = 63 Alt. angles // lines =. = 63 Angles same arc =. Page 18 (other reasons are possible) 50. A = 99 B = 62 The et. angle cyclic quad. = Opp. int. angle. 51. C = 90 Tangent at Rt. angle to radius. D = = 67 Base angles isos are =. F = 67 Base angles isos are =. G = 23 H = J = 57 K = 114 = 21 Angle between a tangent and a radius is a Rt. angle. 54. P = 22 Angle between a tangent and a radius is a Rt. angle. Q = 90 Angle in semi circle 90 and Adj. angles str. line 180. R = 68 Angle sum of a = 180. S = 22 Angle sum of a = T = 107 The et. angle of a cyclic U = 73 Adj. angles str. line 180. Page 19 (other reasons are possible) 56. V = 138 W = X = 110 Y = 111 The et. angle of a cyclic quad. = the Opp. angle. AS 1.6 Year 11 athematics and Statistics Published by NuLake Ltd New Zealand Robert Lakeland & Carl Nugent
3 AS 1.6 Geometric Reasoning 67 Page 19 cont A = 39 Base angles isos are =. B = 102 Angle sum of a = 180. C = 78 D = 60 Angle sum of a = e = 61 BD = 119 c = 119 b = 96 f = 96 The et. angle of a cyclic The et. angle of a cyclic The et. angle of a cyclic The et. angle of a cyclic 60. m = 68 Adj. angles str. line 180. j = 68 k = 90 Angle in semi circle 90. g = 22 Angle sum of a = u = 71 Page t = 68 v = 62 q = 59 y 15 = y = 10 cm = 30 cm 63. w = 20 m z = 30 m 64. j = 48 mm k = 40.5 mm 65. p = 24 m Page q = m (26.7 m 3 sf) 1160 = = 232 mm y = 2.50 m 67. n = 735 mm m = 360 mm 68. q = 2.67 m p = 1440 mm 69. s = 107 cm r = 27 cm Page QP and NP are similar because QR X = NP X Corr. angles // lines =. RQ W = PN W Corr. angles // lines =. QR Y = NP Y Common q = 30 q = 22 m m = 21 m 71. TUS and TVW are similar because TUS X = TVW X Alt. angles // lines =. TSU V = TWV Z Alt. angles // lines =. STU W = WTV W Vertically opp angles =. w = 28 m t = 52.5 m 72. AXD and BXC are similar because DAX X = CBX W Angles same arc =. ADX X = BCX X Angles same arc =. AXD W = BXC W Vertically opp angles =. Page 25 c = 18 m 73. GHF and DF are similar because HGF X = DF W Given GFH V = FD V Common GHF X = DF X Angle sum of a = 180. e = e = 6 mm f = mm (16.7 mm 3sf) 74. KLG and JHG are similar because GKL X = GJH T Given LGK X = HGJ X Common GLK W = GHJ X Angle sum of a = m g = 8 cm h = 20 cm 1.8 m 15.2 m Triangles are similar as both contain a common angle and a right angle. h = 8.55 m h AS 1.6 Year 11 athematics and Statistics Published by NuLake Ltd New Zealand Robert Lakeland & Carl Nugent
4 68 AS 1.6 Geometric Reasoning Page Using triangles QST and SPT we get QST V = TPS W Angle in semi. O = 90 and angle between tangent and radius = 90 STQ W = PTW S Common TQX S = TV SP Angle sum of a = 180. r s + r = = rs ( + r) 77. Using triangles QPS and QST we get QST V = QPS W Angle in semi. O = 90 and angle between tangent and radius = 90 SQT X = PQS X Common QTS W = QSP V Angle sum of a = 180. s+ r = s = s ( s+ r) 78. a) VXW W = YUW X The et. angle of a cyclic Page 28 b) Similar triangles WVX and WYU as VXW W = YUW X Shown in a) VWX Z = YWU Z Common XVW X = UYW W Angle sum of a = 180. c) XVW X = 90 Angle in semi. O = 90. Therefore the corresponding angle in the second similar triangle must also be a right angle so UYW W = 90. Also means UX is a diameter of the larger circle. 79. h = 83.2 mm 80. j = 5.67 m 81. k = 522 mm 82. m = 1.84 km 83. n = 360 mm 84. p = 7.0 m (2 sf) Page Dist. = 139 km 86. h = 116 mm 87. d = = 3360 mm 89. tra = = 443 m 90. = 105 m Page Base diagonal = 59.4 mm height = = 80.7 mm 92. Base diagonal = 86.0 mm diagonal = 87.5 mm 93. ABC W = 90 Angle in a semi circle = Page 31 AC = r = 12.7 mm OÂT = 90 Tangent at 90 to radius. OT 2 = OA 2 + AT 2 (41 + k) 2 = k = 44.9 mm 95. PCR X = 90 Alt. L s // = twice PR = 20.6 km 96. Two right-angled triangles. Angle in a semi circle = 90 Page 32 XY = 55.6 mm YZ = 94.6 mm XZ = mm 97. OAT is a right-angled triangle. Tangent at 90 to radius. OT 2 = OA 2 + AT 2 (7 + r) 2 = r r = 7.66 mm 98. Angles at T, V, U and W are all right angles as the angle in semi circle 90. Therefore the figure is a rectangle. (2r) 2 = a 2 + b 2 4r 2 = a 2 + b 2 r 2 = 0.25(a 2 + b 2 ) r = 0.25(a 2 + b 2 ) T W r V a b U AS 1.6 Year 11 athematics and Statistics Published by NuLake Ltd New Zealand Robert Lakeland & Carl Nugent
5 AS 1.6 Geometric Reasoning 69 Page p = 4.87 m 100. q = 16.7 m 101. r = 55.1 mm 102. s = 8.24 km 103. t = 92.6 mm 104. u = 153 mm (2 sf) 105. v = 73.0 mm 106. w = 6.12 m 107. y = 15.1 m 108. z = 245 cm (2 sf) Page h = 18.4 m 110. h = 8.17 m 111. h = m (4 sf) 112. sh = 4.43 m vh = 1.07 m 113. S = 114 km = 37 km (0 dp) 114. s = 3.45 m Page A = B = C = D = 34.7 Page = 52.1 e = 3.99 m (2 dp) 120. F = 41.5 f = 92 mm (0 dp) 121. G = 33.6 g = 82.9 cm (0.83 m) (3 sf ) 122. H = 50.6 h = 92 mm (0 dp) 123. I = 42.6 J = K = 45.0 L = 45.0 k = 1167 mm (0 dp) 125. = 53.1 m = 7.5 m 126. N = 50.2 P = 39.8 n = 204 cm (0 dp) Page h = = Cable = 93.9 m = Interior angles he. = 120 D = cos 60 D = 40 mm = 2 20 cos 30 = 34.6 mm 129. Adj = 280 cos 24 Adj = cm Above ground h = 45 + ( ) 130. h = 7.16 m Page 39 = 69 cm (0 dp) A 1 = 36 Angle sum of a = 180. Angle h to tail = A 2 A 2 = cos -1 (7.16/7.95) = 25.8 A = ACB = 36 Alt. L s // lines =. = m y = m (4 sf) (4 sf) 132. IHJ = 17 Corr. L s // lines =. a = 1.89 sin 17 = 0.55 m (2 sf) b = 2.45 tan 17 = 0.75 m (2 sf) 133. PQR X = 90 Angle in a semi circle = 90 PQ = 7.02 cm RQ = 6.10 cm 134. TWV Z = 90 Angle in a semi circle = 90 Page 42 TVW X = 11 Angles on same arc = 2r = r = 4.14 m 135. Right angle between two headings. dist. = 358 km (4 sf) Bearing = 40 + tan = Page Right angle between two headings. Dist. = 1778 m Bearing = (0 dp) 137. a) North = 88 km (0 dp) ast = 43 km b) Dist. = 98 km (0 dp) 138. Right angle between two radar directions. Page Dist. = 18.7 km Bearing 122. A = (tan -1 ) = Dist. = 383 km (0 dp) Bearing = 345 (0 dp) 141. Dist. = 755 km (0 dp) Bearing = 181 (0 dp) 142. PN = (126, 150) (0 dp) 143. Dist. = 574 km (0 dp) Page 45 Bearing = 263 (0 dp) 144. BAC = = 27 cos 27 = AX 12 and cos 48 = XC 8.2 AX = , XC = AC = AX + XC = = 16 cm (2 sf) 145. Let ABC = and BC = y ADC =, Angles on same arc =. CAD = 90 Angle in a semi circle = y = 180 Angle sum of a = y = 48 In CB = y + 34 t. angle of a = the two opp. int. angles. y + y + 34 = 48 so y = 7 and = 41. AS 1.6 Year 11 athematics and Statistics Published by NuLake Ltd New Zealand Robert Lakeland & Carl Nugent
6 70 AS 1.6 Geometric Reasoning Page 45 cont YVW and YXZ are similar triangles a 30 = a = 660 a = 17.1 cm b + 15 b = b = 14b b = 210 b = 26.3 cm Page cos34 = 5.2 = sin 34 = y 5.2 y = tan 40 = km 34 z z = Dist. = = 7.8 km 148. tending F and labelling angles v and w. A C v F z = v + w t. angle of a = the two opp. int. angles. = v Corr. angles =, // lines. y = w Vert. opp. angles =. z = + y Substitution y = 360 Angle sum of a point = 360. DAB = 1 2 DCB = 1 2 y DAB + DCB = y z w z 40 y 106 y G 320 B D Page OQ = 5 2 OQ 2 = 9 OQ = 3 tan R = 3 4 R = 36.9 PR 2 = PR = 50 PRO = PRO = 81.9 sin 81.9 = height 50 height = 7.0 m 151. a) 15 = = 120 = 10 2y y + 8 = y + 64 = 24y 16y = 64 y = 4 b) OP is parallel to N LOP is isoceles LN is isoceles Page h 2 = 2 ( ) 2 + ( + 4 ) 2 h 2 = h 2 = h = ( ) 2 + ( ) = = = = 32 = a) Bearing from Y to X is 238. b) Bearing from Z to Y is 136. Distance from Z to Y is (splitting into two right-angled triangles): + y where cos 78 = y and cos 78 = = 27 cos 78 = y = 27 cos 78 = y = 11.2 km c) Bearing from Z to X is 214. = 1 ( + y) 2 = 1 2 (360 ) = 180 AS 1.6 Year 11 athematics and Statistics Published by NuLake Ltd New Zealand Robert Lakeland & Carl Nugent
7 AS 1.6 Geometric Reasoning 71 Pages Practice ternal Assessment Task 1 Question One Geometric Reasoning 1.6 a) i) CB = 1184 A ii) CAB = 46.2 A ii) XCB = 43.8 (A) As ABC = 43.8 Angle sum of a = 180. XCB = 43.8 Alt. angles // lines = b) BK = sin584. c = 14.6 cm c) From start North = 625 cos cos 35 = 1000 km (0 dp) (A) West = 475 sin sin 12 = 143 km (0 dp) () North 1000 km, West 143 km Bearing 352 d) Interior angles of a pentagon = 108 Height = cos cos 54 = m (3 dp) Question Two a) DF = 102 (A) As DF = 39 Base angles Isos. =. DF = 102 Angle sum of a = 180. b) TB 1 B 2 = 46 Alt. angles // lines =. TB 2 B 1 = 46 Base angles Isos. =. B 1 TB 2 = 88 Angle sum of a = 180. Bearing = 134 c) VWY = 75 (A) as ZYW + TYU = 210 Angle sum point = 360. ZYW = TYU Given. ZYW = 105 VWY = 75 Co-int. angles // = 180. d) i) PSQ = PTR Corr. angles // lines =. Sin 41 = y 6 y = 3.94 m (2 sf) ii) Triangles PQS and PRT are similar. Note no right-angled triangle. Therefore = 6 = 3.05 m e) LQR = 36 t. angles = 360. = JQP JQL = 144 Adj. angles / lines = 180. () PJQ = 108 Angles Isos. =. KJQ = 72 Adj. angles / lines =180. = KLQ (symmetry) JKL = 72 Angle sum of a quad. = 360. Question Three a) i) AOB = 136 (A) AOB (refle) = 224 AOB (obtuse) = 136 Angle sum of a point = 360. ii) ODF = f Base angles Isos. =. DOF = 180 2f Angle sum of a = 180. OD = 2f Angle sum of a = 180. () DO = 90 Angle Tang. and radius = 90. DO = 90 2f Angle sum of a = 180. b) DCB = 92 (A) DB = 46 Co-int. angles // = 180. DC = 90 Symmetry. BDC = 44 DBC = 44 Symmetry DCB = 92 Angle sum of a = 180. c) HGX = KJX Alt. angles // lines =. GHX = JKX Alt. angles // lines =. GXH = JXK Vertically opposite angles =. Therefore triangles GHX and JKX are similar 10 = 12 8 = 15 cm Correct answer without proving similar triangles is d) QN = 23 Angles same arc =. Judgement Q = 84 mm Diameter (A) NQ = 90 Angle in semi circle 90. () N = 84 sin 23 = 32.8 mm The grade in brackets is an alternative if the full grade is not earned. In each question you would need at least 2 A for an Achievement, 2 for a erit and 1 for an cellence. The final grade is found by combining the results of the three questions. Achievement Requires two question Achievements or better. erit Requires two question erits or better. cellence Requires two question cellences. AS 1.6 Year 11 athematics and Statistics Published by NuLake Ltd New Zealand Robert Lakeland & Carl Nugent
8 72 AS 1.6 Geometric Reasoning Pages Practice ternal Assessment Task 2 Question One Geometric Reasoning 1.6 a) i) BXC = 44 (A) ii) XBC = 68 Adj. angles / lines =180. BCX = 68 Base angles Isos. =. BXC = 44 Angle sum of a = 180. XBC = 68 Adj. angles / lines = B = tan68 c = 48.5 BC = 97 cm b) Let HG and GF = = = mm Perim. = 4115 mm (0 dp) c) i) N 2 BA = 58 Alt. angles // lines =. Bearing = = 234 A ii) CX = 243 sin 68 CX = m AX = tan58c = m () BX = 243 cos 68 = 91.0 AB = 232 m (0 dp) d) TVO = 34 (A) Question Two OTU = 90 Tangent at 90 to radius. TOU = 68 Angle sum of a = 180. TVO = 34 a) i) DAB = 127 (A) CBA = 53 Angle sum pt. = 360 DAB = 127 Co-int. angles // = 180. ii) B = 21.1 mm () AB = 35.1 mm Pythagoras Perim. = 252 mm (0 dp) b) i) KLR = 78 (A) LRQ = 78 Base angles Isos. =. KLR = 78 Alt. angles // lines =. ii) Form Rt. angled triangle by adding altitude. 70 RL = cos78 c = 337 mm c) SRT = TPQ Alt. angles // lines =. RTS = PTQ Vertically opp. angles =. Similar triangles RTS and PTQ. y = 52 mm Question Three a) i) PQR = 54 (A) JKQ = 72 t angles of polygon = 360. KJQ = 72 t angles of polygon = 360. KQJ = 36 Angle sum of a = 180. KQR = 90 Angle in semi circle 90. PQR = 54 ii) OAB = OBA Base angles Isos. =. OAB = BDC t. angle cyclic quad = Opp. Int. OBA = ACD t. angle cyclic quad = Opp. Int. As OAB = ACD and OBA = BDC and they are corresponding angles them the lines AB and CD are parallel. b) HGK = 113 (A) HOK = 134 Vertically Opp. angles =. HIO = 67 HGK = 113 Opp. angles cyclic quad. = 180. c) DB = 68 (A) DB = 48 Adj. angles / lines =180. DB = 68 Angle sum of a = 180. d) ABC = 120 Int. angles heagon. Judgement y 1 = 30 Base angles Isos. =. y 4 = 30 Symmetry () ACD = 90 As BCA = 30 and BCD = 120 ADC = 60 As half CD and CD = 120 y 2 = 30 Angle sum of a = 180. y 3 = 30 Symmetry The grade in brackets is an alternative if the full grade is not earned. In each question you would need at least 2 A for an Achievement, 2 for a erit and 1 for and cellence. The final grade is found by combining the results of the three questions. Achievement Requires two question Achievements or better. erit Requires two question erits or better. cellence Requires two question cellences. AS 1.6 Year 11 athematics and Statistics Published by NuLake Ltd New Zealand Robert Lakeland & Carl Nugent
SHW 1-01 Total: 30 marks
SHW -0 Total: 30 marks 5. 5 PQR 80 (adj. s on st. line) PQR 55 x 55 40 x 85 6. In XYZ, a 90 40 80 a 50 In PXY, b 50 34 84 M+ 7. AB = AD and BC CD AC BD (prop. of isos. ) y 90 BD = ( + ) = AB BD DA x 60
SOLUTIONS SECTION A [1] = 27(27 15)(27 25)(27 14) = 27(12)(2)(13) = cm. = s(s a)(s b)(s c)
![SOLUTIONS SECTION A [1] = 27(27 15)(27 25)(27 14) = 27(12)(2)(13) = cm. = s(s a)(s b)(s c)](/thumbs/79/80187791.jpg "SOLUTIONS SECTION A [1] = 27(27 15)(27 25)(27 14) = 27(12)(2)(13) = cm. = s(s a)(s b)(s c)")
1. (A) 1 1 1 11 1 + 6 6 5 30 5 5 5 5 6 = 6 6 SOLUTIONS SECTION A. (B) Let the angles be x and 3x respectively x+3x = 180 o (sum of angles on same side of transversal is 180 o ) x=36 0 So, larger angle=3x
2. In ABC, the measure of angle B is twice the measure of angle A. Angle C measures three times the measure of angle A. If AC = 26, find AB.
2009 FGCU Mathematics Competition. Geometry Individual Test 1. You want to prove that the perpendicular bisector of the base of an isosceles triangle is also the angle bisector of the vertex. Which postulate/theorem
Grade 6 Math Circles March 22-23, 2016 Contest Prep - Geometry
Faculty of Mathematics Waterloo, Ontario N2L 3G1 Centre for Education in Mathematics and Computing Grade 6 Math Circles March 22-23, 2016 Contest Prep - Geometry Today we will be focusing on how to solve
TRIANGLES CHAPTER 7. (A) Main Concepts and Results. (B) Multiple Choice Questions
CHAPTER 7 TRIANGLES (A) Main Concepts and Results Triangles and their parts, Congruence of triangles, Congruence and correspondence of vertices, Criteria for Congruence of triangles: (i) SAS (ii) ASA (iii)
21. Prove that If one side of the cyclic quadrilateral is produced then the exterior angle is equal to the interior opposite angle.
21. Prove that If one side of the cyclic quadrilateral is produced then the exterior angle is equal to the interior opposite angle. 22. Prove that If two sides of a cyclic quadrilateral are parallel, then
2012 GCSE Maths Tutor All Rights Reserved
2012 GCSE Maths Tutor All Rights Reserved www.gcsemathstutor.com This book is under copyright to GCSE Maths Tutor. However, it may be distributed freely provided it is not sold for profit. Contents angles
Properties of the Circle
9 Properties of the Circle TERMINOLOGY Arc: Part of a curve, most commonly a portion of the distance around the circumference of a circle Chord: A straight line joining two points on the circumference
Set 5 Paper 2. Set 5 Paper 2. 1 Pearson Education Asia Limited 2017
Set Paper Set Paper. B. C. B. C. C 6. D 7. A. D. A. A. C. C. B. B. C 6. C 7. C. A. B. D. B. D. A. A. B 6. B 7. D. D. C. A. C. D. D. A. D 6. D 7. A. A. C. C. B. D. B. D. A Section A. B ( 7) 7 ( ) 7 ( )
Chapter 7. Geometric Inequalities
4. Let m S, then 3 2 m R. Since the angles are supplementary: 3 2580 4568 542 Therefore, m S 42 and m R 38. Part IV 5. Statements Reasons. ABC is not scalene.. Assumption. 2. ABC has at least 2. Definition
Math 9 Chapter 8 Practice Test
Name: Class: Date: ID: A Math 9 Chapter 8 Practice Test Short Answer 1. O is the centre of this circle and point Q is a point of tangency. Determine the value of t. If necessary, give your answer to the
Fill in the blanks Chapter 10 Circles Exercise 10.1 Question 1: (i) The centre of a circle lies in of the circle. (exterior/ interior) (ii) A point, whose distance from the centre of a circle is greater
NATIONAL SENIOR CERTIFICATE MEMORANDUM MATHEMATICS MEMORANDUM P2 SEPTEMBER 2016 GRADE 12
NATIONAL SENIOR CERTIFICATE MEMORANDUM MATHEMATICS MEMORANDUM P SEPTEMBER 06 GRADE This memo consists of 5 pages Income Maths Memo / P September 06 QUESTION.. 700 answer ().. 700 answer ().. 45 minutes
Revision Question Bank
Revision Question Bank Triangles 1. In the given figure, find the values of x and y. Since, AB = AC C = B [angles opposite to the equal sides are equal] x = 50 Also, the sum of all angles of a triangle
Triangles. Example: In the given figure, S and T are points on PQ and PR respectively of PQR such that ST QR. Determine the length of PR.
Triangles Two geometric figures having the same shape and size are said to be congruent figures. Two geometric figures having the same shape, but not necessarily the same size, are called similar figures.
Similarity of Triangle
Similarity of Triangle 95 17 Similarity of Triangle 17.1 INTRODUCTION Looking around you will see many objects which are of the same shape but of same or different sizes. For examples, leaves of a tree
Topic 2 [312 marks] The rectangle ABCD is inscribed in a circle. Sides [AD] and [AB] have lengths
![Topic 2 [312 marks] The rectangle ABCD is inscribed in a circle. Sides [AD] and [AB] have lengths](/thumbs/86/93150791.jpg "Topic 2 [312 marks] The rectangle ABCD is inscribed in a circle. Sides [AD] and [AB] have lengths")
Topic 2 [312 marks] 1 The rectangle ABCD is inscribed in a circle Sides [AD] and [AB] have lengths [12 marks] 3 cm and (\9\) cm respectively E is a point on side [AB] such that AE is 3 cm Side [DE] is
Proofs. by Bill Hanlon
Proofs by Bill Hanlon Future Reference To prove congruence, it is important that you remember not only your congruence theorems, but know your parallel line theorems, and theorems concerning triangles.
Class IX Chapter 7 Triangles Maths. Exercise 7.1 Question 1: In quadrilateral ACBD, AC = AD and AB bisects A (See the given figure).
Exercise 7.1 Question 1: In quadrilateral ACBD, AC = AD and AB bisects A (See the given figure). Show that ABC ABD. What can you say about BC and BD? In ABC and ABD, AC = AD (Given) CAB = DAB (AB bisects
9. AD = 7; By the Parallelogram Opposite Sides Theorem (Thm. 7.3), AD = BC. 10. AE = 7; By the Parallelogram Diagonals Theorem (Thm. 7.6), AE = EC.
3. Sample answer: Solve 5x = 3x + 1; opposite sides of a parallelogram are congruent; es; You could start b setting the two parts of either diagonal equal to each other b the Parallelogram Diagonals Theorem
SMT 2018 Geometry Test Solutions February 17, 2018
SMT 018 Geometry Test Solutions February 17, 018 1. Consider a semi-circle with diameter AB. Let points C and D be on diameter AB such that CD forms the base of a square inscribed in the semicircle. Given
Assessment Schedule 2011 Mathematics and Statistics: Apply geometric reasoning in solving problems (91031)
NCEA Level 1 Mathematics and Statistics (91031) 2011 page 1 of 5 Assessment Schedule 2011 Mathematics and Statistics: problems (91031) Evidence Statement with Merit with Excellence problems will involve:
0811ge. Geometry Regents Exam BC, AT = 5, TB = 7, and AV = 10.
0811ge 1 The statement "x is a multiple of 3, and x is an even integer" is true when x is equal to 1) 9 2) 8 3) 3 4) 6 2 In the diagram below, ABC XYZ. 4 Pentagon PQRST has PQ parallel to TS. After a translation
Canadian Open Mathematics Challenge
The Canadian Mathematical Society in collaboration with The CENTRE for EDUCATION in MATHEMATICS and COMPUTING presents the Canadian Open Mathematics Challenge Wednesday, November, 006 Supported by: Solutions
Question 1: In quadrilateral ACBD, AC = AD and AB bisects A (See the given figure). Show that ABC ABD. What can you say about BC and BD?
Class IX - NCERT Maths Exercise (7.1) Question 1: In quadrilateral ACBD, AC = AD and AB bisects A (See the given figure). Show that ABC ABD. What can you say about BC and BD? Solution 1: In ABC and ABD,
Exercise 10.1 Question 1: Fill in the blanks (i) The centre of a circle lies in of the circle. (exterior/ interior)
Exercise 10.1 Question 1: Fill in the blanks (i) The centre of a circle lies in of the circle. (exterior/ interior) (ii) A point, whose distance from the centre of a circle is greater than its radius lies
Class IX Chapter 7 Triangles Maths
Class IX Chapter 7 Triangles Maths 1: Exercise 7.1 Question In quadrilateral ACBD, AC = AD and AB bisects A (See the given figure). Show that ABC ABD. What can you say about BC and BD? In ABC and ABD,
11 is the same as their sum, find the value of S.
Answers: (998-99 HKMO Final Events) Created by: Mr. Francis Hung Last updated: July 08 Individual Events I P 4 I a 8 I a 6 I4 a I5 a IS a Q 8 b 0 b 7 b b spare b 770 R c c c c 0 c 57 S 0 d 000 d 990 d
6 CHAPTER. Triangles. A plane figure bounded by three line segments is called a triangle.
6 CHAPTER We are Starting from a Point but want to Make it a Circle of Infinite Radius A plane figure bounded by three line segments is called a triangle We denote a triangle by the symbol In fig ABC has
UKMT UKMT UKMT. IMOK Olympiad. Thursday 16th March Organised by the United Kingdom Mathematics Trust. Solutions
UKMT UKMT UKMT IMOK Olympiad Thursday 16th March 2017 Organised by the United Kingdom Mathematics Trust s These are polished solutions and do not illustrate the process of failed ideas and rough work by
Euclidian Geometry Grade 10 to 12 (CAPS)
Euclidian Geometry Grade 10 to 12 (CAPS) Compiled by Marlene Malan marlene.mcubed@gmail.com Prepared by Marlene Malan CAPS DOCUMENT (Paper 2) Grade 10 Grade 11 Grade 12 (a) Revise basic results established
Class IX - NCERT Maths Exercise (10.1)
Class IX - NCERT Maths Exercise (10.1) Question 1: Fill in the blanks (i) The centre of a circle lies in of the circle. (exterior/interior) (ii) A point, whose distance from the centre of a circle is greater
2007 Shortlist JBMO - Problems
Chapter 1 007 Shortlist JBMO - Problems 1.1 Algebra A1 Let a be a real positive number such that a 3 = 6(a + 1). Prove that the equation x + ax + a 6 = 0 has no solution in the set of the real number.
BHP BILLITON UNIVERSITY OF MELBOURNE SCHOOL MATHEMATICS COMPETITION, 2003: INTERMEDIATE DIVISION
BHP BILLITON UNIVERSITY OF MELBOURNE SCHOOL MATHEMATICS COMPETITION, 00: INTERMEDIATE DIVISION 1. A fraction processing machine takes a fraction f and produces a new fraction 1 f. If a fraction f = p is
Mathematics. A basic Course for beginers in G.C.E. (Advanced Level) Mathematics
Mathematics A basic Course for beginers in G.C.E. (Advanced Level) Mathematics Department of Mathematics Faculty of Science and Technology National Institute of Education Maharagama Sri Lanka 2009 Director
Triangles. Chapter Flowchart. The Chapter Flowcharts give you the gist of the chapter flow in a single glance.
Triangles Chapter Flowchart The Chapter Flowcharts give you the gist of the chapter flow in a single glance. Triangle A plane figure bounded by three line segments is called a triangle. Types of Triangles
Solution Of Class 10 th CBSE SA-II Board (Set-1)Mathematics
L.K. Gupta (Mathematic Classes) www.poineermathematics.com. MOBILE: 98155771, 461771 Solution Of Class 10 th CBSE SA-II Board (Set-1)Mathematics 1. (k 1) k = (k + 1) (k 1) k 1 k = k + 1 k + 1 k 1 = k.
Grade 9 Geometry-Overall
ID : au-9-geometry-overall [1] Grade 9 Geometry-Overall For more such worksheets visit www.edugain.com Answer t he quest ions (1) A chord of a circle is equal to its radius. Find the angle subtended by
Visit: ImperialStudy.com For More Study Materials Class IX Chapter 12 Heron s Formula Maths
Exercise 1.1 1. Find the area of a triangle whose sides are respectively 150 cm, 10 cm and 00 cm. The triangle whose sides are a = 150 cm b = 10 cm c = 00 cm The area of a triangle = s(s a)(s b)(s c) Here
Maharashtra State Board Class X Mathematics Geometry Board Paper 2015 Solution. Time: 2 hours Total Marks: 40
Maharashtra State Board Class X Mathematics Geometry Board Paper 05 Solution Time: hours Total Marks: 40 Note:- () Solve all questions. Draw diagrams wherever necessary. ()Use of calculator is not allowed.
CAPS Mathematics GRADE 11. Sine, Cosine and Area Rules
CAPS Mathematics GRADE Sine, Cosine and Area Rules Outcomes for this Topic. Calculate the area of a triangle given an angle and the two adjacent sides. Lesson. Apply the Sine Rule for triangles to calculate
0811ge. Geometry Regents Exam
0811ge 1 The statement "x is a multiple of 3, and x is an even integer" is true when x is equal to 1) 9 ) 8 3) 3 4) 6 In the diagram below, ABC XYZ. 4 Pentagon PQRST has PQ parallel to TS. After a translation
9 th CBSE Mega Test - II
9 th CBSE Mega Test - II Time: 3 hours Max. Marks: 90 General Instructions All questions are compulsory. The question paper consists of 34 questions divided into four sections A, B, C and D. Section A
ANSWERS. CLASS: VIII TERM - 1 SUBJECT: Mathematics. Exercise: 1(A) Exercise: 1(B)
ANSWERS CLASS: VIII TERM - 1 SUBJECT: Mathematics TOPIC: 1. Rational Numbers Exercise: 1(A) 1. Fill in the blanks: (i) -21/24 (ii) -4/7 < -4/11 (iii)16/19 (iv)11/13 and -11/13 (v) 0 2. Answer True or False:
0110ge. Geometry Regents Exam Which expression best describes the transformation shown in the diagram below?
0110ge 1 In the diagram below of trapezoid RSUT, RS TU, X is the midpoint of RT, and V is the midpoint of SU. 3 Which expression best describes the transformation shown in the diagram below? If RS = 30
Page 9 cont f (x) f (x) f (x) like. Is zero. 7. Gradient
S.7 Calculus ethods 65 Page 5 nswers 1. t x =, m = t x =, m =. t x =, m = Page 6 t x = 1, m = t x = 1, m = 0. a) m = 1 at (,.5) b) m = at (, 5.5). a) m = at x =.5 approx. and x =.5 approx. b) m = 0 at
REVISION EXERCISES ON GEOMETRY END TRIGONOMETRY
REVISION EXERCISES ON GEOMETRY END TRIGONOMETRY 1 The bearing of B from A is 030. Find the bearing of A from B. 2 For triangle ABC, AB = 60 cm, BC = 80 cm and the magnitude of angle ABC is 120. Find the
2 13b + 37 = 54, 13b 37 = 16, no solution
Answers: (999-00 HKMO Final Events) Created by: Mr. Francis Hung Last updated: 6 February 07 Individual Events SI P 6 I P 5 I P 6 I P I P I5 P Q 7 Q 8 Q 8 Q Q Q R R 7 R R 996 R R S 990 S 6 S S 666 S S
NATIONAL SENIOR CERTIFICATE GRADE 11
NATIONAL SENIOR CERTIFICATE GRADE MATHEMATICS P NOVEMBER 05 MARKS: 50 TIME: 3 hours This question paper consists of 5 pages and a 4-page answer book. Mathematics/P DBE/November 05 CAPS Grade INSTRUCTIONS
Class IX Chapter 8 Quadrilaterals Maths
Class IX Chapter 8 Quadrilaterals Maths Exercise 8.1 Question 1: The angles of quadrilateral are in the ratio 3: 5: 9: 13. Find all the angles of the quadrilateral. Answer: Let the common ratio between
Class IX Chapter 8 Quadrilaterals Maths
1 Class IX Chapter 8 Quadrilaterals Maths Exercise 8.1 Question 1: The angles of quadrilateral are in the ratio 3: 5: 9: 13. Find all the angles of the quadrilateral. Let the common ratio between the angles
ICSE QUESTION PAPER Class X Maths (2016) Solution
ICSE QUESTION PAPER Class X Maths (016) Solution SECTION A 1. (a) Let f(x) x x kx 5 Using remainder theorem, f() 7 () () k() 5 7 (8) (4) k() 5 7 16 1 k 5 7 k 16 1 5 7 k 6 k 1 (b) A = 9A + MI A 9A mi...
JEFFERSON MATH PROJECT REGENTS AT RANDOM
JEFFERSON MATH PROJECT REGENTS AT RANDOM The NY Geometry Regents Exams Fall 2008-January 2010 Dear Sir I have to acknolege the reciept of your favor of May 14. in which you mention that you have finished
CONGRUENCE OF TRIANGLES
Congruence of Triangles 11 CONGRUENCE OF TRIANGLES You might have observed that leaves of different trees have different shapes, but leaves of the same tree have almost the same shape. Although they may
PLC Papers. Created For:
PLC Papers Created For: ed by use of accompanying mark schemes towards the rear to attain 8 out of 10 marks over time by completing Circle Theorems 1 Grade 8 Objective: Apply and prove the standard circle
SHW6-R1 1M+1A 1M+1A 1M+1A. 11. (a) 14. (a) With the notations in the figure, With the notations in the figure, AG BH 800 m Consider ACG.
SHW6-R 8@. (a) 4. (a) With the notations in the figure, With the notations in the figure, AG BH Consider G. ΑG tan G tan 50 tan 50 Consider CHG. GH tan H GH tan 70 tan 50 tan 70 GH tan 50 The speed of
SOLUTIONS SECTION A SECTION B
SOLUTIONS SECTION A 1. C (1). A (1) 3. B (1) 4. B (1) 5. C (1) 6. B (1) 7. A (1) 8. D (1) SECTION B 9. 3 3 + 7 = 3 3 7 3 3 7 3 3 + 7 6 3 7 = 7 7 6 3 7 3 3 7 0 10 = = 10. To find: (-1)³ + (7)³ + (5)³ Since
chapter 1 vector geometry solutions V Consider the parallelogram shown alongside. Which of the following statements are true?
chapter vector geometry solutions V. Exercise A. For the shape shown, find a single vector which is equal to a)!!! " AB + BC AC b)! AD!!! " + DB AB c)! AC + CD AD d)! BC + CD!!! " + DA BA e) CD!!! " "
CONGRUENCE AND SIMILARITY
CONGRUENCE ND SIMILRITY 1.1CONGRUENT FIGURES The figures that have the same size and the same shape, i.e. one shape fits exactly onto other is called Congruent figures. CONGRUENT TRINGLES: 1. Two triangles
Udaan School Of Mathematics Class X Chapter 10 Circles Maths
Exercise 10.1 1. Fill in the blanks (i) The common point of tangent and the circle is called point of contact. (ii) A circle may have two parallel tangents. (iii) A tangent to a circle intersects it in
The University of the State of New York REGENTS HIGH SCHOOL EXAMINATION GEOMETRY. Student Name:
GEOMETRY The University of the State of New York REGENTS HIGH SCHOOL EXAMINATION GEOMETRY Wednesday, August 17, 2011 8:30 to 11:30 a.m., only Student Name: School Name: Print your name and the name of
Solutions Key Right Triangles and Trigonometry
CHAPTER 8 Solutions Key Right Triangles and Trigonometry ARE YOU READY? PAGE 515 1. D. C 3. A 4. E 5. PR RT 10 5 QR ; RS 1 6 PRQ TRS by Vert. Thm. yes; PRQ TRS by SAS 6. AB FE 6 4 3 ; BC ED 15 10 3 B E
Question 1
General Mathematics Paper 2,Nov/Dec. 2007 Question 1 (a) Without using four-figure tables or calculators, evaluate 0.0024 x 0.064 0.048 leaving your answer in standard form. (b) (i). What range of values
Shape Booster 6 Similar Shapes
Shape Booster 6 Similar Shapes Check: 85T) The two triangles are similar. 5cm y x 37.8cm 8cm 43.2cm a) Work out the size of x. b) Work out the size of y. a) x = 27cm b) y = 7cm Learn: Maths Watch Reference
2, find the value of a.
Answers: (99-9 HKMO Final Events) reated by: Mr. Francis Hung Last updated: 7 December 05 Individual Events I a I a 6 I a 4 I4 a 8 I5 a 0 b b 60 b 4 b 9 b c c 00 c 50 c 4 c 57 d d 50 d 500 d 54 d 7 Group
0113ge. Geometry Regents Exam In the diagram below, under which transformation is A B C the image of ABC?
0113ge 1 If MNP VWX and PM is the shortest side of MNP, what is the shortest side of VWX? 1) XV ) WX 3) VW 4) NP 4 In the diagram below, under which transformation is A B C the image of ABC? In circle
Core Mathematics 2 Radian Measures
Core Mathematics 2 Radian Measures Edited by: K V Kumaran Email: kvkumaran@gmail.com Core Mathematics 2 Radian Measures 1 Radian Measures Radian measure, including use for arc length and area of sector.
1.1 Exercises, Sample Solutions
DM, Chapter, Sample Solutions. Exercises, Sample Solutions 5. Equal vectors have the same magnitude and direction. a) Opposite sides of a parallelogram are parallel and equal in length. AD BC, DC AB b)
CIRCLES, CHORDS AND TANGENTS
NAME SCHOOL INDEX NUMBER DATE CIRCLES, CHORDS AND TANGENTS KCSE 1989 2012 Form 3 Mathematics Working Space 1. 1989 Q24 P2 The figure below represents the cross section of a metal bar. C A 4cm M 4cm B The
VAISHALI EDUCATION POINT (QUALITY EDUCATION PROVIDER)
BY:Prof. RAHUL MISHRA Class :- X QNo. VAISHALI EDUCATION POINT (QUALITY EDUCATION PROVIDER) CIRCLES Subject :- Maths General Instructions Questions M:9999907099,9818932244 1 In the adjoining figures, PQ
Mathematics Class X Board Paper 2011
Mathematics Class X Board Paper Solution Section - A (4 Marks) Soln.. (a). Here, p(x) = x + x kx + For (x-) to be the factor of p(x) = x + x kx + P () = Thus, () + () k() + = 8 + 8 - k + = k = Thus p(x)
CBSE MATHEMATICS (SET-2)_2019
CBSE 09 MATHEMATICS (SET-) (Solutions). OC AB (AB is tangent to the smaller circle) In OBC a b CB CB a b CB a b AB CB (Perpendicular from the centre bisects the chord) AB a b. In PQS PQ 4 (By Pythagoras
Maharashtra State Board Class IX Mathematics Geometry Board Paper 1 Solution
Maharashtra State Board Class IX Mathematics Geometry Board Paper Solution Time: hours Total Marks: 40. i. Let the measure of each interior opposite angle be x. Since, Sum of two interior opposite angles
MEMO MATHEMATICS: PAPER II
MEMO CLUSTER PAPER 2016 MATHEMATICS: PAPER II Time: 3 hours 150 marks PLEASE READ THE FOLLOWING INSTRUCTIONS CAREFULLY 1. This question paper consists of 28 pages and an Information Sheet of 2 pages(i-ii).
(b) Write down the largest integer, y, which satisfies -7 < 2y + 4 < 12.
4E/N Prelim 010 Paper 1 1. (.1 Evaluate, giving your answer in standard form and correct to 0.79 7.1 significant figures. ( Simplify p (p 1). Answer ( [] ( [1]. ( Simplify x. ( Given that 64 1 k, find
RMT 2013 Geometry Test Solutions February 2, = 51.
RMT 0 Geometry Test Solutions February, 0. Answer: 5 Solution: Let m A = x and m B = y. Note that we have two pairs of isosceles triangles, so m A = m ACD and m B = m BCD. Since m ACD + m BCD = m ACB,
LLT Education Services
8. The length of a tangent from a point A at distance 5 cm from the centre of the circle is 4 cm. Find the radius of the circle. (a) 4 cm (b) 3 cm (c) 6 cm (d) 5 cm 9. From a point P, 10 cm away from the
CBSE CLASS X MATH -SOLUTION Therefore, 0.6, 0.25 and 0.3 are greater than or equal to 0 and less than or equal to 1.
CBSE CLASS X MATH -SOLUTION 011 Q1 The probability of an event is always greater than or equal to zero and less than or equal to one. Here, 3 5 = 0.6 5% = 5 100 = 0.5 Therefore, 0.6, 0.5 and 0.3 are greater
IB MYP Unit 6 Review
Name: Date: 1. Two triangles are congruent if 1. A. corresponding angles are congruent B. corresponding sides and corresponding angles are congruent C. the angles in each triangle have a sum of 180 D.
Maharashtra Board Class IX Mathematics (Geometry) Sample Paper 3 Solution
Maharashtra Board Class IX Mathematics (Geometry) Sample Paper Solution Time: hours Total Marks: 40. i. Three sides of an equilateral triangle are congruent. Perimeter of an equilateral triangle = side
Revision Pack. Edexcel GCSE Maths (1 9) Geometry. Edited by: K V Kumaran
Edexcel GCSE Maths (1 9) Revision Pack Geometry Edited by: K V Kumaran kvkumaran@gmail.com 07961319548 www.kumarmaths.weebly.com kumarmaths.weebly.com 1 Q1. Diagram NOT accurately drawn The diagram shows
IB Math SL 1: Trig Practice Problems: MarkScheme Circular Functions and Trig - Practice Problems (to 07) MarkScheme
IB Math SL : Trig Practice Problems: MarkScheme Circular Functions and Trig - Practice Problems (to 07) MarkScheme. (a) Evidence of using the cosine rule p + r q eg cos P Qˆ R, q p + r pr cos P Qˆ R pr
"32 4"2; therefore HJ EF.
Answers for Lesson 5-, pp. 6 64 Exercises. 9. 7 3. 4 4. 3 5. 6. 7. 40 8. 50 9. 60 0. 80. UW TX; UY VX; YW TV. GJ FK; JL HF; GL HK 3. a. ST PR; SU QR; UT PQ b. 4. FE 5. FG 6. 7. EG 8. AC 9. 0. a. 050 ft
CCE PR Revised & Un-Revised
D CCE PR Revised & Un-Revised 560 00 KARNATAKA SECONDARY EDUCATION EXAMINATION BOARD, MALLESWARAM, BANGALORE 560 00 08 S.S.L.C. EXAMINATION, JUNE, 08 :. 06. 08 ] MODEL ANSWERS : 8-K Date :. 06. 08 ] CODE
EXERCISE 10.1 EXERCISE 10.2
NCERT Class 9 Solved Questions for Chapter: Circle 10 NCERT 10 Class CIRCLES 9 Solved Questions for Chapter: Circle EXERCISE 10.1 Q.1. Fill in the blanks : (i) The centre of a circle lies in of the circle.
PRACTICE QUESTIONS CLASS IX: CHAPTER 4 LINEAR EQUATION IN TWO VARIABLES
PRACTICE QUESTIONS CLASS IX: CHAPTER 4 LINEAR EQUATION IN TWO VARIABLES 1. Find the value of k, if x =, y = 1 is a solution of the equation x + 3y = k.. Find the points where the graph of the equation
COMMON UNITS OF PERIMITER ARE METRE
MENSURATION BASIC CONCEPTS: 1.1 PERIMETERS AND AREAS OF PLANE FIGURES: PERIMETER AND AREA The perimeter of a plane figure is the total length of its boundary. The area of a plane figure is the amount of
Answers: ( HKMO Final Events) Created by Mr. Francis Hung Last updated: 13 December Group Events
Individual Events SI a 900 I1 P 100 I k 4 I3 h 3 I4 a 18 I5 a 495 b 7 Q 8 m 58 k 6 r 3 b p R 50 a m 4 M 9 4 x 99 q 9 S 3 b 3 p Group Events 15 16 w 1 Y 109 SG p 75 G6 x 15 G7 M 5 G8 S 7 G9 p 60 G10 n 18
STRAND J: TRANSFORMATIONS, VECTORS and MATRICES
Mathematics SKE, Strand J STRAND J: TRANSFORMATIONS, VECTORS and MATRICES J4 Matrices Text Contents * * * * Section J4. Matrices: Addition and Subtraction J4.2 Matrices: Multiplication J4.3 Inverse Matrices:
Question Bank Tangent Properties of a Circle
Question Bank Tangent Properties of a Circle 1. In quadrilateral ABCD, D = 90, BC = 38 cm and DC = 5 cm. A circle is inscribed in this quadrilateral which touches AB at point Q such that QB = 7 cm. Find
RD Sharma Solutions for Class 10 th
RD Sharma Solutions for Class 10 th Contents (Click on the Chapter Name to download the solutions for the desired Chapter) Chapter 1 : Real Numbers Chapter 2 : Polynomials Chapter 3 : Pair of Linear Equations
Mathematics Revision Guide. Shape and Space. Grade C B
Mathematics Revision Guide Shape and Space Grade C B 1 A of = b h 2 Area 6cm = 10 6 2 = 60 2 8cm = 30cm 2 6cm 12cm A of = (a+b) h 2 = (6+12) 5 2 = (18) 5 2 = 90 2 = 4 2 7cm 1 6cm A of = π r r = π 6 6 =
5. Using a compass and straightedge, construct a bisector of the angle shown below. [Leave all construction marks.]
![5. Using a compass and straightedge, construct a bisector of the angle shown below. [Leave all construction marks.]](/thumbs/88/116790473.jpg "5. Using a compass and straightedge, construct a bisector of the angle shown below. [Leave all construction marks.]")
Name: Regents Review Session Two Date: Common Core Geometry 1. The diagram below shows AB and DE. Which transformation will move AB onto DE such that point D is the image of point A and point E is the
AQA Level 2 Certificate in Further Mathematics. Worksheets - Teacher Booklet
AQA Level Certificate in Further Mathematics Worksheets - Teacher Booklet Level Specification Level Certificate in Further Mathematics 860 Worksheets - Teacher Booklet Our specification is published on
Grade 9 Circles. Answer t he quest ions. For more such worksheets visit
ID : th-9-circles [1] Grade 9 Circles For more such worksheets visit www.edugain.com Answer t he quest ions (1) ABCD is a cyclic quadrilateral such that AB is a diameter of the circle circumscribing it
Geometry Midterm Exam Review 3. Square BERT is transformed to create the image B E R T, as shown.
1. Reflect FOXY across line y = x. 3. Square BERT is transformed to create the image B E R T, as shown. 2. Parallelogram SHAQ is shown. Point E is the midpoint of segment SH. Point F is the midpoint of
Math 5 Trigonometry Fair Game for Chapter 1 Test Show all work for credit. Write all responses on separate paper.
Math 5 Trigonometry Fair Game for Chapter 1 Test Show all work for credit. Write all responses on separate paper. 12. What angle has the same measure as its complement? How do you know? 12. What is the
Name. 9. Find the diameter and radius of A, B, and C. State the best term for the given figure in the diagram.
Name LESSON 10.1 State the best term for the given figure in the diagram. 9. Find the diameter and radius of A, B, and C. 10. Describe the point of intersection of all three circles. 11. Describe all the
Triangles. 3.In the following fig. AB = AC and BD = DC, then ADC = (A) 60 (B) 120 (C) 90 (D) none 4.In the Fig. given below, find Z.
Triangles 1.Two sides of a triangle are 7 cm and 10 cm. Which of the following length can be the length of the third side? (A) 19 cm. (B) 17 cm. (C) 23 cm. of these. 2.Can 80, 75 and 20 form a triangle?
10. Circles. Q 5 O is the centre of a circle of radius 5 cm. OP AB and OQ CD, AB CD, AB = 6 cm and CD = 8 cm. Determine PQ. Marks (2) Marks (2)
10. Circles Q 1 True or False: It is possible to draw two circles passing through three given non-collinear points. Mark (1) Q 2 State the following statement as true or false. Give reasons also.the perpendicular