Chapter 18 Exercise 18.1

Size: px
Start display at page:

Download "Chapter 18 Exercise 18.1"

Transcription

1 hapter 18 Eercise 18.1 Q. 1. (i) = 143 ( = 143 ) (ii) = 63 ( = 63 ) = 90 (y = 90 ) (iii) = = 135 (iv) = y 90 = y = = y 66 + ( + y) + 47 = y = 67 = 67 y = = nswer: + y = 67 y = 90 Q.. (i) = 53 Opposite angles = = 17 Interior angles = 53 lternate angles (ii) = = 68 (Interior angles) = = 46 (Straight line) = = 134 (Interior angles) nswer: = 68 = 134 = 46 (iii) = 73 orresponding angles = 41 Opposite angles = = 66 Straight line D = 66 orresponding angles E = E = 114 Straight line Q. 3. (iv) = 10 orresponding E = 63 Opposite D = 10 Opposite = 180 E (180 ) = (78) = 39 = = 141 Straight line (v) = 115 (Opposite) = = 65 (Interior) = = 105 (Straight line) D = 75 (lternate) (i) line is a straight line that goes on forever in both directions; it has no endpoints. ray is part of a line that has one endpoint; the other end goes on forever. (ii) Points that lie on the same plane are coplanar. Points that lie on the same line are collinear. (iii) acute angle is one that is less than 90. n ordinary angle is one that is less than 180. (iv) n aiom is a statement we accept as true even though there is no proof. theorem is a statement we accept as true, as there is a proof. Q. 4. (i) = = 137 = = 54 (ii) = = 117 = 117 = 58.5 = = 70.5 = = 38.5 ctive Maths (Strands 1 5): h 18 Solutions 1

2 (iii) = = 130 = 130 = 65 = = 69 = = 69 = 69 = 69 = 65 (iv) = = 47 = 5 (orresponding) = = 81 Q. 5. (i) = 3 (lternate angles) = 58 (Opposite angles/rules of parallelogram) : ( + 3) + (58) = = = = 198 = 99 OR ( + 3) + 58 = = 1 = 1 3 = 99 (ii) = 84 (orresponding) = = 84 (Opposite) ( + 14) + = = 180 = = 8 (iii) = 31 lternate 31 y 31 = 31 (isosceles triangle) by alternate angles = 31 Similarly y = c (by isosceles) 180 = y 180 = = 59 = = = = 90 (iv) 10 y = = 78 by isosceles triangles = = 78 y = 180 = 180 (78) y = 4 = = 78 = = 78 Q. 6. (i) y opposite angles = 49 (ii) = = 131 (iii) = 180 = 49 = 4.5 Q. 7. (ii) 3 and 7 are corresponding (iii) 8 and 7 make up 180 on a straight line (iv) 5 and 4 lternate angles ctive Maths (Strands 1 5): h 18 Solutions

3 Q. 8. (a) (i) Triangle and triangle FDE are congruent by SSS (all sides on are the same as those on FDE). (ii) Though it looks like a rhombus we can t assume, as they haven t given us the information that sides are equal. Q. 9. not congruent: TS = QT and PT is common but PQ PS and no angles are similar. (iii) ongruent: F = FE (Since = ) = ED DFE = F (opposite) FDE = F (alternating) FED = F y SS or S (iv) ΔE is congruent to ΔDE E is common to both D = E = ED (as E is the bisector of D) by SSS [could also be proved by SS or S] (b) Diagram (ii) ΔPQT ΔSTR ΔQTR ΔPTS Diagram (iv) all triangles are congruent. i.e. ΔE ΔE ΔE ΔED = 70 l m 110 opposite opposite Opposite Straight line = = 94 n = 110 p = 110 Lines a and b are parallel Internal angles sum to 180 ( ) orresponding angles (70 and 110 ) lternate angles (70 and 110 ) Lines l and p also show the same features. b a Q. 10. (i) rhombus (all sides equal) (ii) Three n equilateral triangle (iii) Four square (all sides equal) lternative method: (i) Rectangle has symmetries (ii) Triangle has 3 symmetries (iii) Square has 4 symmetries ctive Maths (Strands 1 5): h 18 Solutions 3

4 Q. 11. (i) (ii) Q. 1. (i) = Translation = entral symmetry = ial symmetry (ii) = entral symmetry = ial symmetry = Translation (iii) = ial symmetry = Translation = entral symmetry (iii) No centre of symmetry Q. 13. (a) y (iii) (iv) (i) 3 4 (ii) (b) (i) ( 1, 1), ( 4, 1), ( 4, 3), ( 1, 3) (ii) (1, 1), (4, 1), (4, 3), (1, 3) (iii) (1,1), (4,1), (4,3), (1,3) (v) No centre of symmetry Q. 14. (a) (i) y (iii) (ii) (vi) No centre of symmetry (b) 3 4 (i) (,1), ( 5,1), ( 4,3) (ii) (, 1), ( 5, 1), ( 4, 3) s (iii) (0,1), (3,1), (,3) No ais of symmetry 4 ctive Maths (Strands 1 5): h 18 Solutions

5 Q. 15. (a) Q. 16. (b) 5 (i) 4 3 y 5 4 (ii) (iii) (i) ( 1,1), ( 3,1), ( 4,3), (,3) (ii) ( 1,3), (1,3), (,5), (0,5) (iii) (1, 1), (3, 1), (4, 3), (, 3) 5 y q 4 3 (ii) p (i) 3 (b) r (iii) (i) (4, 1), (6,0), (8, ) (ii) (,3), ( 4,), ( 6,4) (iii) ( 1, ), (0, 4), (, 6) Q. 17. (i) QP = RS = 0 (ii) PS = QP = 16 (iii) MQ = MS = 15 (iv) SQ = MS + MQ = 30 Q. 18. (i) + 30 = + y = y + 6 y = 4 ( + 30) + ( + y + 16) = y + 46 = = 180 = = 86 = 43 (ii) + y + 33 = 5 4y y = 46 simultaneous eq. (5 4y 13) + ( + 3y + 4) = y = 189 simultaneous eq y = y = = 899 = 9 y = = y = 14 Q. 19. (i) Equation 1: 4a + b = 3a b + 11 (Diagonals) a + b = 13 Equation : a + b 1 = a + b 3 (Diagonals) a b = a + b = 13 a b = a + b = a 5 = 3b = 15 a = + 5 b = 5 a = 3 (ii) (a + b) + (4a + 0) = 180 6a + b = (5a + 70) + (4a + 0) = 180 9a = 90 a = 10 b = 160 6(10) b = 100 a = 3, b = 5 ctive Maths (Strands 1 5): h 18 Solutions 5

6 Eercise 18. Q. 1. (i) = 7.5 theorem 11 (ii) = 3 theorem 11 Q.. (i) = 5 y = 6 theorem 11 (ii) = 15 y = 0 theorem 11 (iii) + 5 = 8 theorem 11 = 3 y 3 = 10 theorem 11 y = 13 Q. 3. (i) + D = 56 Q. 4. (i) = 56 theorem 11 8 = 56 = 7 cm (ii) DE + GH = 40 cm (ii) (iii) ( + ) + ( + ) = 40 cm theorem 11 T 3 X 4 U 4 = 40 = 10 cm XU = 16 9 = 7 T 6 Y 8 V YV = = 8 (OR 7 ) T 9 1 Eercise 18.3 Q. 1. (i) bottom length bottom length = top length top length 4 = 1 = (ii) 4 = 1 3 = 4 3 = (iii) 3 = 3 (iv) Q.. (i) = 9 = 4.5 top length top length = total length total length 7 10 = 11 = = 7.7 bottom length total length y 4 = 3 5 y = 1 5 =.4 y (ii) 3.75 = 3 5 y = 11.5 =.5 5 (iii) y 6 = 1 5 y = 6 5 = 1. (iv) y 7 = 6 16 y = 4 16 bottom length = total length y = 1 8 = 5 8 =.65 Z W TW = 1 (by similar triangles), so ZW = = 63 (OR 3 7 ) 6 ctive Maths (Strands 1 5): h 18 Solutions

7 Q. 3. (i) PS.5 = 1 15 PS = = 18 (ii) ST =.5 18 = 4.5 (iii) PS : ST = 18 : 4.5 = 4 : 1 Q. 4. (i) : D = D D : D = 5 : = 3 : (ii) E 15 = 3 E = 30 3 = 10 No: only the ratios are given, not the absolute values Q. 5. (i) DE 3.75 = D 9.5 = DE = = (ii) FG 3.75 = 9.5 FG = = 5 Q. 6. (i) 5 : (ii) : E = : D = 7 : 5 (iii) E : = D : = : 7 Q. 7. (i) 3 4 (ii) Y Y = X X = 3 4 (iii) Y = X = 4 7 Q. 8. : =.5 : 0.75 = 3 : 1 = D : D E. nswer: Yes = Q. 9. Y : Y = 3.5 :.5 = 7 : 5 X : X = 3 : Q. 10. Y : Y X : X XY i.e. not parallel. nswer: No E D 35 (i) 50 = = 8150 = = 191 m Total distance = = 441m (ii) = 14.7 = 14 minutes 4 seconds Eercise 18.4 Q. 1. (i) 3 = 1 = 3 = 1.5 (ii) 6 = 8 10 = = 4.8 (iii) 10 = 3 4 = 30 4 = 7.5 (iv) 10 = 1 9 = 10 9 = Q.. (i) y 4 = 4 5 y = 16 5 = 3. ctive Maths (Strands 1 5): h 18 Solutions 7

8 (ii) y 9 = 0 15 y = = 1 y (iii) 1 = 1 9 y = 5 9 = 8 y (iv) 4 = = 3 y 4 = 3 y = 84 3 = 8 Q. 3. (i) 1 = 4 6 = 48 6 = 8 y 3 = 4 y 3 = y = 6 (ii) 8 = 4 10 = = 19. (iii) 16.8 y 16.8 = y = = 4y y = = = = = 1 y = y = 9 (iv) 4 = = 30 3 = 10 y 0 = 4 10 y = = 8 Q. 4. (i) RST = PQR = 90 SRT is common (ii) T STR = RPQ RST and PQR are similar 6 S P 1 Q 10 Let PR = 10 = = = 0 PR = 0 (iii) Let RQ = y 0 = 1 + y 400 = y 56 = y R 16 = y RQ = 16 (iv) Let RS = z z 16 = 10 0 z 16 = 1 z = 8 RS = 8 Q. 5. (i) EG = EG = 0.75 (ii) EF = 7 (4) = GH 3.5 = 5 4 GH = = Q. 6. (i) + 6 = = 100 = 64 = 8 R 8 ctive Maths (Strands 1 5): h 18 Solutions

9 (ii) ED 6 = 4 10 ED = 4 10 =.4 (iii) ED + D = E (.4) + D = (4) D = 16 D = 10.4 D = 3. (iv) area Δ : area ΔED 1 (8)(6) : 1 (3.)(.4) 4 : : : 1 5 : 4 Q Is ΔPST similar to ΔPQR? PQ PS = = 4. PR PT = 6.4 = 3. s PQ PS PS PT i.e. the lengths of matching sides are not in proportion The triangles are not similar. No: ΔPST is not similar to ΔPQR 11.5 z Q. 7. (i) 8 E D 4 1 Is Δ similar to ΔDE? E = 1 = DE = 4 8 = 1 s E DE No: Δ is not similar to ΔDE (ii) P Q y 1 = y = 15 y = m.1 m pole building ( 1) h = height of building h 1 =.1 h =.8 m 0.75 The building is.8 m tall. h y Q P 3 R S T ctive Maths (Strands 1 5): h 18 Solutions 9

10 Q. 10. Q. 1. (i) Q m h = height of tree in metres. h 1.7 = ( 1.7) h = h = 7.4 m The tree is 7.4 m tall P P 1 cm R Model R 15 cm ctual 3 m (300 cm) P R 1 = P R = 15 P R = 40 cm =.4 m Q Q The length of PR on the actual frame is.4 m. h Q. 13. Q. 14. (ii) h 1. m 1.45 m girl 4. m h 5.4 = h = 1. h = 6.55 m h = 65.5 cm h Flagpole ( 5.4) The flagpole is 653 cm high to the nearest cm m m 5 m h 1.45 = h = 5 h = 6.38 m The school building is 6.38 m high. 5 m 1 m m 5 = = 1 = m The river is m wide 8 m 10 ctive Maths (Strands 1 5): h 18 Solutions

11 Q. 15. (i) m m 7 km y 1 m l h 1.7 m 1.7 m 60 m Q. 16. (i) The height of the kite = h + 1.7m h 0.3 = 60 1 h = h = 18 m Height of kite = = 19.7 m (ii) l = length of kite string Using Pythagoras Theorem: l = l = 3,94 l = 3,94 l = m l = 6, cm The length of the kite is 6,64 cm or 6.64 m to the nearest centimetre. y Park km = distance between shop and park = 7 = 14 = 14 = km Distance between shop and park is 3,74 m to the nearest m. (ii) Let y = distance between school and park Using Pythagoras Theorem, y = 7 + ( 14 ) y = y = 63 y = 63 y = km 1000 y = m The distance between the school and park is 7,937 m to the nearest m School 7 km Shop km Home Q. 17. (i) 1.85 m.15 James Sign 1 m 38 m House ctive Maths (Strands 1 5): h 18 Solutions 11

12 (ii) 1 m 0.3 = 50 1 = = 0.3 m = 50 m = 1.5 m Height of house = 1.5 m m = 3.1 m The house is 3.1 m high. Q. 18. (i), and must be collinear also, E and D must be collinear, [E] [D] (ii) E + 48 = = 57( + 48) 133 = = 736 = 36 The distance between the trees is 36 m. (iii) E = 45 let y be the distance between and D y 40 = y = 36 y = 50 m (iv) To ensure [E] was parallel to [D] students could have inserted a peg F into the ground, such that EF is a parallelogram where = FE = 48 m and E = F = 57 m y D 133 m Let = distance between the two trees E 133 m D D Eercise 18.5 Q. 1. (i) h = a + b h = 13, a = 1, b = 13 = = = 5 = 5 (ii) h = a + b h =, a = 1, b = 16 = = = 400 = 0 1 ctive Maths (Strands 1 5): h 18 Solutions

13 (iii) h = a + b h =, a = 11, b = 5 = ( 11 ) + 5 = = 36 = 6 (iv) h = a + b h = 5, a = 4, b = 5 = = = 49 = 7 (v) h = a + b h =, a = 15, b = 7 = ( 15 ) + 7 = = 64 = 8 (vi) = = = 656 = 656 (vii) 4 = = = = 119 (viii) = + ( 3 ) = = 16 = 4 Q.. (i) = = = 65 = 65 ( 65 ) = 8 + y 65 = 64 + y y = 1 y = 1 (ii) = = = 5 = = 15 + y 89 = 5 + y 64 = y y = 8 (iii) 3 = = 5 = 16 = 4 y = 4 + ( 33 ) y = = 49 y = 7 (iv) = 1 + = = 5 = 5 y = ( 5 ) + y = = 9 y = 3 Q. 3. (i) = = = 44 = 44 = = 15.6 to 3 significant figures (ii) = = = 30 = 30 = = 17.9 to 3 significant figures ctive Maths (Strands 1 5): h 18 Solutions 13

14 (iii) 8 3 y 11 y = y = y = 73 y = 73 + ( 73 ) = 11 = 11 ( 73 ) = = 48 = 48 = = 6.93 to 3 significant figures (iv) 5 (v) = 5 = = 1.75 = 1.75 = = 3.57 to 3 significant figures 9 (vi) 17 = = = 08 = = 14.4 (to three significant figures) a 1 1st triangle a = a = a = b a nd triangle. b = 1 + a 1 1 b = b = c 1 15 y b 8 y = y = 89 y = 89 y = 17 3rd triangle. c = 1 + b c = c = ctive Maths (Strands 1 5): h 18 Solutions

15 Following this pattern on the 7th triangle = = 8 = =.83 to 3 significant figures. Q. 4. (i) Triangle with sides 7, 70 and 1. 7 = 5, = 4, = 5,341 No: since this is not a right-angled triangle (ii) Triangle with sides 8.9, 8 and = = = 79.1 Yes: since = 8.9 this is a right-angled triangle (iii) Triangle with sides 16, 134 and = 6, = 17, ,404 = 8,360 No: since this is not a right-angled triangle (iv) Triangle with sides 113, 11 and = 1, = 1, = 1,769 Yes: since = 113 this is a right-angled triangle Q. 5. (i) Side lengths 60, 63, 87 Longest side squared must equal the sum of the squares of the two shorter sides. 87 = 7, = 3, ,969 = 7,569 Yes: these lengths will form a rightangled triangle since = 87 (ii) Side lengths 39, 80, = 7, = 1,51 + 6,400 = 7,91 Yes: these lengths will form a rightangled triangle since = 89 (iii) Side lengths 55, 130, = 1, = 3, ,900 = 19,95 No: these lengths will not form a right-angled triangle since (iv) Side lengths 64, 10, = 18, = 4, ,400 = 18,496 Yes: these side lengths will form a right-angled triangle since = 136 Q = 30 = =,500 =,500 = 50 The distance is ctive Maths (Strands 1 5): h 18 Solutions 15

16 (ii) = = 16,900 = Q. 7. (i) () + (3) = = 1, = 1,600 = 1, ,600 = 13 = = to d.p. (ii) + ( + 8) = =, ,704 = ,640 = 0 a = 5, b = 3, c =,640 b 4ac = (3) 4(5)(,640) = 53,84 Using the quadratic formula: = 3 ± 53,84 5 = ,84 10 OR = 3 53,84 10 = 0 OR = 6.4 Since > 0 reject = 6.4 = 0 (iii) ( 1) + ( ) = = = 0 ( 1)( 5) = 0 1 = 0 OR 5 = 0 = 1 OR = 5 if = 1 then one of the sides 1 = 1 1 = 0 Since we cannot have a side of zero length, = 1 is rejected. = 5 (iv) ( 1) + ( + 1) = ( + 5) = = = 0 a = 1, b = 10, c = 3 b 4ac = ( 10) 4(1)( 3) = = 19 Using the quadratic formula: ( 10) ± 19 = = OR = = OR = Since > 0, = is rejected = to d.p. Q. 8. (i) = distance of foot of ladder from the wall =.5 = =.505 =.505 = m = cm The foot of the ladder is 158 cm (to the nearest cm) from the wall. 16 ctive Maths (Strands 1 5): h 18 Solutions

17 (ii) 5 cm 90 cm h There are no whole number roots to this equation (p has a decimal value on using the quadratic formula) s p is not a whole number {p + 1, p + 6, p 3} cannot be positive integers and do not form a Pythagorean triple. h = new height of ladder against the wall h = 5 h = h = 14,55 h = 14,55 Slip = original height new height = 160 cm 14,55 cm = cm The ladder slipped 39 cm (to the nearest cm). Q. 9. (i) (p ) + (p + 1) = (p + 4) p 4p p + p + 1 = p + 8p + 16 p p + 5 p 8p 16 = 0 p 10p 11 = 0 (p + 1)(p 11) = 0 p + 1 = 0 OR p 11 = 0 p = 1 OR p = 11 if p = 1 then p + 1 = = 0 annot have side of zero length p = 1 is rejected p = 11 (ii) Show that {p + 1, p + 6, p 3} does not form a Pythagorean triple. (p + 1) + (p 3) = (p + 6) p + p p 6p + 9 = p + 1p + 36 p 4p + 10 p 1p 36 = 0 p 16p 6 = 0. Q. 10. Q. 11. Q cm l 39 cm Diameter of drum = 39 = 78 cm l = length of wire l = l = 1,484 l = 1,484 l = The longest wire is cm to 1 d.p. 40 cm = 1, = 1,441,600 = 1,441,600 = 1, cm The insects are 1,01 cm apart to the nearest cm. Diagonal 3 y 3 3 ctive Maths (Strands 1 5): h 18 Solutions 17

18 1st find the length of the diagonal of the base of the cube = = = 18 nd use Pythagoras Theorem again (on the blue triangle). y = 3 + ( 18 ) y = y = 7 y = cm The diagonal is 5. cm to 1 d.p. Q. 14. ube of side lengths 1 cm. = 10 mm Radius of cone = 10 = 60 mm Height of cone = 10 mm let l = slant height of cone h l Q. 13. (i) 6 m w r 4.8 m w = width of garden w = 6 w = w = 1.96 w = 1.96 w = 3.6 m Width of garden is 3.6 m (ii) Perimeter = l + w = = 16.8 m. Perimeter of garden is 16.8 m (iii) (3 + 5) m l (1 5) m l = length of drainage pipe l = (3 + 5 ) + (1 5 ) l = l = l = l = m l = cm The pipe is 68 cm to the nearest cm. l = h + r l = l = 18,000 l = 18,000 l = mm r = 60 mm h = 10 mm The slant height to the nearest mm is 134 mm Q cm 18 cm 6 cm 3 6 cm Let = height of truncated cone + 3 = 6 7 = 6 3 = 7 = h 9 18 ctive Maths (Strands 1 5): h 18 Solutions

19 Two similar triangles in the cone where h = height of original cone. h = h = h = cm h = mm Height of original cone is 156 mm to nearest mm. Q. 16. G 5 P 5 F 8 3 m H E 8 m 10 m D (i) E = E = 164 E = 164 E = m E = 1,80.6 cm E = 1.81 m (to nearest cm) (ii) F = F + 3 F = F = 173 F = 173 F = m F = 1, cm F = m (to nearest cm) (iii) P = P = 89 P = 89 P = m P = cm P = 9.43 m (to nearest cm) (iv) P = ( 89 ) + 3 P = P = 98 P = m P = cm P = 9.90 m (to nearest cm) Q. 17. Note diagonals of a square bisect each other at right angles. + = 0 l = 400 = 00 = cm (Pyramid height) l = length of wire l = 30 + ( 00 ) l = l = 1,100 l = 1,100 l = cm Length of wire to connect point to is 33 cm to nearest cm Q km 9 km (i) H = km H 4 km H = 405 H = 405 H = km 1 km 8 km Ship is 0 km from the harbour to significant figures. 0 ctive Maths (Strands 1 5): h 18 Solutions 19

20 (ii) H = H = 169 H = 169 H = 13 km Ship is 13 km from the harbour. (iii) 18 1 = = 14 = = 3 = 3 = km Ships and are 15 km apart to significant figures. (iv) = = 401 = 401 = km Ships and are 0 km apart to significant figures. (v) 5 4 = = = 13 = = 845 = 845 = km = 0 Ships and are 9 km apart to significant figures. Eercise 18.6 Q. 1. (i) = 80 angle at the center of a circle = 40 (ii) = (50 ) = 100 angle at the centre of a circle (iii) = 90 angle in a semi circle (iv) = 1 (110 ) = 55 angle at the centre of a circle (v) = 1 (66 ) = 33 angle at the centre of a circle (vi) = 1 (80 ) = 40 angle at the centre of a circle (vii) Smaller angle at 0 = = 100 = 1 (100 ) = 50 (viii) Larger angle at 0 = = 80 = 1 (80 ) = ctive Maths (Strands 1 5): h 18 Solutions

21 Q.. (i) = 34 angles standing on the same arc. = 56 angles standing on the same arc. (ii) = (110 ) = 0 angle at the centre of a circle = = 140 angles at a point (iii) = = 144 opposite angle in cyclic quadrilateral = = 73 (iv) = = 19 opposite angle in cyclic quadrilateral = (19 ) = 58 angle at the centre of a circle (v) = = 76 straight angle = = 104 opposite angle is a cyclic quadrilateral (vi) = 48 angles standing on the same arc = = 13 opposite angle is a cyclic quadrilateral Q. 3. (i) = (44 ) = 88 angle at the centre of a circle = 180 angle is a Δ, isosceles Δ = 9 = 46 (ii) angle at a point 360 O = angle at the centre of a circle (360 ) + 0 = 360 quadrilateral = 360 = 50 = 100 (iii) = 140 = 70 dding in the radius makes life easier (mark in the equal radius and you ll see the isosceles triangles) O = 70 = 34 (iv) = 90 3 = 58 angle is a semi circle, isosceles Δ = 180 (58 ) angles is a Δ = = 64 ctive Maths (Strands 1 5): h 18 Solutions 1

22 (v) same arc 44 O angle in a semi circle (vi) = 180 ( ) = 46 angles in a = 46 angles standing on the same arc. standing on the same arc O 30 = = 30 angle is a semicircle and angles standing on the same arc. = 30 standing on the same arc Q. 4. (i) = 41 as both are angles at the circle being subtended by the same arc. = (41 ) = 8 (ii) = (5 ) = 50 = = 5 as both are angles at the circumference subtended by the same arc. (iii) Reproduce diagram & label points P, Q, R, S, T & U as shown. QO = RO as both radii OQR = ORQ as base angles of a Δ. ORU = ORS = 90 as US is a tangent OQT = OQS as TS is a tangent. OQS = ORS = QRS ΔQRS is isosceles So = = 119 = 59.5 OQR = = 30.5 = 180 ()(30.5 ) = = 119 = 1 (119 ) = 59.5 P T O 90 Q U R 61 S ctive Maths (Strands 1 5): h 18 Solutions

23 (iv) = + = = = 90 = = = (90 ) = = = 180 = 58 = (58 ) = 116 = = 3 (v) s shown in (iv) = (90 ) = = = = 51 = = (51 ) = 10 O O 38 Q. 5. (i) Yes. s opposite angles add to 180 i.e = 180 and = 180 (ii) No. s opposite angles do not add to 180 i.e = 170 and = 190 (iii) No. s opposite angles don t add to 180 i.e = 170 (iv) Yes. s opposite angles add to 180 i.e = 180 Q. 6. (i) D = 38 (both angles at circle subtended by same arc) (ii) D = 90 (angle subtended by diameter) (iii) D = = 14 = 71 (iv) D = = 5 Q. 7. (i) O ; ΔOE ΔOE by RHS. OD = OD OD = = 59 O = (59 ) = 118 (ii) ΔO is isosceles as O = O O = = O = = 9.5 ctive Maths (Strands 1 5): h 18 Solutions 3

24 (iii) s in (ii) ΔO is isosceles O = O = = 9.5 (iv) D is a cyclic quadrilateral D + = 180 = (9.5 ) = 59 D = = 11 Q. 8. (i) QOS = (58 ) = 116. (ii) QPSR is a cyclic quadrilateral QRS = = 1 (iii) ΔOSQ is isosceles OSQ = = 3 ΔRSQ is isosceles RSQ = = 9 RSO = = 61 (iv) P, O, R are collinear PR is a diameter as O is centre PQR = 90 PQO = 90 RQO = = 9 Q. 9. parallelogram inscribed in a circle must have opposite angles which add to 180. Therefore every angle must be 90, as opposite angles in a parallelogram are equal. Therefore, only a rectangle or a square can be inscribed in a circle. Q. 10. Join the centre of the circle to each point of the star. This gives 5 equal angles which are 7 each. ngle is the angle at the circle being subtended by the same arc as the 7 angle at the centre and is therefore 1 (7 ) = 36. Q. 11. Row S T G E Seats Daniel could sit in Row, Seat 1 or Row, Seat 7 and have the same viewing angle. Reasoning: two angles subtended by the same arc (stage), touching the circumference, are equal in measure. nother possible answer: Row 5, seat 5. Revision Eercises Q. 1. (a) (i) 3 = 150 vertically opposite = 50 y = = 30 straight angle (ii) = = 50 straight angle y = = 70 straight angle 4 ctive Maths (Strands 1 5): h 18 Solutions

25 (iii) = = 80 straight angle = 40 4y = 100 corresponding angle y = 5 (iv) = 180 straight angle 5 = 90 = 18 y = = 85 straight angle (v) = 180 angle in a Δ, isosceles Δ = 65 = 3.5 y isosceles triangles 115º y 30º (3.5 + y) + (3.5 + y) + 30 = 180 angles in a Δ y + 95 = 180 y = 85 y = 4.5 (b) : Y Y + Y : Y X + X : X : 5 8 : 5 Q.. (a) (i) = 180 angles in a Δ y = + 3 eterior angle 1 = 180 = 15 y = 5 = 75 (ii) + + y = 180 angles in a 3 + y = 180 y + 0 = + y eterior angle y = y = = 00 3(40 ) + y = 180 = 40 y = 60 ctive Maths (Strands 1 5): h 18 Solutions 5

26 (iii) ( + y) + ( + ) + = 180 angles in a Δ 3 + y = y = + y + + eterior angle + 9y = ( 3) 3 + y = y = y = = 158 8y = 4 3 = 150 y = 8 = 50 (iv) 4y y + 6y 4 = 180 angles in a Δ + 11y = y 4 = 180 straight angle + 6y = y = y = (11 ) = 9 8y = 88 (nd 1st) + 33 = 9 y = 11 = 59 (b) (i) b = 8, a = 6 (ii) a = 7.5 b = 5 Q. 3. (a) (i) + 65 = + 30 alternate = y = 180 straight angle y = 180 y = 80 y = 40 (ii) + y = 9 y opposite angles 8 4y = 0 + y y = 180 parallelogram 8 + 5y = y = (0 ) = 180 9y = = 80 y = 0 = 10 (iii) 7 + 4y y = 180 cyclic quadrilateral y = y y = 180 cyclic quadrilateral y = y = (9) = y = = 180 0y = = 7 y = 9 = 7. 6 ctive Maths (Strands 1 5): h 18 Solutions

27 (b) (i) (y + ) + (4y) = (5y) right-angled Δ 4y + 8y y = 5y 5y 8y 4 = 0 (5y + )(y ) = 0 5y + = 0 or y = 0 y 5 y = Height of flagpole = 4y = 8 (ii) ( + 1) = () + ( 1) = = 0 0 = Height = 1 = 3 (iii) No, as 3 1 (8) Q. 4. (a) (i) = 50 isosceles Δ = 180 ( ) = 80 angles in a Δ = = 130 eterior angle (ii) = 4 corresponding = = 14 corresponding angle and straight angle = 14 alternate (iii) = = 118 straight angle ngles in isosceles Δ = 1 ( ) = 75 = = 105 = 360 ( ) quadrilateral + vertically opposite = 88 angle (iv) + 75 = 180 parallelogram = = 105 opposite = 40 = alternate = 40 (v) = = 66 straight line = 180 (66 ) = 48 angles in a Δ + vertically opposite = 90 straight angle and square ctive Maths (Strands 1 5): h 18 Solutions 7

28 (b) (i) 3 3 D 4 E Q. 5. (a) (i) 7 = 9 6 (ii) = DE corresponding = E corresponding is common Δ and ΔDE are equiangular. (iii) E 3 = 1 E = 3 = 1.5 (iv) = 3 4 = 1 = 6 (ii) (iii) = = 8 10 = 50 1 = 14 8 = (iv) 8 = 10 6 = 80 6 y 1 = = 10.5 y = 9 = 8 y = = 5 y = = 11 y 3 = = 1 y = 6 = 7 y = 10 6 y + 6 = 16 3 y + 6 = = y = 10 3 (b) h 15 = 5 5 h = 75 5 = 3 m 8 ctive Maths (Strands 1 5): h 18 Solutions

29 Q. 6. (a) (i) Is 5 = 8.4?.5 = 3.3? No DE (ii) Is 3 5 = ? 3 Is 0.6 = 0.6? Yes RQ TS (b) (i) Is = ? 5 1 = 5 1? Yes Triangles are similar. (ii) Is 11 = 10 15? 11 = 3? No Δs are not similar. (c) (i) Question 1 is incorrect as Δs are congruent and therefore angles marked 50 and 55 should be equal. (ii) Question : = 60 6 = 34. Q. 7. (a) (i) = 1 (180 4 ) radii + angles in a Δ = 1 (138 ) = 69 = = 1 angles in a semicircle and isosceles Δ (ii) = 180 (41 ) = 98 angles in a Δ and isosceles Δ. Top angle = 1 (98 ) = 49 angle at the centre 49º 41º 41º 0º + 0 = 49 = 9 (iii) = = 1 (110 ) isosceles Δ ctive Maths (Strands 1 5): h 18 Solutions 9

30 + 0 = 55 = = = 180 = 75 (iv) 1 51º O 15º 1 = = 19 straight angle = 1 ( ) = 5.5 isosceles Δ (b) Similarly = (15 ) = = 10 4 = 1.6 ( 10 4 ) = 4 m Q. 8. (a) (i) +7 = ( + 5) = y + 49 = (9) = y = = y = 4 y = 890 y = 890 (ii) + 3 = 7 y = ( 40 ) + 9 = 49 y = 40 = 40 y = = 40 (iii) + 5 = ( 41 ) () + 5 = y + 5 = = y = = y = 4 89 = y y = ctive Maths (Strands 1 5): h 18 Solutions

31 (iv) = + Label hypotenuse in middle = = 8 triangle as z: = 8 z = + z = z = 1 + z = y = y y = 16 y = 4 (b) h 650 m 1 Slope = rise run = 1 5 Ratio = h : 5h Using Pythagoras Theorem: (h) + (5h) = (650) h + 5h = 4,500 6h = 4,500 h = 16,50 h = h 17 m Q. 9. (a) (i) No, as no two strips are equal (ii) No, as opposite sides of a parallelogram are equal and no two strips are equal. (iii) (iv) In a right-angled triangle, Pythagoras Theorem holds. Is 5 = 4 + 7? 65 = ? 65 = 65? True 4 h 0 ctive Maths (Strands 1 5): h 18 Solutions 31

32 (b) Missing hypotenuse h = h = h = 976 h = 976 h = 31.4 cm D E 00 m 10 m 80 m (i) ΔE is similar to ΔD we could use the similar triangles theorem to calculate ED as = (ii) 80(10 + ) = 00(10) = 4, = 14,400 = 180 m Q. 10. (a) r = (b) = 35 8 = 441 = 1 Depth of oil = 35 1 = 14 cm (i) = cm (ii) (9) + (16) = = 10, = 10, ctive Maths (Strands 1 5): h 18 Solutions

33 = = Length = = = 89 to nearest cm Height = = = 50 to nearest cm (iii) (4) + (3) = = 10, = 10,3.56 = = 0.3 Length = = 81.8 = 81 to nearest cm Height = = = 61 to nearest cm (81 61) (89 50) = 491cm The area of the second TV screen (units aspect ratio 4 : 3) is 491 cm larger. Q. 11. (a) (i) OR (ii) (any equilateral triangle) any (rectangle) (iii) (any square) (b) - central symmetry - aial symmetry - translation D - aial symmetry (c) a = 180 a = 180 a = 47 a = 3.5 ctive Maths (Strands 1 5): h 18 Solutions 33

34 a + b + 73 = b + 73 = 180 b = 180 b = 83.5 l 1 l a and g are alternate g = a g = 3.5 (d) (i) = 11 + = = 15 = 15 = 5 5 (ii) D = D + () = (5 5 ) + 4 = 15 5 = 15 = 5 = 5 34 ctive Maths (Strands 1 5): h 18 Solutions

Name: GEOMETRY: EXAM (A) A B C D E F G H D E. 1. How many non collinear points determine a plane?

Name: GEOMETRY: EXAM (A) A B C D E F G H D E. 1. How many non collinear points determine a plane? GMTRY: XM () Name: 1. How many non collinear points determine a plane? ) none ) one ) two ) three 2. How many edges does a heagonal prism have? ) 6 ) 12 ) 18 ) 2. Name the intersection of planes Q and

More information

Exercise. and 13x. We know that, sum of angles of a quadrilateral = x = 360 x = (Common in both triangles) and AC = BD

Exercise. and 13x. We know that, sum of angles of a quadrilateral = x = 360 x = (Common in both triangles) and AC = BD 9 Exercise 9.1 Question 1. The angles of quadrilateral are in the ratio 3 : 5 : 9 : 13. Find all the angles of the quadrilateral. Solution Given, the ratio of the angles of quadrilateral are 3 : 5 : 9

More information

Time : 2 Hours (Pages 3) Max. Marks : 40. Q.1. Solve the following : (Any 5) 5 In PQR, m Q = 90º, m P = 30º, m R = 60º. If PR = 8 cm, find QR.

Time : 2 Hours (Pages 3) Max. Marks : 40. Q.1. Solve the following : (Any 5) 5 In PQR, m Q = 90º, m P = 30º, m R = 60º. If PR = 8 cm, find QR. Q.P. SET CODE Q.1. Solve the following : (ny 5) 5 (i) (ii) In PQR, m Q 90º, m P 0º, m R 60º. If PR 8 cm, find QR. O is the centre of the circle. If m C 80º, the find m (arc C) and m (arc C). Seat No. 01

More information

Plane geometry Circles: Problems with some Solutions

Plane geometry Circles: Problems with some Solutions The University of Western ustralia SHL F MTHMTIS & STTISTIS UW MY FR YUNG MTHMTIINS Plane geometry ircles: Problems with some Solutions 1. Prove that for any triangle, the perpendicular bisectors of the

More information

MT - MATHEMATICS (71) GEOMETRY - PRELIM II - PAPER - 6 (E)

MT - MATHEMATICS (71) GEOMETRY - PRELIM II - PAPER - 6 (E) 04 00 Seat No. MT - MTHEMTIS (7) GEOMETRY - PRELIM II - (E) Time : Hours (Pages 3) Max. Marks : 40 Note : ll questions are compulsory. Use of calculator is not allowed. Q.. Solve NY FIVE of the following

More information

Chapter 19 Exercise 19.1

Chapter 19 Exercise 19.1 hapter 9 xercise 9... (i) n axiom is a statement that is accepted but cannot be proven, e.g. x + 0 = x. (ii) statement that can be proven logically: for example, ythagoras Theorem. (iii) The logical steps

More information

MATHEMATICS. Unit 2. Part 2 of 2. Relationships

MATHEMATICS. Unit 2. Part 2 of 2. Relationships MTHEMTIS Unit Part of Relationships ngles Eercise 1 opy the following diagrams into your jotter and fill in the sizes of all the angles:- 1) 50 ) 50 60 3) 4) 5) 85 6) 7) 7 54 7 8) 56 9) 70 Maths Department

More information

Year 9 Term 3 Homework

Year 9 Term 3 Homework Yimin Math Centre Year 9 Term 3 Homework Student Name: Grade: Date: Score: Table of contents 5 Year 9 Term 3 Week 5 Homework 1 5.1 Geometry (Review)................................... 1 5.1.1 Angle sum

More information

MT - GEOMETRY - SEMI PRELIM - II : PAPER - 5

MT - GEOMETRY - SEMI PRELIM - II : PAPER - 5 017 1100 MT MT - GEOMETRY - SEMI PRELIM - II : PPER - 5 Time : Hours Model nswer Paper Max. Marks : 40.1. ttempt NY FIVE of the following : (i) X In XYZ, ray YM bisects XYZ XY YZ XM MZ Y Z [Property of

More information

QUESTION BANK ON STRAIGHT LINE AND CIRCLE

QUESTION BANK ON STRAIGHT LINE AND CIRCLE QUESTION BANK ON STRAIGHT LINE AND CIRCLE Select the correct alternative : (Only one is correct) Q. If the lines x + y + = 0 ; 4x + y + 4 = 0 and x + αy + β = 0, where α + β =, are concurrent then α =,

More information

Geometry Final Review. Chapter 1. Name: Per: Vocab. Example Problems

Geometry Final Review. Chapter 1. Name: Per: Vocab. Example Problems Geometry Final Review Name: Per: Vocab Word Acute angle Adjacent angles Angle bisector Collinear Line Linear pair Midpoint Obtuse angle Plane Pythagorean theorem Ray Right angle Supplementary angles Complementary

More information

Math 9 Unit 8: Circle Geometry Pre-Exam Practice

Math 9 Unit 8: Circle Geometry Pre-Exam Practice Math 9 Unit 8: Circle Geometry Pre-Exam Practice Name: 1. A Ruppell s Griffon Vulture holds the record for the bird with the highest documented flight altitude. It was spotted at a height of about 11 km

More information

Review for Grade 9 Math Exam - Unit 8 - Circle Geometry

Review for Grade 9 Math Exam - Unit 8 - Circle Geometry Name: Review for Grade 9 Math Exam - Unit 8 - ircle Geometry Date: Multiple hoice Identify the choice that best completes the statement or answers the question. 1. is the centre of this circle and point

More information

MATHEMATICS. S2 Level 3/4 Course -1- Larkhall Maths Department Academy

MATHEMATICS. S2 Level 3/4 Course -1- Larkhall Maths Department Academy MTHEMTIS S2 Level 3/4 ourse -1- Larkhall Maths Department cademy 17 cm The ircle Eercise 1() Find the circumference ( 1) 2) ) of the following circles 3) 4) 1 12 cm 5 cm 28 m 5) 6) 7) 3 2 cm 8) 15 m 22

More information

Section A Pythagoras Theorem Grade C

Section A Pythagoras Theorem Grade C Name: Teacher ssessment Section Pythagoras Theorem Grade 1. support for a flagpole is attached at a height of 3 m and is fixed to the ground at a distance of 1.2 m from the base. Not to scale x 3 m 1.2

More information

Solve problems involving tangents to a circle. Solve problems involving chords of a circle

Solve problems involving tangents to a circle. Solve problems involving chords of a circle 8UNIT ircle Geometry What You ll Learn How to Solve problems involving tangents to a circle Solve problems involving chords of a circle Solve problems involving the measures of angles in a circle Why Is

More information

Triangles. Exercise 4.1

Triangles. Exercise 4.1 4 Question. xercise 4. Fill in the blanks using the correct word given in brackets. (i) ll circles are....(congruent, similar) (ii) ll squares are....(similar, congruent) (iii) ll... triangles are similar.

More information

Practice Test Student Answer Document

Practice Test Student Answer Document Practice Test Student Answer Document Record your answers by coloring in the appropriate bubble for the best answer to each question. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26

More information

MT - GEOMETRY - SEMI PRELIM - II : PAPER - 4

MT - GEOMETRY - SEMI PRELIM - II : PAPER - 4 017 1100 MT.1. ttempt NY FIVE of the following : (i) In STR, line l side TR S SQ T = RQ x 4.5 = 1.3 3.9 x = MT - GEOMETRY - SEMI RELIM - II : ER - 4 Time : Hours Model nswer aper Max. Marks : 40 4.5 1.3

More information

GM1.1 Answers. Reasons given for answers are examples only. In most cases there are valid alternatives. 1 a x = 45 ; alternate angles are equal.

GM1.1 Answers. Reasons given for answers are examples only. In most cases there are valid alternatives. 1 a x = 45 ; alternate angles are equal. Cambridge Essentials Mathematics Extension 8 GM1.1 Answers GM1.1 Answers Reasons given for answers are examples only. In most cases there are valid alternatives. 1 a x = 45 ; alternate angles are equal.

More information

2. A diagonal of a parallelogram divides it into two congruent triangles. 5. Diagonals of a rectangle bisect each other and are equal and vice-versa.

2. A diagonal of a parallelogram divides it into two congruent triangles. 5. Diagonals of a rectangle bisect each other and are equal and vice-versa. QURILTERLS 1. Sum of the angles of a quadrilateral is 360. 2. diagonal of a parallelogram divides it into two congruent triangles. 3. In a parallelogram, (i) opposite sides are equal (ii) opposite angles

More information

PRACTICE TEST 1 Math Level IC

PRACTICE TEST 1 Math Level IC SOLID VOLUME OTHER REFERENCE DATA Right circular cone L = cl V = volume L = lateral area r = radius c = circumference of base h = height l = slant height Sphere S = 4 r 2 V = volume r = radius S = surface

More information

A. 180 B. 108 C. 360 D. 540

A. 180 B. 108 C. 360 D. 540 Part I - Multiple Choice - Circle your answer: 1. Find the area of the shaded sector. Q O 8 P A. 2 π B. 4 π C. 8 π D. 16 π 2. An octagon has sides. A. five B. six C. eight D. ten 3. The sum of the interior

More information

Name Score Period Date. m = 2. Find the geometric mean of the two numbers. Copy and complete the statement.

Name Score Period Date. m = 2. Find the geometric mean of the two numbers. Copy and complete the statement. Chapter 6 Review Geometry Name Score Period Date Solve the proportion. 3 5 1. = m 1 3m 4 m = 2. 12 n = n 3 n = Find the geometric mean of the two numbers. Copy and complete the statement. 7 x 7? 3. 12

More information

Unit 2 Review. Determine the scale factor of the dilation below.

Unit 2 Review. Determine the scale factor of the dilation below. Unit 2 Review 1. oes the graph below represent a dilation? Why or why not? y 10 9 8 7 (0, 7) 6 5 4 3 (0, 3.5) 2 1 (5, 7) (5, 3.5) -10-9 -8-7 -6-5 -4-3 -2-1 0 1 2 3 4 5 6 7 8 9 10-1 F -2 (5, 0) -3-4 -5-6

More information

Reteaching , or 37.5% 360. Geometric Probability. Name Date Class

Reteaching , or 37.5% 360. Geometric Probability. Name Date Class Name ate lass Reteaching Geometric Probability INV 6 You have calculated probabilities of events that occur when coins are tossed and number cubes are rolled. Now you will learn about geometric probability.

More information

The Pythagorean Theorem

The Pythagorean Theorem The Pythagorean Theorem Geometry y now, you know the Pythagorean Theorem and how to use it for basic problems. The onverse of the Pythagorean Theorem states that: If the lengths of the sides of a triangle

More information

Trigonometric ratios:

Trigonometric ratios: 0 Trigonometric ratios: The six trigonometric ratios of A are: Sine Cosine Tangent sin A = opposite leg hypotenuse adjacent leg cos A = hypotenuse tan A = opposite adjacent leg leg and their inverses:

More information

A. leg B. hipponamoose C. hypotenuse D. Big Guy. A. congruent B. complementary C. supplementary D. cute little things

A. leg B. hipponamoose C. hypotenuse D. Big Guy. A. congruent B. complementary C. supplementary D. cute little things 3 rd quarter Review Name: Date: 1.] The longest side of a right triangle is called the.. leg. hipponamoose. hypotenuse D. ig Guy 2.] The acute angles of a right triangle are always.. congruent. complementary.

More information

Answers. Chapter10 A Start Thinking. and 4 2. Sample answer: no; It does not pass through the center.

Answers. Chapter10 A Start Thinking. and 4 2. Sample answer: no; It does not pass through the center. hapter10 10.1 Start Thinking 6. no; is not a right triangle because the side lengths do not satisf the Pthagorean Theorem (Thm. 9.1). 1. (3, ) 7. es; is a right triangle because the side lengths satisf

More information

1. Town A is 48 km from town B and 32 km from town C as shown in the diagram. A 48km

1. Town A is 48 km from town B and 32 km from town C as shown in the diagram. A 48km 1. Town is 48 km from town and 32 km from town as shown in the diagram. 32km 48km Given that town is 56 km from town, find the size of angle (Total 4 marks) Â to the nearest degree. 2. The diagram shows

More information

Common Core Readiness Assessment 4

Common Core Readiness Assessment 4 ommon ore Readiness ssessment 4 1. Use the diagram and the information given to complete the missing element of the two-column proof. 2. Use the diagram and the information given to complete the missing

More information

Chapter (Circle) * Circle - circle is locus of such points which are at equidistant from a fixed point in

Chapter (Circle) * Circle - circle is locus of such points which are at equidistant from a fixed point in Chapter - 10 (Circle) Key Concept * Circle - circle is locus of such points which are at equidistant from a fixed point in a plane. * Concentric circle - Circle having same centre called concentric circle.

More information

"Full Coverage": Pythagoras Theorem

Full Coverage: Pythagoras Theorem "Full Coverage": Pythagoras Theorem This worksheet is designed to cover one question of each type seen in past papers, for each GCSE Higher Tier topic. This worksheet was automatically generated by the

More information

Preliminary chapter: Review of previous coursework. Objectives

Preliminary chapter: Review of previous coursework. Objectives Preliminary chapter: Review of previous coursework Objectives By the end of this chapter the student should be able to recall, from Books 1 and 2 of New General Mathematics, the facts and methods that

More information

Prove that a + b = x + y. Join BD. In ABD, we have AOB = 180º AOB = 180º ( 1 + 2) AOB = 180º A

Prove that a + b = x + y. Join BD. In ABD, we have AOB = 180º AOB = 180º ( 1 + 2) AOB = 180º A bhilasha lasses lass- IX ate: 03- -7 SLUTIN (hap 8,9,0) 50 ob no.-947967444. The sides and of a quadrilateral are produced as shown in fig. rove that a + b = x + y. Join. In, we have y a + + = 80º = 80º

More information

Circles-Tangent Properties

Circles-Tangent Properties 15 ircles-tangent roperties onstruction of tangent at a point on the circle. onstruction of tangents when the angle between radii is given. Tangents from an external point - construction and proof Touching

More information

Geometry Simulation Test 2014 Region 1

Geometry Simulation Test 2014 Region 1 Geometry Simulation Test 2014 Region 1 1 etermine the inverse of Mark Twain s quote If you tell the truth, you don t have to remember anything. You don t have to remember anything if you tell the truth.

More information

Theorem 1.2 (Converse of Pythagoras theorem). If the lengths of the sides of ABC satisfy a 2 + b 2 = c 2, then the triangle has a right angle at C.

Theorem 1.2 (Converse of Pythagoras theorem). If the lengths of the sides of ABC satisfy a 2 + b 2 = c 2, then the triangle has a right angle at C. hapter 1 Some asic Theorems 1.1 The ythagorean Theorem Theorem 1.1 (ythagoras). The lengths a b < c of the sides of a right triangle satisfy the relation a + b = c. roof. b a a 3 b b 4 b a b 4 1 a a 3

More information

Geometry Honors Final Exam Review June 2018

Geometry Honors Final Exam Review June 2018 Geometry Honors Final Exam Review June 2018 1. Determine whether 128 feet, 136 feet, and 245 feet can be the lengths of the sides of a triangle. 2. Casey has a 13-inch television and a 52-inch television

More information

Int. Geometry Units 1-6 Review 1

Int. Geometry Units 1-6 Review 1 Int. Geometry Units 1-6 Review 1 Things to note about this review and the Unit 1-6 Test: 1. This review packet covers major ideas of the first six units, but it does not show examples of all types of problems..

More information

Chapter 1. Some Basic Theorems. 1.1 The Pythagorean Theorem

Chapter 1. Some Basic Theorems. 1.1 The Pythagorean Theorem hapter 1 Some asic Theorems 1.1 The ythagorean Theorem Theorem 1.1 (ythagoras). The lengths a b < c of the sides of a right triangle satisfy the relation a 2 + b 2 = c 2. roof. b a a 3 2 b 2 b 4 b a b

More information

MT EDUCARE LTD. MATHEMATICS SUBJECT : Q L M ICSE X. Geometry STEP UP ANSWERSHEET

MT EDUCARE LTD. MATHEMATICS SUBJECT : Q L M ICSE X. Geometry STEP UP ANSWERSHEET IS X MT UR LT. SUJT : MTHMTIS Geometry ST U NSWRSHT 003 1. In QL and RM, LQ MR [Given] LQ RM [Given] QL ~ RM [y axiom of similarity] (i) Since, QL ~ RM QL M L RM QL RM L M (ii) In QL and RQ, we have Q

More information

2. In the diagram, PQ and TS are parallel. Prove that a + b + c = 360. We describe just two of the many different methods that are possible. Method 1

2. In the diagram, PQ and TS are parallel. Prove that a + b + c = 360. We describe just two of the many different methods that are possible. Method 1 s to the Olympiad Cayley Paper 1. The digits p, q, r, s and t are all different. What is the smallest five-digit integer pqrst that is divisible by 1, 2, 3, 4 and 5? Note that all five-digit integers are

More information

AREAS RELATED TO CIRCLES

AREAS RELATED TO CIRCLES HPTER 1 Points to Remember : RES RELTE T IRLES 1. circle is a collection of points which moves in a plane in such a way that its distance from a fixed point always remains the same. The fixed point is

More information

Time : 2 Hours Preliminary Model Answer Paper Max. Marks : 40. [Given] [Taking square roots]

Time : 2 Hours Preliminary Model Answer Paper Max. Marks : 40. [Given] [Taking square roots] .P. SET CODE MT - w 05 00 - MT - w - MTHEMTICS (7) GEOMETRY - (E) Time : Hours Preliminary Model nswer Paper Max. Marks : 40.. ttempt NY FIVE of the following : (i) BC ~ PQ [Given] ( BC) ( PQ) BC PQ [reas

More information

VAISHALI EDUCATION POINT (QUALITY EDUCATION PROVIDER)

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

More information

CHAPTER 1 BASIC ARITHMETIC

CHAPTER 1 BASIC ARITHMETIC CHAPTER 1 BASIC ARITHMETIC EXERCISE 1, Page 4 1. Evaluate 67 kg 8 kg + 4 kg without using a calculator 67 kg 8 kg + 4 kg = 67 kg + 4 kg 8 kg = 101 kg 8 kg = 19 kg. Evaluate 851 mm 7 mm without using a

More information

Intermediate Math Circles Wednesday October Problem Set 3

Intermediate Math Circles Wednesday October Problem Set 3 The CETRE for EDUCTI in MTHEMTICS and CMPUTIG Intermediate Math Circles Wednesday ctober 24 2012 Problem Set 3.. Unless otherwise stated, any point labelled is assumed to represent the centre of the circle.

More information

Question 1 ( 1.0 marks) places of decimals? Solution: Now, on dividing by 2, we obtain =

Question 1 ( 1.0 marks) places of decimals? Solution: Now, on dividing by 2, we obtain = Question 1 ( 1.0 marks) The decimal expansion of the rational number places of decimals? will terminate after how many The given expression i.e., can be rewritten as Now, on dividing 0.043 by 2, we obtain

More information

CHAPTER 11 AREAS OF PLANE FIGURES

CHAPTER 11 AREAS OF PLANE FIGURES CHAPTER 11 AREAS OF PLANE FIGURES EXERCISE 45, Page 106 1. Find the angles p and q in diagram (a) below. p = 180-75 = 105 (interior opposite angles of a parallelogram are equal) q = 180-105 - 40 = 35.

More information

CHAPTER 10 TRIGONOMETRY

CHAPTER 10 TRIGONOMETRY CHAPTER 10 TRIGONOMETRY EXERCISE 39, Page 87 1. Find the length of side x in the diagram below. By Pythagoras, from which, 2 25 x 7 2 x 25 7 and x = 25 7 = 24 m 2. Find the length of side x in the diagram

More information

Prerequisite Skills. y x =

Prerequisite Skills. y x = Prerequisite Skills BLM 1 1... Solve Equations 1. Solve. 2x + 5 = 11 x 5 + 6 = 7 x 2 = 225 d) x 2 = 24 2 + 32 2 e) 60 2 + x 2 = 61 2 f) 13 2 12 2 = x 2 The Pythagorean Theorem 2. Find the measure of the

More information

GEOMETRY. Similar Triangles

GEOMETRY. Similar Triangles GOMTRY Similar Triangles SIMILR TRINGLS N THIR PROPRTIS efinition Two triangles are said to be similar if: (i) Their corresponding angles are equal, and (ii) Their corresponding sides are proportional.

More information

1 = 1, b d and c d. Chapter 7. Worked-Out Solutions Chapter 7 Maintaining Mathematical Proficiency (p. 357) Slope of line b:

1 = 1, b d and c d. Chapter 7. Worked-Out Solutions Chapter 7 Maintaining Mathematical Proficiency (p. 357) Slope of line b: hapter 7 aintaining athematical Proficienc (p. 357) 1. (7 x) = 16 (7 x) = 16 7 x = 7 = 7 x = 3 x 1 = 3 1 x = 3. 7(1 x) + = 19 = 7(1 x) = 1 7(1 x) 7 = 1 7 1 x = 3 1 = 1 x = x 1 = 1 x = 3. 3(x 5) + 8(x 5)

More information

Chapter 6. Worked-Out Solutions. Chapter 6 Maintaining Mathematical Proficiency (p. 299)

Chapter 6. Worked-Out Solutions. Chapter 6 Maintaining Mathematical Proficiency (p. 299) hapter 6 hapter 6 Maintaining Mathematical Proficiency (p. 99) 1. Slope perpendicular to y = 1 x 5 is. y = x + b 1 = + b 1 = 9 + b 10 = b n equation of the line is y = x + 10.. Slope perpendicular to y

More information

So, PQ is about 3.32 units long Arcs and Chords. ALGEBRA Find the value of x.

So, PQ is about 3.32 units long Arcs and Chords. ALGEBRA Find the value of x. ALGEBRA Find the value of x. 1. Arc ST is a minor arc, so m(arc ST) is equal to the measure of its related central angle or 93. and are congruent chords, so the corresponding arcs RS and ST are congruent.

More information

(D) (A) Q.3 To which of the following circles, the line y x + 3 = 0 is normal at the point ? 2 (A) 2

(D) (A) Q.3 To which of the following circles, the line y x + 3 = 0 is normal at the point ? 2 (A) 2 CIRCLE [STRAIGHT OBJECTIVE TYPE] Q. The line x y + = 0 is tangent to the circle at the point (, 5) and the centre of the circles lies on x y = 4. The radius of the circle is (A) 3 5 (B) 5 3 (C) 5 (D) 5

More information

Similar Triangles, Pythagorean Theorem, and Congruent Triangles.

Similar Triangles, Pythagorean Theorem, and Congruent Triangles. ay 20 Teacher Page Similar Triangles, Pythagorean Theorem, and ongruent Triangles. Pythagorean Theorem Example 1: circle has a radius of 20 units. triangle is formed by connecting a point on the perimeter

More information

0809ge. Geometry Regents Exam Based on the diagram below, which statement is true?

0809ge. Geometry Regents Exam Based on the diagram below, which statement is true? 0809ge 1 Based on the diagram below, which statement is true? 3 In the diagram of ABC below, AB AC. The measure of B is 40. 1) a b ) a c 3) b c 4) d e What is the measure of A? 1) 40 ) 50 3) 70 4) 100

More information

8-6. a: 110 b: 70 c: 48 d: a: no b: yes c: no d: yes e: no f: yes g: yes h: no

8-6. a: 110 b: 70 c: 48 d: a: no b: yes c: no d: yes e: no f: yes g: yes h: no Lesson 8.1.1 8-6. a: 110 b: 70 c: 48 d: 108 8-7. a: no b: yes c: no d: yes e: no f: yes g: yes h: no 8-8. b: The measure of an exterior angle of a triangle equals the sum of the measures of its remote

More information

THOUGHTS AND CROSSES. TOPIC: volume and surface area

THOUGHTS AND CROSSES. TOPIC: volume and surface area THOUGHTS ND CROSSES TOPIC: volume and surface area * * block of wood, 9cm by 11cm by 12cm, has a hole of radius 2.5cm drilled out. Calculate the mass of the wood if the density is pepper pot consists of

More information

6 CHAPTER. Triangles. A plane figure bounded by three line segments is called a triangle.

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

More information

Pre-Algebra Chapter 9 Spatial Thinking

Pre-Algebra Chapter 9 Spatial Thinking Pre-Algebra Chapter 9 Spatial Thinking SOME NUMBERED QUESTIONS HAVE BEEN DELETED OR REMOVED. YOU WILL NOT BE USING A CALCULATOR FOR PART I MULTIPLE-CHOICE QUESTIONS, AND THEREFORE YOU SHOULD NOT USE ONE

More information

OBJECTIVE TEST. Answer all questions C. N3, D. N3, Simplify Express the square root of in 4

OBJECTIVE TEST. Answer all questions C. N3, D. N3, Simplify Express the square root of in 4 . In a particular year, the exchange rate of Naira (N) varies directly with the Dollar ($). If N is equivalent to $8, find the Naira equivalent of $6. A. N8976 B. N049 C. N40. D. N.7. If log = x, log =

More information

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 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

More information

Geometry Final Exam 2014 Study Guide. Name Date Block

Geometry Final Exam 2014 Study Guide. Name Date Block Geometry Final Exam 014 Study Guide Name Date Block The final exam for Geometry will take place on June 5. The following study guide will help you prepare for the exam. Everything we have covered is fair

More information

Chapter 3 Cumulative Review Answers

Chapter 3 Cumulative Review Answers Chapter 3 Cumulative Review Answers 1a. The triangle inequality is violated. 1b. The sum of the angles is not 180º. 1c. Two angles are equal, but the sides opposite those angles are not equal. 1d. The

More information

Cumulative Test 1. Name Date. In Exercises 1 5, use the diagram at the right. Answers

Cumulative Test 1. Name Date. In Exercises 1 5, use the diagram at the right. Answers Name Date umulative Test In Eercises 5, use the diagram at the right.. Name the intersection of ED @##$ and @##$ D. E. 2. Name the intersection of plane D and plane E. 3. re points,, and D collinear? 2.

More information

TOPIC-1 Rational Numbers

TOPIC-1 Rational Numbers TOPI- Rational Numbers Unit -I : Number System hapter - : Real Numbers Rational Number : number r is called a rational number, if it can be written in the form p/q, where p and q are integers and q 0,

More information

It is known that the length of the tangents drawn from an external point to a circle is equal.

It is known that the length of the tangents drawn from an external point to a circle is equal. CBSE -MATHS-SET 1-2014 Q1. The first three terms of an AP are 3y-1, 3y+5 and 5y+1, respectively. We need to find the value of y. We know that if a, b and c are in AP, then: b a = c b 2b = a + c 2 (3y+5)

More information

MT - w A.P. SET CODE MT - w - MATHEMATICS (71) GEOMETRY- SET - A (E) Time : 2 Hours Preliminary Model Answer Paper Max.

MT - w A.P. SET CODE MT - w - MATHEMATICS (71) GEOMETRY- SET - A (E) Time : 2 Hours Preliminary Model Answer Paper Max. .P. SET CODE.. Solve NY FIVE of the following : (i) ( BE) ( BD) ( BE) ( BD) BE D 6 9 MT - w 07 00 - MT - w - MTHEMTICS (7) GEOMETRY- (E) Time : Hours Preliminary Model nswer Paper Max. Marks : 40 [Triangles

More information

Skills Practice Skills Practice for Lesson 11.1

Skills Practice Skills Practice for Lesson 11.1 Skills Practice Skills Practice for Lesson.1 Name ate Riding a Ferris Wheel Introduction to ircles Vocabulary Identify an instance of each term in the diagram. 1. circle X T 2. center of the circle H I

More information

[Class-X] MATHEMATICS SESSION:

[Class-X] MATHEMATICS SESSION: [Class-X] MTHEMTICS SESSION:017-18 Time allowed: 3 hrs. Maximum Marks : 80 General Instructions : (i) ll questions are compulsory. (ii) This question paper consists of 30 questions divided into four sections,

More information

JEFFERSON MATH PROJECT REGENTS AT RANDOM

JEFFERSON MATH PROJECT REGENTS AT RANDOM JEFFERSON MATH PROJECT REGENTS AT RANDOM The NY Geometry Regents Exams Fall 2008-August 2009 Dear Sir I have to acknolege the reciept of your favor of May 14. in which you mention that you have finished

More information

Circle Geometry. This booklet belongs to:

Circle Geometry. This booklet belongs to: Circle Geometry This booklet belongs to: LESSON # DATE QUESTIONS FROM NOTES 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. Pg. Pg. Pg. Pg. Pg. Pg. Pg. Pg. Pg. Pg. REVIEW TEST Questions that I find difficult Find

More information

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.

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

More information

4.! ABC ~ DEF,! AC = 6 ft, CB = 3 ft, AB = 7 ft, DF = 9 ft.! What is the measure of EF?

4.! ABC ~ DEF,! AC = 6 ft, CB = 3 ft, AB = 7 ft, DF = 9 ft.! What is the measure of EF? Name:!!!!!!!!!!!!! Geo(2) GEOMETRY (2) REVIEW FOR FINAL EXAM #2 1. If ABC is similar to ADE, then AB AD =? AE. Which replaces the? to make the statement true? A. AC!! B. AE!! C. DE!! D. BC 2. In ABC,

More information

Indicate whether the statement is true or false.

Indicate whether the statement is true or false. PRACTICE EXAM IV Sections 6.1, 6.2, 8.1 8.4 Indicate whether the statement is true or false. 1. For a circle, the constant ratio of the circumference C to length of diameter d is represented by the number.

More information

0610ge. Geometry Regents Exam The diagram below shows a right pentagonal prism.

0610ge. Geometry Regents Exam The diagram below shows a right pentagonal prism. 0610ge 1 In the diagram below of circle O, chord AB chord CD, and chord CD chord EF. 3 The diagram below shows a right pentagonal prism. Which statement must be true? 1) CE DF 2) AC DF 3) AC CE 4) EF CD

More information

CONGRUENCE AND SIMILARITY

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

More information

Euclidian Geometry Grade 10 to 12 (CAPS)

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

More information

Chapter-wise questions

Chapter-wise questions hapter-wise questions ircles 1. In the given figure, is circumscribing a circle. ind the length of. 3 15cm 5 2. In the given figure, is the center and. ind the radius of the circle if = 18 cm and = 3cm

More information

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

More information

BOARD ANSWER PAPER :OCTOBER 2014

BOARD ANSWER PAPER :OCTOBER 2014 BRD NSWER PPER :CTBER 04 GEETRY. Solve any five sub-questions: BE i. BE ( BD) D BE 6 ( BD) 9 ΔBE (ΔBD) ----[Ratio of areas of two triangles having equal base is equal to the ratio of their corresponding

More information

2012 GCSE Maths Tutor All Rights Reserved

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

More information

b) Parallelogram Opposite Sides Converse c) Parallelogram Diagonals Converse d) Opposite sides Parallel and Congruent Theorem

b) Parallelogram Opposite Sides Converse c) Parallelogram Diagonals Converse d) Opposite sides Parallel and Congruent Theorem Chapter 7 1. State which theorem you can use to show that the quadrilateral is a parallelogram. a) Parallelogram Opposite Angles Converse b) Parallelogram Opposite Sides Converse c) Parallelogram Diagonals

More information

1. Draw and label a diagram to illustrate the property of a tangent to a circle.

1. Draw and label a diagram to illustrate the property of a tangent to a circle. Master 8.17 Extra Practice 1 Lesson 8.1 Properties of Tangents to a Circle 1. Draw and label a diagram to illustrate the property of a tangent to a circle. 2. Point O is the centre of the circle. Points

More information

Honors Geometry Mid-Term Exam Review

Honors Geometry Mid-Term Exam Review Class: Date: Honors Geometry Mid-Term Exam Review Multiple Choice Identify the letter of the choice that best completes the statement or answers the question. 1. Classify the triangle by its sides. The

More information

1. Use. What are the vertices of A.,, B.,, C.,, D.,,

1. Use. What are the vertices of A.,, B.,, C.,, D.,, 1. Use. What are the vertices of A.,, B.,, C.,, D.,, 2. Given, how are the distances to the origin from each image point related to the distance to the origin from each corresponding preimage point? A.

More information

G.C.E.(O.L.) Support Seminar

G.C.E.(O.L.) Support Seminar - 1 - G..E.(O.L.) Support Seminar - 015 Mathematics I Two Hours Part nswer all questions on this question paper itself. 1. If 50 rupees is paid as rates for a quarter for a certain house, find the value

More information

COMMON UNITS OF PERIMITER ARE METRE

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

More information

Chapter 6. Worked-Out Solutions AB 3.61 AC 5.10 BC = 5

Chapter 6. Worked-Out Solutions AB 3.61 AC 5.10 BC = 5 27. onstruct a line ( DF ) with midpoint P parallel to and twice the length of QR. onstruct a line ( EF ) with midpoint R parallel to and twice the length of QP. onstruct a line ( DE ) with midpoint Q

More information

H. Math 2 Benchmark 1 Review

H. Math 2 Benchmark 1 Review H. Math 2 enchmark 1 Review Name: ate: 1. Parallelogram C was translated to parallelogram C. 2. Which of the following is a model of a scalene triangle?.. How many units and in which direction were the

More information

MAHESH TUTORIALS. Time : 1 hr. 15 min. Q.1. Solve the following : 3

MAHESH TUTORIALS. Time : 1 hr. 15 min. Q.1. Solve the following : 3 S.S.. MHESH TUTRILS Test - II atch : S Marks : 30 Date : GEMETRY hapter : 1,, 3 Time : 1 hr. 15 min..1. Solve the following : 3 The areas of two similar triangles are 18 cm and 3 cm respectively. What

More information

MATHEMATICS. IMPORTANT FORMULAE AND CONCEPTS for. Summative Assessment -II. Revision CLASS X Prepared by

MATHEMATICS. IMPORTANT FORMULAE AND CONCEPTS for. Summative Assessment -II. Revision CLASS X Prepared by MATHEMATICS IMPORTANT FORMULAE AND CONCEPTS for Summative Assessment -II Revision CLASS X 06 7 Prepared by M. S. KUMARSWAMY, TGT(MATHS) M. Sc. Gold Medallist (Elect.), B. Ed. Kendriya Vidyalaya GaCHiBOWli

More information

Final Exam Review. Multiple Choice Identify the choice that best completes the statement or answers the question.

Final Exam Review. Multiple Choice Identify the choice that best completes the statement or answers the question. ( Final Exam Review Multiple hoice Identify the choice that best completes the statement or answers the question. 1. is an isosceles triangle. is the longest side with length. = and =. Find. 4 x + 4 7

More information

Math 9 Chapter 8 Practice Test

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

More information

SSC EXAMINATION GEOMETRY (SET-A)

SSC EXAMINATION GEOMETRY (SET-A) GRND TEST SS EXMINTION GEOMETRY (SET-) SOLUTION Q. Solve any five sub-questions: [5M] ns. ns. 60 & D have equal height ( ) ( D) D D ( ) ( D) Slope of the line ns. 60 cos D [/M] [/M] tan tan 60 cos cos

More information

Name: Class: Date: 5. If the diagonals of a rhombus have lengths 6 and 8, then the perimeter of the rhombus is 28. a. True b.

Name: Class: Date: 5. If the diagonals of a rhombus have lengths 6 and 8, then the perimeter of the rhombus is 28. a. True b. Indicate whether the statement is true or false. 1. If the diagonals of a quadrilateral are perpendicular, the quadrilateral must be a square. 2. If M and N are midpoints of sides and of, then. 3. The

More information