Chapter 18 Exercise 18.1
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- Arnold Henry
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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
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