hapter 18 Eercise 18.1 Q. 1. (i) 180 37 = 143 ( = 143 ) (ii) 180 117 = 63 ( = 63 ) 180 90 = 90 (y = 90 ) (iii) + + 3 + 45 = 180 4.5 = 135 (iv) 180 90 = y 90 = y = 30 45 = y 66 + ( + y) + 47 = 180 + y = 67 = 67 y = 67 45 = nswer: + y = 67 y = 90 Q.. (i) = 53 Opposite angles = 180 53 = 17 Interior angles = 53 lternate angles (ii) = 180 11 = 68 (Interior angles) = 180 134 = 46 (Straight line) = 180 46 = 134 (Interior angles) nswer: = 68 = 134 = 46 (iii) = 73 orresponding angles = 41 Opposite angles = 180 73 41 = 66 Straight line D = 66 orresponding angles E = 180 66 E = 114 Straight line Q. 3. (iv) = 10 orresponding E = 63 Opposite D = 10 Opposite = 180 E (180 ) = 180 63 (78) = 39 = 180 39 = 141 Straight line (v) = 115 (Opposite) = 180 115 = 65 (Interior) = 180 75 = 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) = 180 43 = 137 = 180 43 83 = 54 (ii) = 180 63 = 117 = 117 = 58.5 = 180 51 58.5 = 70.5 = 180 71 70.5 = 38.5 ctive Maths (Strands 1 5): h 18 Solutions 1
(iii) = 180 50 = 130 = 130 = 65 = 180 46 65 = 69 = = 69 = 69 = 69 = 65 (iv) = 180 5 81 = 47 = 5 (orresponding) = 180 47 5 = 81 Q. 5. (i) = 3 (lternate angles) = 58 (Opposite angles/rules of parallelogram) : ( + 3) + (58) = 360 + 46 = 360 116 = 44 46 = 198 = 99 OR ( + 3) + 58 = 180 + 3 = 1 = 1 3 = 99 (ii) = 84 (orresponding) = = 84 (Opposite) ( + 14) + = 180 + 14 + 84 = 180 = 180 98 = 8 (iii) = 31 lternate 31 y 31 = 31 (isosceles triangle) by alternate angles = 31 Similarly y = c (by isosceles) 180 = 31 + + + y 180 = 6 + 118 = 59 = = 180 31 = 180 59 31 = 90 (iv) 10 y = 180 10 = 78 by isosceles triangles = 180 10 = 78 y = 180 = 180 (78) y = 4 = 180 10 = 78 = 180 78 4 = 78 Q. 6. (i) y opposite angles = 49 (ii) = 180 49 = 131 (iii) + 131 = 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
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 180 110 = 70 l m 110 opposite 70 70 110 opposite Opposite Straight line 110 86 94 86 70 86 94 180 110 = 70 180 86 = 94 n 180 70 = 110 p 180 70 = 110 Lines a and b are parallel Internal angles sum to 180 (70 + 110) 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
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 5 4 3 1 (iii) 4 3 1 0 1 3 4 1 (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) 5 4 3 1 4 3 1 y (iii) 0 1 3 4 5 1 (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
Q. 15. (a) Q. 16. (b) 5 (i) 4 3 y 5 4 (ii) 3 1 1 0 1 1 3 4 5 3 4 5 (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 1 0 6 5 4 3 1 0 1 3 4 5 6 7 8 9 1 (i) 3 (b) r (iii) 4 5 6 (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 + 6 30 = y + 6 y = 4 ( + 30) + ( + y + 16) = 180 + y + 46 = 180 + 48 + 46 = 180 = 180 94 = 86 = 43 (ii) + y + 33 = 5 4y 13 4 + 5y = 46 simultaneous eq. (5 4y 13) + ( + 3y + 4) = 180 7 y = 189 simultaneous eq. 4 + 5y = 46 35 5y = 945 31 = 899 = 9 y = 7 189 = 03 189 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 = 160 1 (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
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) + + 3 + 3 = 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 = 64 36 = 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 = 1 1 3 (iii) 3 = 3 (iv) Q.. (i) = 9 = 4.5 top length top length = total length total length 7 10 = 11 = 77 10 = 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 = 144 81 = 63 (OR 3 7 ) 6 ctive Maths (Strands 1 5): h 18 Solutions
Q. 3. (i) PS.5 = 1 15 PS = 70 15 = 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 = 9.5 6 9.5 DE = 83.15 = 8.75 9.5 (ii) FG 3.75 = 9.5 FG = 47.5 9.5 = 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 = 3.5 9.5 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 50 45 D 35 (i) 50 = 35 45 45 = 8150 = 191.176... = 191 m Total distance = 50 + 191 = 441m (ii) 441 30 = 14.7 = 14 minutes 4 seconds Eercise 18.4 Q. 1. (i) 3 = 1 = 3 = 1.5 (ii) 6 = 8 10 = 48 10 = 4.8 (iii) 10 = 3 4 = 30 4 = 7.5 (iv) 10 = 1 9 = 10 9 = 3 1 3 Q.. (i) y 4 = 4 5 y = 16 5 = 3. ctive Maths (Strands 1 5): h 18 Solutions 7
(ii) y 9 = 0 15 y = 180 15 = 1 y (iii) 1 = 1 9 y = 5 9 = 8 y (iv) 4 = 44 66 = 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 10 = 19. (iii) 16.8 y 16.8 = 10 4 403. 4y = 168 35. = 4y y = 9.8 4.5 = + 3.5 1 4.5 = 1 + 3.5 4.5 = + 3.5 = 1 y = 4.5 1 y = 9 (iv) 4 = 7.5 3 = 30 3 = 10 y 0 = 4 10 y = 80 10 = 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 = 1 6 10 = = 0 PR = 0 (iii) Let RQ = y 0 = 1 + y 400 = 144 + 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 = 1 3 4 EG = 0.75 (ii) EF = 7 (4) = 3.5 8 GH 3.5 = 5 4 GH = 17.5 4 = 4.375 Q. 6. (i) + 6 = 10 + 36 = 100 = 64 = 8 R 8 ctive Maths (Strands 1 5): h 18 Solutions
(ii) ED 6 = 4 10 ED = 4 10 =.4 (iii) ED + D = E (.4) + D = (4) 5.76 + D = 16 D = 10.4 D = 3. (iv) area Δ : area ΔED 1 (8)(6) : 1 (3.)(.4) 4 : 3.84 6.5 : 1 5 4 : 1 5 : 4 Q. 8. 15 Is ΔPST similar to ΔPQR? PQ PS = 1.6 3 = 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 = 6 11. DE = 4 8 = 1 s E DE No: Δ is not similar to ΔDE (ii) P 1.6 6.4 Q. 9. 1 y 1 = 11.5 15 1 11.5 y = 15 y = 15.75 1 0.75 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
Q. 10. Q. 1. (i) Q. 11..5 1.7 10.5 m h = height of tree in metres. h 1.7 = 10.5.5 ( 1.7) h = 10.5 1.7.5 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 = 300 15 1 300 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 = 1.45 1. 1.45 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. 1.45 m m 5 m h 1.45 = 5 1.45 h = 5 h = 6.38 m The school building is 6.38 m high. 5 m 1 m m 5 = 8 1 8 5 = 1 = 58 1 3 m The river is 58 1 3 m wide 8 m 10 ctive Maths (Strands 1 5): h 18 Solutions
Q. 15. (i) 0.3 1.7 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 = 60 0.3. h = 18 m Height of kite = 18 + 1.7 = 19.7 m (ii) l = length of kite string Using Pythagoras Theorem: l = 18 + 60 l = 3,94 l = 3,94 l = 6.641... m l = 6,64.1... 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 = 3.7416.. 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 = 49 + 14 y = 63 y = 63 y = 7.937... km 1000 y = 7937... 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
(ii) 1 m 0.3 = 50 1 =.15 1.85 = 0.3 m 38 + 1 = 50 m 50 0.3 1 = 1.5 m Height of house = 1.5 m + 1.85 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) 48 57 E + 48 = 57 133 133 = 57( + 48) 133 = 57 + 736 76 = 736 = 36 The distance between the trees is 36 m. (iii) 36 40 E 36 + 9 = 45 let y be the distance between and D y 40 = 45 36 45 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. + 48 57 E 133 m D D Eercise 18.5 Q. 1. (i) h = a + b h = 13, a = 1, b = 13 = 1 + 169 = 144 + = 5 = 5 (ii) h = a + b h =, a = 1, b = 16 = 1 + 16 = 144 + 56 = 400 = 0 1 ctive Maths (Strands 1 5): h 18 Solutions
(iii) h = a + b h =, a = 11, b = 5 = ( 11 ) + 5 = 11 + 5 = 36 = 6 (iv) h = a + b h = 5, a = 4, b = 5 = 4 + 65 = 576 + = 49 = 7 (v) h = a + b h =, a = 15, b = 7 = ( 15 ) + 7 = 15 + 49 = 64 = 8 (vi) = 0 + 16 = 400 + 56 = 656 = 656 (vii) 4 = 10 + 576 = 100 + 476 = = 119 (viii) = + ( 3 ) = 4 + 1 = 16 = 4 Q.. (i) = 4 + 7 = 16 + 49 = 65 = 65 ( 65 ) = 8 + y 65 = 64 + y y = 1 y = 1 (ii) = 9 + 1 = 81 + 144 = 5 = 15 17 = 15 + y 89 = 5 + y 64 = y y = 8 (iii) 3 = + 5 9 + = 5 = 16 = 4 y = 4 + ( 33 ) y = 16 + 33 = 49 y = 7 (iv) = 1 + = 1 + 4 = 5 = 5 y = ( 5 ) + y = 5 + 4 = 9 y = 3 Q. 3. (i) = 10 + 1 = 100 + 144 = 44 = 44 = 15.60... = 15.6 to 3 significant figures (ii) 16 8 8 16 = 8 + 16 = 64 + 56 = 30 = 30 = 17.888... = 17.9 to 3 significant figures ctive Maths (Strands 1 5): h 18 Solutions 13
(iii) 8 3 y 11 y = 8 + 3 y = 64 + 9 y = 73 y = 73 + ( 73 ) = 11 = 11 ( 73 ) = 11 73 = 48 = 48 = 6.98... = 6.93 to 3 significant figures (iv) 5 (v) 3.5 3.5 7 + 3.5 = 5 = 5 3.5 = 1.75 = 1.75 = 3.5707... = 3.57 to 3 significant figures 9 (vi) 17 = + 9 89 = + 81 08 = 08 = = 14.4 (to three significant figures) a 1 1st triangle a = 1 + 1 a = 1 + 1 a = b a nd triangle. b = 1 + a 1 1 b = 1 + 1 + 1 b = 1 + 1 + 1 c 1 15 y b 8 y = 15 + 8 y = 89 y = 89 y = 17 3rd triangle. c = 1 + b c = 1 + 1 + 1 + 1 c = 1 + 1 + 1 + 1 14 ctive Maths (Strands 1 5): h 18 Solutions
Following this pattern on the 7th triangle = 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 = 8 =.884... =.83 to 3 significant figures. Q. 4. (i) Triangle with sides 7, 70 and 1. 7 = 5,184 70 + 1 = 4,900 + 441 = 5,341 No: since 70 + 1 7 this is not a right-angled triangle (ii) Triangle with sides 8.9, 8 and 3.9 8.9 = 79.1 8 + 3.9 = 64 + 15.1 = 79.1 Yes: since 8 + 3.9 = 8.9 this is a right-angled triangle (iii) Triangle with sides 16, 134 and 10 16 = 6,44 134 + 10 = 17,956 + 10,404 = 8,360 No: since 134 + 10 16 this is not a right-angled triangle (iv) Triangle with sides 113, 11 and 15. 113 = 1,769 11 + 15 = 1,544 + 5 = 1,769 Yes: since 11 + 15 = 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 = 7,569 60 + 63 = 3,600 + 3,969 = 7,569 Yes: these lengths will form a rightangled triangle since 60 + 63 = 87 (ii) Side lengths 39, 80, 89 89 = 7,91 39 + 80 = 1,51 + 6,400 = 7,91 Yes: these lengths will form a rightangled triangle since 39 + 80 = 89 (iii) Side lengths 55, 130, 148 148 = 1,904 55 + 130 = 3,05 + 16,900 = 19,95 No: these lengths will not form a right-angled triangle since 55 + 130 148 (iv) Side lengths 64, 10, 136 136 = 18,496 64 + 10 = 4,096 + 14,400 = 18,496 Yes: these side lengths will form a right-angled triangle since 64 + 10 = 136 Q. 6. 75 75 45 = 30 = 30 + 40 =,500 =,500 = 50 The distance is 50. 40 40 45 ctive Maths (Strands 1 5): h 18 Solutions 15
(ii) 74 66 38 = 66 + 11 = 16,900 = 130 11 Q. 7. (i) () + (3) = 40 4 + 9 = 1,600 13 = 1,600 = 1,600 13 1,600 = 13 = 11.094... = 11.09 to d.p. (ii) + ( + 8) = 5 + 4 + 3 + 64 =,704 5 + 3 + 64,704 = 0 5 + 3,640 = 0 a = 5, b = 3, c =,640 b 4ac = (3) 4(5)(,640) = 53,84 Using the quadratic formula: = 3 ± 53,84 5 = 3 + 53,84 10 OR = 3 53,84 10 = 0 OR = 6.4 Since > 0 reject = 6.4 = 0 (iii) ( 1) + ( ) = + 1 + 4 + 4 = 0 6 + 5 = 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) + 1 + + + 1 = + 10 + 5 + 10 5 = 0 10 3 = 0 a = 1, b = 10, c = 3 b 4ac = ( 10) 4(1)( 3) = 100 + 9 = 19 Using the quadratic formula: ( 10) ± 19 = 1 10 + 19 = OR 10 19 = = 11.98... OR = 1.98... Since > 0, = 1.98... is rejected = 11.93 to d.p. Q. 8. (i) = distance of foot of ladder from the wall + 1.6 =.5 =.5 1.6 =.505 =.505 = 1.5819... m = 158.19... 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
(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 + 190 = 5 h = 5 190 h = 14,55 h = 14,55 Slip = original height new height = 160 cm 14,55 cm = 39.480... cm The ladder slipped 39 cm (to the nearest cm). Q. 9. (i) (p ) + (p + 1) = (p + 4) p 4p + 4 + 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 = 1 + 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 + 1 + p 6p + 9 = p + 1p + 36 p 4p + 10 p 1p 36 = 0 p 16p 6 = 0. Q. 10. Q. 11. Q. 1. 80 cm l 39 cm Diameter of drum = 39 = 78 cm l = length of wire l = 78 + 80 l = 1,484 l = 1,484 l = 111.7 3... The longest wire is 111.7 cm to 1 d.p. 40 cm = 1,00 + 40 = 1,441,600 = 1,441,600 = 1,00.66... 100 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
1st find the length of the diagonal of the base of the cube = 3 + 3 = 9 + 9 = 18 nd use Pythagoras Theorem again (on the blue triangle). y = 3 + ( 18 ) y = 9 + 18 y = 7 y = 5.196... 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 + 4.8 = 6 w = 6 4.8 w = 1.96 w = 1.96 w = 3.6 m Width of garden is 3.6 m (ii) Perimeter = l + w = 4.8 + 3.6 = 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 = 9 + 6 5 + 5 + 1 4 5 + 4 5 l = 5 + 35 l = 5 + 35 l = 6.8... m l = 68... cm The pipe is 68 cm to the nearest cm. l = h + r l = 10 + 60 l = 18,000 l = 18,000 l = 134.16... mm r = 60 mm h = 10 mm The slant height to the nearest mm is 134 mm Q. 15. 6 cm 18 cm 6 cm 3 6 cm Let = height of truncated cone + 3 = 6 7 = 6 3 = 7 = 7 6 3 h 9 18 ctive Maths (Strands 1 5): h 18 Solutions
Two similar triangles in the cone where h = height of original cone. h = 9 7 3 h = 9 7 3 h = 15.588... cm h = 155.88... 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 = 8 + 10 E = 164 E = 164 E = 1.806... m E = 1,80.6 cm E = 1.81 m (to nearest cm) (ii) F = F + 3 F = 164 + 9 F = 173 F = 173 F = 13.15... m F = 1,315... cm F = 13.15 m (to nearest cm) (iii) P = 5 + 8 P = 89 P = 89 P = 9.433... m P = 943.3... cm P = 9.43 m (to nearest cm) (iv) P = ( 89 ) + 3 P = 89 + 9 P = 98 P = 9.899...m P = 989.9... 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 = 00 00 30 cm (Pyramid height) l = length of wire l = 30 + ( 00 ) l = 900 + 00 l = 1,100 l = 1,100 l = 33.166... cm Length of wire to connect point to is 33 cm to nearest cm Q. 18. 18 km 9 km (i) H = 18 + 9 5 km H 4 km H = 405 H = 405 H = 0.1... 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
(ii) H = 5 + 1 H = 169 H = 169 H = 13 km Ship is 13 km from the harbour. (iii) 18 1 = 6 9 + 5 = 14 = 6 + 14 = 3 = 3 = 15.3... km Ships and are 15 km apart to significant figures. (iv) = 1 + 0 = 401 = 401 = 0.0... km Ships and are 0 km apart to significant figures. (v) 5 4 = 1 18 + 8 = 6 9 + 4 = 13 = 6 + 13 = 845 = 845 = 9.06... km 1 + 8 = 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 = 360 60 = 100 = 1 (100 ) = 50 (viii) Larger angle at 0 = 360 80 = 80 = 1 (80 ) = 140 0 ctive Maths (Strands 1 5): h 18 Solutions
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 = 360 0 = 140 angles at a point (iii) = 180 36 = 144 opposite angle in cyclic quadrilateral = 180 107 = 73 (iv) = 180 51 = 19 opposite angle in cyclic quadrilateral = (19 ) = 58 angle at the centre of a circle (v) = 180 104 = 76 straight angle = 180 76 = 104 opposite angle is a cyclic quadrilateral (vi) = 48 angles standing on the same arc = 180 48 = 13 opposite angle is a cyclic quadrilateral Q. 3. (i) = (44 ) = 88 angle at the centre of a circle + + 88 = 180 angle is a Δ, isosceles Δ = 9 = 46 (ii) angle at a point 360 O 30 30 = angle at the centre of a circle + 30 + (360 ) + 0 = 360 quadrilateral + 410 = 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) 36 36 O 140 + 36 = 70 = 34 (iv) = 90 3 = 58 angle is a semi circle, isosceles Δ = 180 (58 ) angles is a Δ = 180 116 = 64 ctive Maths (Strands 1 5): h 18 Solutions 1
(v) same arc 44 O angle in a semi circle (vi) = 180 (90 + 44 ) = 46 angles in a = 46 angles standing on the same arc. standing on the same arc 60 60 O 30 = 90 60 = 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. 180 61 So = = 119 = 59.5 OQR = 90 59.5 = 30.5 = 180 ()(30.5 ) = 180 61 = 119 = 1 (119 ) = 59.5 P T O 90 Q U 90 90 90 R 61 S ctive Maths (Strands 1 5): h 18 Solutions
(iv) = + = 180 + = 180 + = 90 = 90. + 34 + + 4 + = 180 + 58 + = 180 + 58 + (90 ) = 180 + 58 + 180 = 180. 38 = 180 = 58 = (58 ) = 116 = 90 58 = 3 (v) s shown in (iv) = 90 + 13 + 38 + (90 ) = 180 + 13 + 38 + 180 = 180 13 + 38 = 0 13 + 38 = 51 = = (51 ) = 10 O 34 13 4 O 38 Q. 5. (i) Yes. s opposite angles add to 180 i.e. 50 + 130 = 180 and 85 + 95 = 180 (ii) No. s opposite angles do not add to 180 i.e. 80 + 90 = 170 and 115 + 75 = 190 (iii) No. s opposite angles don t add to 180 i.e. 10 + 50 = 170 (iv) Yes. s opposite angles add to 180 i.e. 108 + 7 = 180 Q. 6. (i) D = 38 (both angles at circle subtended by same arc) (ii) D = 90 (angle subtended by diameter) 180 38 (iii) D = = 14 = 71 (iv) D = 180 90 38 = 5 Q. 7. (i) O ; ΔOE ΔOE by RHS. OD = OD OD = 180 90 31 = 59 O = (59 ) = 118 (ii) ΔO is isosceles as O = O O = 180 59 = 11 180 11 O = = 9.5 ctive Maths (Strands 1 5): h 18 Solutions 3
(iii) s in (ii) ΔO is isosceles O = 11 180 11 O = = 9.5 (iv) D is a cyclic quadrilateral D + = 180 = (9.5 ) = 59 D = 180 59 = 11 Q. 8. (i) QOS = (58 ) = 116. (ii) QPSR is a cyclic quadrilateral QRS = 180 58 = 1 (iii) ΔOSQ is isosceles 180 116 OSQ = = 3 ΔRSQ is isosceles 180 1 RSQ = = 9 RSO = 3 + 9 = 61 (iv) P, O, R are collinear PR is a diameter as O is centre PQR = 90 PQO = 90 RQO = 90 61 = 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 1 3 4 5 6 Seats 1 3 4 5 6 7 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 = 180 150 = 30 straight angle (ii) = 180 130 = 50 straight angle y = 180 110 = 70 straight angle 4 ctive Maths (Strands 1 5): h 18 Solutions
(iii) = 180 100 = 80 straight angle = 40 4y = 100 corresponding angle y = 5 (iv) 4 + + 90 = 180 straight angle 5 = 90 = 18 y = 180 95 = 85 straight angle (v) + 115 = 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 3 + 5 : 5 8 : 5 Q.. (a) (i) 7 + + 3 = 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 = 0 3 + y = 180 5 = 00 3(40 ) + y = 180 = 40 y = 60 ctive Maths (Strands 1 5): h 18 Solutions 5
(iii) ( + y) + ( + ) + = 180 angles in a Δ 3 + y = 158 + 10y = + y + + eterior angle + 9y = ( 3) 3 + y = 158 3 + 7y = 66 3 + y = 158 3 + 8 = 158 8y = 4 3 = 150 y = 8 = 50 (iv) 4y + 4 + + y + 6y 4 = 180 angles in a Δ + 11y = 180 + 6y 4 = 180 straight angle + 6y = 184 + 3y = 9 + 11y = 180 + 3(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 = 35 + 65 + y = 180 straight angle 35 + 65 + y = 180 y = 80 y = 40 (ii) + y = 9 y opposite angles 8 4y = 0 + y + 7 + 3y = 180 parallelogram 8 + 5y = 180 8 + 4y = 0 8 + 5(0 ) = 180 9y = 180 8 = 80 y = 0 = 10 (iii) 7 + 4y + 8 + 4y = 180 cyclic quadrilateral 15 + 8y = 180 1 +8y +8 + 4y = 180 cyclic quadrilateral 10 + 1y = 180 30 + 16y = 360 1 10 + 1(9) = 180 30 + 36y = 540 10 + 108 = 180 0y = 180 10 = 7 y = 9 = 7. 6 ctive Maths (Strands 1 5): h 18 Solutions
(b) (i) (y + ) + (4y) = (5y) right-angled Δ 4y + 8y + 4 + 16y = 5y 5y 8y 4 = 0 (5y + )(y ) = 0 5y + = 0 or y = 0 y 5 y = Height of flagpole = 4y = 8 (ii) ( + 1) = () + ( 1) 4 + 4 + 1 = 4 + 4 4 + 1 4 8 = 0 0 = Height = 1 = 3 (iii) No, as 3 1 (8) Q. 4. (a) (i) = 50 isosceles Δ = 180 (50 + 50 ) = 80 angles in a Δ = 50 + 80 = 130 eterior angle (ii) = 4 corresponding = 180 56 = 14 corresponding angle and straight angle = 14 alternate (iii) = 180 6 = 118 straight angle ngles in isosceles Δ = 1 (180 30 ) = 75 = 180 75 = 105 = 360 (105 + 105 + 6 ) quadrilateral + vertically opposite = 88 angle (iv) + 75 = 180 parallelogram = 105 + 65 = 105 opposite = 40 = alternate = 40 (v) = 180 114 = 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
(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) = 63 6 6.5 = 8 10 = 50 1 = 14 8 = 168 8 (iv) 8 = 10 6 = 80 6 y 1 = 6 9 7 = 10.5 y = 9 = 8 y 13.75 = 8 10 10 = 5 y = 110 10 = 11 y 3 = 14 6 4 = 1 y = 6 = 7 y + 6 10 = 10 6 y + 6 = 16 3 y + 6 = 100 6 = 13 1 3 y = 10 3 (b) h 15 = 5 5 h = 75 5 = 3 m 8 ctive Maths (Strands 1 5): h 18 Solutions
Q. 6. (a) (i) Is 5 = 8.4?.5 = 3.3? No DE (ii) Is 3 5 = 8 13 1? 3 Is 0.6 = 0.6? Yes RQ TS (b) (i) Is 7.5 31.5 = 1 50.4? 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 = 90 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) 180 70 = 110 + 0 = 1 (110 ) isosceles Δ ctive Maths (Strands 1 5): h 18 Solutions 9
+ 0 = 55 = 35 + 35 + 70 = 180 + 105 = 180 = 75 (iv) 1 51º O 15º 1 = 180 51 = 19 straight angle = 1 (180 19 ) = 5.5 isosceles Δ (b) Similarly = (15 ) = 30 1.6 = 10 4 = 1.6 ( 10 4 ) = 4 m Q. 8. (a) (i) +7 = 5 7 + ( + 5) = y + 49 = 65 7 + (9) = y = 576 49 + 841 = y = 4 y = 890 y = 890 (ii) + 3 = 7 y + 4.5 = ( 40 ) + 9 = 49 y + 0.5 = 40 = 40 y = 19.75 = 40 (iii) + 5 = ( 41 ) () + 5 = y + 5 = 41 8 + 5 = y = 16 64 + 5 = y = 4 89 = y y = 89 30 ctive Maths (Strands 1 5): h 18 Solutions
(iv) = + Label hypotenuse in middle = 4 + 4 = 8 triangle as z: = 8 z = + z = 4 + 8 z = 1 + z = y 4 + 1 = 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 = 17.475 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) 5 7 4 (iv) In a right-angled triangle, Pythagoras Theorem holds. Is 5 = 4 + 7? 65 = 576 + 49? 65 = 65? True 4 h 0 ctive Maths (Strands 1 5): h 18 Solutions 31
(b) Missing hypotenuse h = 4 + 0 h = 576 + 400 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 10 10 + = 80 00. (ii) 80(10 + ) = 00(10) 9600 + 80 = 4,000 80 = 14,400 = 180 m Q. 10. (a) r = 35 8 8 (b) = 35 8 = 441 = 1 Depth of oil = 35 1 = 14 cm (i) 40.54 = 101.6 cm (ii) (9) + (16) = 101.6 81 + 56 = 10,3.56 337 = 10,3.56 3 ctive Maths (Strands 1 5): h 18 Solutions
= 30.63 = 5.534 Length = 16 5.534 = 88.55 = 89 to nearest cm Height = 9 5.534 = 49.81 = 50 to nearest cm (iii) (4) + (3) = 101.6 16 + 9 = 10,3.56 5 = 10,3.56 = 41.904 = 0.3 Length = 4 0.3 = 81.8 = 81 to nearest cm Height = 3 0.3 = 60.96 = 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 + 73 + 60 = 180 a + 133 = 180 a = 47 a = 3.5 ctive Maths (Strands 1 5): h 18 Solutions 33
a + b + 73 = 180 3.5 + b + 73 = 180 b + 96.5 = 180 b = 83.5 l 1 l a and g are alternate g = a g = 3.5 (d) (i) = 11 + = 11 + 4 = 15 = 15 = 5 5 (ii) D = D + () = (5 5 ) + 4 = 15 5 = 15 = 5 = 5 34 ctive Maths (Strands 1 5): h 18 Solutions