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hapter 8 ercise 8.1 Q. 1. (i) 17.5 = 10 8 = 10 10 = (17.5) = 80 = (17.5) 10 = 0 = 7 (ii) 1 = 10 9 15 = 10 9 9 = 10 9 = 150 = 0 = 50 (iii) = 1 = 16 = 6 = 6 = 6 = 6 (iv) ll three triangles are similar. = 10 also 10 = 1 = 0 = 10 = 0 = 10 (v) c =.5 1.5 7 b c.5 c = c = a 1.5 a.5 = 7 b 1.5 = 7 a = 17.5 b = 10.5 a = 8.75 b = 5.5 = 8.75 1.5 = 5.5.5 (vi) 5 = 7.5 =.75 5 5 0 5 = = 0 + 5 = 0 = 0 = 5 = 5 ctive Maths ook (Strands 1 5): h 8 Solutions 1

(vii) 7 7 5 7 = 7 5 = 7 = 9 = 1.5 = 8 = 8.75 Q.. (i) Is XZ = YZ? Q.. 10 9 6 5 Δs are not similar. (ii) Is 6 9 = 8 1? = Δs are similar. = 90 = 90 = common Δs have same measure for angles. Similar (ii) = 10 = 1 6 = 0 (iii) = + 0 = 1 + = 0 1 = 56 = 16 (iv) = 16 = 6 1 = 16 ( 1 ) = 8 Q.. (i) 10 = 0 16 = 10 ( 0 16 ) = 18.75 (ii) F 1 = 0 16 F = 1 ( 0 16 ) F =.5 ctive Maths ook (Strands 1 5): h 8 Solutions

(iii) osine Rule a = b + c bc cos 16 = 10 + 1 (10)(1)cos cos = 1 0 = cos 1 ( 1 0 ) = 9.8660 rea = 1 (1)(10) sin 9.8660 = 59.950 59.9 units (iv) rea = 1 (.5)(18.75) sin 9.8660 = 10.677 10.67 units (v) Perimeter ΔF.5 + 18.75 + 0 = 71.5 Perimeter = 10 + 1 + 16 = 8 Ratio of perimeters of the triangles is the same ratio as their individual sides. Q. 5. (i) = = is common Similar Δs (ii) 10 = 16 1 = 10 ( 16 1 ) = 0 (iii) = 10 = 0 10 = 10 (iv) 0 = 1 16 = 15 = 0 15 = 5 Q. 6. F 16 16 (i) F = F alternate angles F = F verticall opposite F = F Similar triangles (ii) = F F 16 = 5 = 80 = + = 80 + 16 = 18 Q. 7. (i) PQ SQ = PR ST PQ 1 = 6 1 PQ = 6 PS = 6 1 = 1 (ii) QT TR = 1 1 TR = 1 QT = 1 (iii) QR = 1 + 1 = 6 Q. 8. (i) 10 : (ii) : 7 (iii) 7 : 10 ctive Maths ook (Strands 1 5): h 8 Solutions

Q. 9. F onsider similar Δs + h = +. 1 h = + onsider similar Δs + h 6 = + 6 h = + quate + = 6 + = 6 Sub into 1 h = + _ h = + h = 5 h = _ 5 h =. = 6 h F 6 F h ercise 8. Q. 1. (i) = 16 + 1 = 56 + 1 = 00 = 0 (ii) = 10 + 1 = = (iii) 6 = + = = (iv) 1.17 = + 1.08 = 1.17 1.08 = 0.05 = 0.5 Q.. (i) = (8 + ) + 1 576 1 = (8 + ) = 19 + 16 + + 16 0 = 0 (ii) ( + 0 )( ) = 0 = 0 = = 1 + ( ) = 1 + 16() = 19 = 8 1 1 8 50 8 50 ctive Maths ook (Strands 1 5): h 8 Solutions

(iii) 1 = + = 1 8 = (50 ) + 8 = (50 ) + 1,0 =,500 100 + + 196,0 =,500 100 + 196 100 =,500 + 196,0 100 = 9 =.9 = 1 (.9) = 180.66 = 1. OR 1 bh = rea 1 (1)(8) = 1 ()(50) 6 = 5 1. = z = + 7 = 16 + 9 = 65 = 65 z = 65 + = 65 + 16 = 81 z = 81 = 9 = 9 + 8 = 81 + 6 = 15 = 15 (iv) = 9.5 5 =,50.5 1,5 = 1,5.5 = 1 9 1,5.5 = = 5 + 7 = 1,7 = _ 1,7 = 7 6 Q.. (i) ( + ) = () + ( ) Q.. + + = + + + + = + + (ii) 10 + 1 = 101 10 1 = 99 (10)(1) = 0 ns = 101, 99, 0 85 Perimeter = 18 + = 85 + = 18 85 + = 97 = 97 (97 ) + = 85 9,09 19 + + = 7,5 19 +,18 = 0 97 + 1,09 = 0 8 1 = 8 or = 1 = 8 = 1 = 1 or = 8 Given > = 8 = 1 ctive Maths ook (Strands 1 5): h 8 Solutions 5

Q. 5. + + + + 6 + 9 = + + 1 = 0 + + 1 = 0 1 = 0 b ± b ac a ± ( 1) () ± + 10.0981 or.0981 ns.10 Q. 6. (i) = + = + = (ii) Fill in diagram as shown: a a = a + a = a a. ( + ). (a + a)(a) LHS = RHS (a)(a) a Q. 7. (i) = 1 + 10 = 96 = 96 = 7 (ii) F = 16 + 1 (iii) = 5 F = 5 = 11 56 10 16 = 56 + 1 = 55 = 55 = 10 + 16 = 56 = 56 Q. 8. (i) = 0 + 0 = 800 = 800 = 0 (ii) F = (10 ) + 5 F = 00 + 65 = 5 F = 5 F = 5 17 (iii) M = M = 5 10 = 55 M = 5 1 6 ctive Maths ook (Strands 1 5): h 8 Solutions

Q. 9. (i) (iv) + + 10 + + 10 (ii) + 1 9 ()() sin 60 = + + 10 ercise 8. h + 1 + ( + 1) + h = ( + + 10 ) + + 1 + h = + + 10 1 + h = 10 h = 9 h = Q. 1. (i) 1 (7.)(.1) = 7.665 units (ii) 1 (8.6)(.1) = 17.6 units (iii) 1 (.6)(.5) = 8.05 units (iv) 1 5 = 65 units (v).9 = 1.7 units (vi) 18 = units Q.. (i) _ 5 7 = 5 = 8.75 ns 9 units (ii) [ 1 ()() sin 60 ] = 16 sq units ns 8 units (iii) 1 (1)(18) = 8 units 1 (1 7) = units 90 units 7 1.5 187 Shaded area _ = 187 9 = 6.590 ns 6 units (v) ( 7) + (1.5)(7) = 1 + 10.5 = 1.5 units (vi) 1 (0) 1 = 10 units (vii) Find r r + r = 00 r = 00 r = 00 r = 10 rea of circle pr = p(10 ) (1.5)( 187 ) _ = 187 = 68.185 rea of square = 00 Shaded area = 68.185 00 = 8.185 ns 8 units ctive Maths ook (Strands 1 5): h 8 Solutions 7

Q.. (i) 1 (6.)() = 1 (9)(.8) = 9(.8) = 6.75 (ii) 1 (6)() = 1 (10.7)(.) 10.7 (.) = 6 =.8 (iii) 1 ()(6.7) = 1 (7)(.95) = 7(.95) 6.7 = 5.1716 5.17 Q.. (i) 5 cm (ii) 5 = 15 cm Q. 5. rea of square rea of triangle 1 Y X = 1 ( 1 ( ) 1 ( ) ) = 1 ( 1. ) = 1 8. ercise 8. Q. 1. (i) = 110 = 55 180 110 = _ = 5 (ii) = 10 = 5 + + 0 + 75 + 90 = 60 5 + + 05 = 60 = 10 90 + 0 + + 0 + 90 + 75 = 60 = 60 5 = 5 (iii) = 8 = 8 8º 8º Q. 6. ns 8 : 1 96º º º 10º 0º 0º 1 () sin 10 = sin 10 = 8 = 16 = 16 = (iv) = 0 = 180 100 0 = 0 (v) = 105 = 5.5 (vi) = 60 8 90 90 = 98 180 98 = = 1 8 ctive Maths ook (Strands 1 5): h 8 Solutions

(vii) (iv) 56º º 11º º P 0º 10º 0º 70º 0º Q º 0º = = 90 Q.. (i) O = 11 (ii) = 180 70 0 = 80 (iii) = 100 (iv) = 80 opposite angles in a cclic quadrilateral Q.. (i) 15 + + 0 + 100 = 5 + 0 + 10 = 100 = 10 (ii) 5 0 + 16 = + 10 + 5 + 16 + + 16 5 0 + 16 = 17 + + 1 8 8 5 = 0 a = 8 b = 8 c = 5 8 ± ( 8) (8)( 5) (8) _ 8 ± 75 = 16 = 10.56 QT = 0 (v) 0 (vi) 0 Q. 5. (a) (i) RS (ii) QRP (iii) QPS Q. 6. O r r T S ommon Given: ircle with tangents at T and S. To prove: PT = PS onstruction: Join O to P Proof: onsider Δs OTP and OSP OP = OP.. common OS = OT.. radii OSP = OTP right angles tangents. ΔOTP ΔOSP PT = PS T RHS P Q.. (i) 90 (ii) 0 (iii) 70 ctive Maths ook (Strands 1 5): h 8 Solutions 9

Q. 7. (i) RX = 7 Q. 9. (ii) QX = OQ OX = 5 = 1 OX = 5 7 OX = 576 = ns: QX = 1 9 1 + (i) = 1 (ii) 15 9 15 9 = 1 (iii) PX = 5 + = 9 (iv) PS = 9 + 7 = 50 PS = 5 (iii) = 17 1 Q. 8. N 1 O 1 M N 0 5 cm O (i) N = 5 0 N = 5 N = 15 = N = 0 cm (ii) O M = 5 M = 9 M = 7 5 = M = (7) M = 1 cm O = 1 + 1 = 88 O = 88 O = 1 Q. 10. The circles, and z are shown. z O 1 1 R a a 0 0 The radius length of circle is 1 cm, and the radius length of circle is 0 cm. If ROP = 90, find the radius length of circle z. P 10 ctive Maths ook (Strands 1 5): h 8 Solutions

+ (1 + a) = (0 + a) 1,0 + 1 + a + a = 00 + 0a + a 1,168 + a = 00 + 0a 16a = 768 a = 8 Radius of z = 8 cm Q. 11. (i) a = b + c bc cos cos = b + c a bc = 10 + 9 9 = 15 (10)(9) = cos 1 ( 15 ) = 9.9 ns 9.9 (ii) rea = 1 (10)(9) sin 9.9 = 18.7 sq units (iii) 1 (10)r + 1 (9)r + 1 (9r) = 18.7 15r = 18.7 r = 7.581 To prove: [M] bisects Proof: onsider Δs M and M M = M. common M = M. isosceles M = M.. given Δs are congruent. SS M = M M bisects Q.. (i) Proof: Δ and Δ = given = given = Δ isosceles Q.. Δ Δ SS (ii) Proof: Δ and Δ = given = given = Δ isosceles. Δ Δ SS 1 180º 1 ercise 8.5 9 60 9 = 60 + 86 = 600 + 86 = 69.7 Q.. 180º 180 1 + 180 + + = 60 60 1 + + = 60 1 + = + G H Q. 1. Given Δ M = Midpoint of. F Δ H is congruent to Δ FG (SS) ommon angle H = G = M = F = ( + ) ctive Maths ook (Strands 1 5): h 8 Solutions 11

H = GF 1 and H = FG From 1 G = FH Δ G is congruent to Δ FH (S) G = FH = } G = FH = F G = HF Δ is congruent to Δ F (SSS) = F } = F = ( + ) ommon side = F = Q. 5. =. rectangle X = X given X = X 90 Δs congruent. SS qual areas. Q.. Q. 6. The perpendicular from the centre to a chord bisects the chord. Perpendicular bisector of [] Q. 8. Q. 9. Q. 10. = Isosceles Δ ut = alternate angles = bisects G Z Y X F onsider Δs ZY and GXY. GYX = ZY. v. opp angles GY = Y.. given YZ = YGX. alternate Δs congruent S. rea of FG = rea ZFX. Y Z entre O Q. 7. (i) Let the distance from l to k = rea of Δ = 1 () Δ = 1 () reas are equal. (ii) rea of ΔX = rea Δ rea ΔX ΔX = rea Δ rea ΔX (i) rea of Δ = rea Δ rea of ΔX = rea ΔX. W onstruction: Produce X to Y. Produce X to Z. Label W. Join Z to W parallel to. Proof: onsider the parallelogram WZ. WZ = Z alternate ΔZ is isosceles. = Z WZ is a rhombus. Z and W intersect at right angles.(*) 1 ctive Maths ook (Strands 1 5): h 8 Solutions

Q. 11. onsider ΔY: Y = Y alternate ut W = Y opposite angles of parallelogram W Y iagonals of rhombus bisect. X = angle at intersection of diagonals of WZ (*) X = 90 R (ii) + = 5 [ + ] + [ + ] = 5[ + ] + 16 + + 16 = 0 + 0 0 + 0 = 0 + 0 Q.. Q. 15. (i) ngles standing on same arc. (ii) Δs P and Q are congruent. reas are equal. height P = height Q Q P Q. 16. R S P Q ( ) = 180 = 5 PQR = 90 T O Q. 1. Y = X given X = Y given common ongruent S. Q. 1. + = + Q. 1. LHS + = + + ( + ) = + LHS = RHS (i) = + = + = + = + RST = 90 angle sitting on diameter. TO = 90 tangent. RST + STO + SRT = 180 STO + SRT = 90 1 ST + STO = 90 STO + SRT = ST + STO STR = ST Q. 17. (i) onsider. = 60 opposite angles cclic quadilateral onsider ΔO O = O radii ΔO is isosceles. O = 60 ΔO is equilateral. ctive Maths ook (Strands 1 5): h 8 Solutions 1

(ii) = Q. 18. P 0º O 0º 10º 60º 60º 0º 60º 10º onstruction: Join to. onsider Δs O and : = common O = both 10 O = both 0 isosceles Δs Δs are congruent. S O = and O =. O is a parallelogram. Similarl, = O. O = O. 0º Q X Y T onstruction: raw lines from X, T and Y. Intersection centre point. PR = PT + TR = PT + YR. as ΔOTR and OYR congruent PQ = PX + XQ = PX + QY as ΔOXQ ΔOYQ RQ = YR + QY given PR + RQ > PQ PT + YR + YR + QY > PX + QY ut PT = PX ΔOPX ΔOTP R PX + YR + QY > PX + QY Q. 19. s T is the tangent to the circle at, drawing a perpendicular line to T cuts the circle at. Join. We have: T = 90 and = 90 + 1 = + 5 T = ut =... angles on the same arc 1 = 5 } T O 1 5 = 5 ( 1 = ) Δ is an isosceles triangle. = Q.. 1 ctive Maths ook (Strands 1 5): h 8 Solutions

ercise 8.6 Q. 1. (i) onsider Δ and Δ. = verticall opposite = alternate = alternate Δs similar. PX RX = SX QX PX. QX = SX. RX PX. QX = XR. XS Q.. (i) YR RX = YS SZ RS XZ Theorem 1. (ii) (ii) Y = corresponding sides. Q.. ΔGF and HF F = F common. GF = HF standing an same arc Q.. Q Remaining angles are equal Δs are similar. P (i) onsider Δs QXR and PXS. QXR = PXS v. opp. XPS = QRX alternate XSP = RQX alternate (ii) Δs similar R X Q P R X S S X R ctive Maths ook (Strands 1 5): h 8 Solutions T S First: onsider Δs YRZ and YSX. YR = YS given Y = Y common YX = YZ given Δs are congruent. s Δ YRS is isosceles given YRS = YSR SRT = RST XRT = ZST (*) Now consider ΔRXT and ΔSTZ. XR = ZS. given RTX = STZ v. opp. XRT = ZST. b (*) Z Δs are congruent. also similar. (iii) ΔXTZ s proven in (ii): SRT = RST ut SRT = TZX alternate angle RST = TXZ alternate angle ΔXTZ is isosceles. 15

Q. 5. Δs F and = F F 1 Δs F and = F F 1 = = Q. 6. In Δs PTS N PUQ: PT TU = PS SQ In Δs PSU and PQR: PS SQ = PU UR. PT TU = PU UR PT : TU = PU : UR Q. 7. onsider Δ Q. 8. F F = F onsider Δ _ G G = F F = _ G G FG Y X Theorem 1 Theorem 1 Mark on [] a point X Such the X is parallelogram. = Y YX ut Y = F YX = F parallelogram = F F F Q. 9. (i) Δ (i) onsider Δs and : = given = common Δs are equiangular. (ii) = =. Q. 10. (i) onsider Δs and. = common = 90 given Δs are equiangular. (ii) = =. 16 ctive Maths ook (Strands 1 5): h 8 Solutions

Q. 11. onsider Δs and. Q. 1. Δs and = 90 angle on diameter. Supplementar angles. =. common. equiangular = =. Q. 1. (i) To Prove: P is a parallelogram. X Given: X = P P 1 P P = Y Y Proof: onsider Δ. 1 [XP] [] [P] [] onsider Δ. [PY] [] [P] [] P is a parallelogram. opposite sides parallel. F onsider Δs and F. =. common = F. 90 given = F Δs are equiangular. (ii) Similarl it can be shown that Δs and F are equiangular. (ii) To Prove: P = P = Prove that XY. Proof: X X = P P P P = Y Y X X = Y Y Sides are in proportion. XY and are similar. XY ctive Maths ook (Strands 1 5): h 8 Solutions F = F = F 1 From (i) = F = F. F =. 1. F =.. F =. =. F =. =. 17

Revision ercises Q.. (a) (i) 7 5 Q. 1. (a) 9 (b) (i) pr r = rea = pr = p(16) = 50.655 = 50.7 cm (ii) rea of Δ: 1 ()() sin 60 = 6.98 Heagon: 6.98 6 = 1.569 1.57 cm (c) (i) Radius of s is the diagonal of the square: H = + H = H = r = (ii) Radius of circle p = 1 (iii) rea of S = πr = π ( ) = π rea of P = πr = π ( 1 ) area = 1 ()(7) = 8 sq units (ii) entre point lies on diameter. r = 1.5 (b) (i) = 6 + 1.9 =,1.61 =,1.61 = 8.1 rea = 1 (1.9)(6) + 1 (60)(8.1) = 57. + 1 =,017. m (ii) 5,017. 10,086 (iii) rea of triangle : Need to calculate area: Find first. Using Sine Rule: sin = _ sin 90 6 8.1 sin = 6 8.1 = sin 1 6 8.1 8.555 So = 90 + 8.555 18.6 So, area of Δ: 1.9 m 18.6 60 m Ratio = π : π ( 1 ) 1 : 1 : 1 1 (1.9)(60) sin 18.6 6.6 units 18 ctive Maths ook (Strands 1 5): h 8 Solutions

(c) (i) (ii) (iii) rea of Δ =,017. 6.6 = 1,8.57 units Length of fence [] Use the osine rule = (1.9) + (60) (1.9) (60) cos 18.6 = 78.8098 = 86.505 Length of fence [] = 86.505 m c d c a d a d a = d d = a c d a c a = a c c = a a a b b = a b b = a c b b b a b d d c Q.. (a) (i) (iv) b d = c a (a ) = (a a) a a = a a Q. =.5 1.8 =. (.5 1.8 ) =.6 ns =.6 + 1 =.6 m (ii) 1 (.5)(.6) + 1 (1.8)(.) =.6 +.16 ns = 6.96 m (b) PS = 1 + = 1 + 9 = 10 PS = 10 PQRS is also a square. area = 10 10 ns = 10 units (c) (i) Find. 0º 0º 1 10º 1 sin 10 = _ 1 sin 0 1 sin 10 = sin 0 = 1 (ii) rea = 1 (1 )(1 ) sin 60 = 187.0615 units ctive Maths ook (Strands 1 5): h 8 Solutions 19

Q.. (a) Proof using congruenc. (b) 1 1 6 m (i) = 6 + (ii) (c) (i) = 6 + 16 = 5 = 1 h d 1 h = 1 1 h = ns: h = 8 8 d 1 ( 1 ) = 8 + d 08 = 6 + d d = 1 d = 1 m istance between tree and house is 1 + 6 = 18 m. 0 100 0 + = h 15 + = h 0 + = h 15 + = h 0 = 15 = 0 15 = 6 1 15 h = Q. 5. (a) 15 ( ) + = 0 7 = 60 7 8.57 m (ii) + = 100 + = 100 7 = 100 = 00 7.86 57.1 istance to : 57.1 m istance to :.86 m (iii) Regardless of how far apart the plons are, the distance between and is irrelevant. In our calculation of h, the distance between and is not used. R onsider Δs OPR and OTR: OR. common line PO = OT radii RTO = RPO. 90 angles (at tangent) Δs are congruent. (b) = F + F RHS P T O = F + + F F Q 0 ctive Maths ook (Strands 1 5): h 8 Solutions

Q. 6. (a) (c) (i) F = (7 ) = (9 1 + ) = 1 9 = 7( 7) (ii) F = 9 + = 9 + = 9 = 9 (iii) 1 (9)() =.5 units R O Q. 7. (a) (i) = + P = + P P = + P. Q. (ii) P = Q PQ = + Q Q P P Q X P 1 5 Q Y (b) (c) T onstruction: tend the line RO to produce [RT]. onstruct the ΔRPT and label angles 1 to 5. Proof: RPQ = 90 angle in semicircle 5 + = 90 angle in Δ + = 90 tangent 5 = but 5 = same arc = but 1 = isosceles = 1 1. = 5 = 1. ( 5 ) RS PQ = 7 cm YX = 90 angle on diameter XY = 90. straight line Since XY is a cclic quadrilateral = 90 opposite angles sum to 180 (b) 1 : (c) (i) = 5 7 = 5 7 5 = (ii) 1 ( 7 5 ) () + 1 (7)(5 ) 7 10 + 5 7 7 + (5 )(5 7) 10 _ = 7 + 175 70 + 7 10 1 70 + 175 _ 10 P Q ctive Maths ook (Strands 1 5): h 8 Solutions 1

Q. 8. (a) (i) ΔPQS and ΔPQR QS = PR given PS = QR. given PQ common congruent SSS (ii) ΔPXQ ΔSXR (iii) Proof: onsider Δs PQS and PQR. PQ common QS = PR given PS = QR given ΔPQS ΔPQR reas are equal as both triangles have base. PQ heights are the same. PQ SR. (b) PT = (r) r = r r = r PT = r = r (b) (i) (c) cos 0 = 1 h S Q P h cos 0 = 1 _ 1 h = cos 0 = 1.86 T TQR = STQ alternate TRQ = PTS compound ut PTS = STQ given R TQR = TRQ ΔTQR is an isosceles triangle (ii) PS SQ = PT TR PS SQ = PT TQ as TR = TQ PS. TQ = PT. SQ (c) (i) 1 r + 1 r + 1 r Q. 9. (a) = rea Δ 1 r ( + + ) = rea Δ = 1 r (perimeter) Q (ii) 500 cm = 1 (8) perimeter 500 = perimeter Perimeter = 15 cm F (i) Proof: onsider Δs F = F v. opp. F = alternate F = F alternate Δs similar (ii) = + = 5 = 5 0º 1 ctive Maths ook (Strands 1 5): h 8 Solutions

(iii) = 1 5 (c) (i) Q. 10. (a) = 1 5 = 1 5 = 5 + () = 5 + 5 = +.5 = 9 + 1.5 = 9 = 6 height = ( ) + h = h = h h =? 9 (ii) Hpotenuse = r 0º = djacent (b) (i).5 (Height) = 9.5 (ii) = 60.75 Height 7.79 r r? r O Shortest distance = r (iii) cos 0 = r r cos 0 = r = cos 0 r = O r r ( ( ) ) r = r + r + r = r + = r = r = r r = ctive Maths ook (Strands 1 5): h 8 Solutions

(iv) rea of incircle: p ( ) rea of circumcircle: p ( ) Ratio: p ( ) p ( ) 1 ns = : 1 ctive Maths ook (Strands 1 5): h 8 Solutions

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