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Maths problems [mixed 2]
http://i226.photobucket.com/albums/dd260/wingreally/maths-1.jpg http://i226.photobucket.com/albums/dd260/wingreally/maths1-1.jpg Can anyone calculate them in details? thx
最佳解答:
(19) Let DX = 2m and CX = m, then △XDY ~ △ABY for the reason of AAA and their side ratios are 2:3 with YX:AY = 2:3. So let the height of △XDY and △ABY be 2h and 3h respectively, then area of ABCD = (3m)(5h) = 15mh Now, Area of △ADX = 5mh, so area of △DAY is 3/5 of it since AY:YX = 3:2. Hence area of △DAY = 3mh Also area of △XDY and △ABY are 2mh and 9mh/2 respectively. So area of XYBC = 15mh - 3mh - 2mh - 9mh/2 = 11mh/2 Therefore area of △DAY:XYBC = 3:11/2 = 6:11 (21) Vertical height of A above the ground = 2x sin (90 - θ) = 2x cos θ Vertical height of C above A = x sin θ Thus ans = 2x cos θ + x sin θ (22) From P and R, draw perp. lines to the building and let the feet of perp. formed then be C and D respectively. From the given, we can let, and also have the followings: AC = x, CD = h, DB = h, BQ = s tan 25 = x/s tan 50 = (x + 2h)/s = tan 25 + 2(h/s), giving h/s = (tan 50 - tan 25)/2 So the angle of elevation of A from R is ∠ARD where tan ∠ARD = (x + h)/s = tan 25 + (tan 50 - tan 25)/2 = (tan 50 + tan 25)/2 ∠ARD = 39.7 (48) From O, drop perp. lines to AB and CD and let the feet then formed be E and F respectively. Then E and F are mid-points of AB and CD resp. So OE = OF = √(36 - 25) = √11 cm Hence OEMF is a square with OM = √22 cm (50) OD and OB are perp. to CE and AC respectively. So ∠BCD = 360 - 90 - 90 - 85 = 95 Thus ∠AFE = 180 - 95 = 85 since ACEF is a cyclic quad. (44) AB = BC tan 20 and BG = BC tan 65 By Pyth. thm: AG = √(AB2 + BG2) = BC √(tan2 20 + tan2 65) = 12√(tan2 20 + tan2 65) = 26 cm (46) Let O be the centre, then after joining BO, we have ∠BOC = 80, i.e. arc BC subtends an angle of 80 at centre. So arc ACB subtends an angle of 260 at centre and therefore its length is 6 (260/80) = 19.5 2009-06-27 22:56:15 補充： (51) With ∠ODE = ∠OFE = 90, ODEF is a cyclic quad. AD is perp. to CE since OD is perp. to CE Also ∠CAD = ∠EAD for tangent properties. ∠ADC = ∠ADE = 90 So △ADC and △ADE are congurent (ASA) 2009-06-27 22:56:25 補充： (51 cont'd) So CD = DE and since BC = CD and EF = DE for tangent properties, BC = EF. ∠AOF = ∠AOB (Tangent properties) So, ∠BOE = 2∠AOF ∠BOE = 2∠BDF (Angle at centre = Twice angle at circumference) So ∠AOF = ∠BDF 2009-06-27 22:56:40 補充： (52) ∠ABX = ∠DCX and ∠XAB = ∠XDC (Angles in the same segment) So, △ABX ~ △DCX ∠ABX = ∠ACD (Angles in the same segment) ∠XAB = ∠DAC (Given) 2009-06-27 22:56:54 補充： (52 cont'd) So, △ABX ~ △ACD ∠ADX = ∠ACB (Angles in the same segment) ∠DAX = ∠CAB (Given) So, △ADX ~ △ACB 2009-06-27 22:57:11 補充： (54) Join the centres and the line's length will be r + 4 Also, the horizontal distance between the centres = 25 - r - 4 = 21 - r And the vertical distance between the centres = r - 4 So by Pyth. thm: (r - 4)^2 + (21 - r)^2 = (r + 4)^2 2009-06-27 22:57:29 補充： (54 cont'd) 2r^2 - 50r + 457 = r^2 + 8r + 16 r^2 - 58r + 441 = 0 (r - 49)(r - 9) = 0 r = 49 (rejected) or 9
其他解答:
好詳細啊! thx so much!|||||Too small, cannot read!!!
Maths problems [mixed 2]
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發問:http://i226.photobucket.com/albums/dd260/wingreally/maths-1.jpg http://i226.photobucket.com/albums/dd260/wingreally/maths1-1.jpg Can anyone calculate them in details? thx
最佳解答:
(19) Let DX = 2m and CX = m, then △XDY ~ △ABY for the reason of AAA and their side ratios are 2:3 with YX:AY = 2:3. So let the height of △XDY and △ABY be 2h and 3h respectively, then area of ABCD = (3m)(5h) = 15mh Now, Area of △ADX = 5mh, so area of △DAY is 3/5 of it since AY:YX = 3:2. Hence area of △DAY = 3mh Also area of △XDY and △ABY are 2mh and 9mh/2 respectively. So area of XYBC = 15mh - 3mh - 2mh - 9mh/2 = 11mh/2 Therefore area of △DAY:XYBC = 3:11/2 = 6:11 (21) Vertical height of A above the ground = 2x sin (90 - θ) = 2x cos θ Vertical height of C above A = x sin θ Thus ans = 2x cos θ + x sin θ (22) From P and R, draw perp. lines to the building and let the feet of perp. formed then be C and D respectively. From the given, we can let, and also have the followings: AC = x, CD = h, DB = h, BQ = s tan 25 = x/s tan 50 = (x + 2h)/s = tan 25 + 2(h/s), giving h/s = (tan 50 - tan 25)/2 So the angle of elevation of A from R is ∠ARD where tan ∠ARD = (x + h)/s = tan 25 + (tan 50 - tan 25)/2 = (tan 50 + tan 25)/2 ∠ARD = 39.7 (48) From O, drop perp. lines to AB and CD and let the feet then formed be E and F respectively. Then E and F are mid-points of AB and CD resp. So OE = OF = √(36 - 25) = √11 cm Hence OEMF is a square with OM = √22 cm (50) OD and OB are perp. to CE and AC respectively. So ∠BCD = 360 - 90 - 90 - 85 = 95 Thus ∠AFE = 180 - 95 = 85 since ACEF is a cyclic quad. (44) AB = BC tan 20 and BG = BC tan 65 By Pyth. thm: AG = √(AB2 + BG2) = BC √(tan2 20 + tan2 65) = 12√(tan2 20 + tan2 65) = 26 cm (46) Let O be the centre, then after joining BO, we have ∠BOC = 80, i.e. arc BC subtends an angle of 80 at centre. So arc ACB subtends an angle of 260 at centre and therefore its length is 6 (260/80) = 19.5 2009-06-27 22:56:15 補充： (51) With ∠ODE = ∠OFE = 90, ODEF is a cyclic quad. AD is perp. to CE since OD is perp. to CE Also ∠CAD = ∠EAD for tangent properties. ∠ADC = ∠ADE = 90 So △ADC and △ADE are congurent (ASA) 2009-06-27 22:56:25 補充： (51 cont'd) So CD = DE and since BC = CD and EF = DE for tangent properties, BC = EF. ∠AOF = ∠AOB (Tangent properties) So, ∠BOE = 2∠AOF ∠BOE = 2∠BDF (Angle at centre = Twice angle at circumference) So ∠AOF = ∠BDF 2009-06-27 22:56:40 補充： (52) ∠ABX = ∠DCX and ∠XAB = ∠XDC (Angles in the same segment) So, △ABX ~ △DCX ∠ABX = ∠ACD (Angles in the same segment) ∠XAB = ∠DAC (Given) 2009-06-27 22:56:54 補充： (52 cont'd) So, △ABX ~ △ACD ∠ADX = ∠ACB (Angles in the same segment) ∠DAX = ∠CAB (Given) So, △ADX ~ △ACB 2009-06-27 22:57:11 補充： (54) Join the centres and the line's length will be r + 4 Also, the horizontal distance between the centres = 25 - r - 4 = 21 - r And the vertical distance between the centres = r - 4 So by Pyth. thm: (r - 4)^2 + (21 - r)^2 = (r + 4)^2 2009-06-27 22:57:29 補充： (54 cont'd) 2r^2 - 50r + 457 = r^2 + 8r + 16 r^2 - 58r + 441 = 0 (r - 49)(r - 9) = 0 r = 49 (rejected) or 9
其他解答:
好詳細啊! thx so much!|||||Too small, cannot read!!!
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