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標題:

Maths question 好深 希望各位幫手

發問:

ABCD is a rectangle, E is a point of CD, ∠AEB=90°, AD=4, CE=3, find ED 圖片URL: http://www.mediamax.com/cytiger2003/Hosted/maths.GIF

最佳解答:

AB=CD=4 Let BE be X AE^2 = 4^2+X^2 = 16+X^2 CE^2=4^2+3^2 = 16+9 =25 AC=BE+ED= X+3 AC^2= AE^2+CE^2 (X+3)^2 = 16+X^2+25 X^2+6X+9=41+X^2 6X=41+9 6X=50 X=8.33 Therefore, BE=8.33

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As you question you mentioned is different from the picture, I would like to solve what the question asked only. Since ABCD is a rectangle, BC = AD = 4 tan ∠BEC = 4 / 3 ∠BEC = 53.13 (correct to 2 d.p.) ∠AED + ∠AEB + ∠BEC = 180 (adj. ∠s on st. line) ∠AED + 90 + 53.13 = 180 ∠AED = 36.87 DE = 4 / tan ∠AED = 16 / 3 2007-01-13 14:31:55 補充： The proof of AK is wrong.There is no reason of SAA to prove 2 triangle are congruent.Also, AK you mixed up the angles. Read it again carefully.|||||From the provided picture, find BE, I think my solution is the simplest... Since angle AEC forms right-angle triangle, we can imagine AEC inside a circle as shown in http://www.acad.polyu.edu.hk/~06900500r/NewFolder/untitled.gif Therefore, (r-3)^2+4^2 = r^2 solve for r, we have r = 25/6 = 4.1667 BE = 2r-3 = 8.3333 -3 = 5.3333 ( I may remove the image soon... Hope this solution can help you ...)|||||呢條問題係冇可能成立~~ ｃ３ｅｄ ︳￣￣￣╱╲￣￣￣︳ ︳╱╲︳ ︳╱╲︳４ ︳╱??????╲︳ ｂａ ∠bec+∠aed+90=180(adj. ∠ on st line.) ∴∠bec+∠aed=90 ∠bce=90(known) ∠bce+∠cbe+∠bec=180(∠ sum of △) ∴∠cbe+∠bec=90 ∵∠aed+∠bec =90 ∴∠cbe=∠aed ∠ade=90(known) ∠ade+∠dae+∠aed=180(∠ sum of △) ∴∠dae+∠aed=90 ∵∠bec+∠aed =90 ∴∠dae=∠bec ∵bc=ad(known) ∠bec=∠dae ∠cbe=∠aed ∴△ceb congruent △dae(saa) ∴ce=da But the description of the question is ce=3.da=4 ∴this question is not come to existence.|||||你幅圖同你既描述有出入喎.... 我以幅圖為準呀! < 代表 angle tri. 代表 triangle * 2007-01-07 00:15:28 補充： 唔知點解顯示唔晒.....去呢個網睇下啦:http://www.cccmkc.edu.hk/~kei31182/math_yahoo.htm|||||A B |---------------| | | |---------------| D E C Please try to draw the rectangle ABCD and join AE and BE for easy thinking Remeber the sequence of A, B, C, D should be labelled clockwise Let DE be X IN triangle-BDE, CE^2 + BC^2 = BE^2 => 3^2 + 4^2 =BE^2 (since, AD=BC, properties of rectangle) => BE = 5 IN triangle-AEB, AE^2 + BE^2 = AB^2 => AE^2 = AB^2 - BE^2 => AE^2 = (DE+CE)^2 - BE^2 (since, AB = DC = DE + CE) => AE^2 = (X + 3)^2 -25 ------------------------------------------------------(1) IN triangle-ADE, AD^2 + DE^2 = AE^2 => DE^2 = AE^2 - AD^2 => X^2 = (X + 3)^2 -25 - 4^2 (Since AD=4 and (1) above) => X^2 = (X^2 + 6X + 9) -25 - 16 => X^2 = X^2 + 6X + 9 -25 - 16 => 6X = 32 => X=5.3333333|||||You mean BE? Let BE be x. AE^2 = AB^2 + BE ^2 ( Pyth. theorem) AE^2 = 4^2 + x ^2 AE^2 = 16 + x ^2 AB = CD = 4 (Pro. of rectangle) ED^2 + CD^2 = CE ^2 (Pyth. theorem) 3^2 + 4^2 = CE ^2 CE ^2 = 25 CE = 5 AC = BD = BE + ED = x+3 (Pro. of rectangle) Therfore AE^2 + CE^2 = AC^2 (Pyth. theorem) (16 + x ^2) + 5^2 = (x+3 )^2 16+25+x^2 = x^2 + 6x+9 6x= 32 x=16/3 Therefore BE = 16/3.|||||諗左好耐,,終於諗到! AB=CD=4 tan x=4/3 tanx=53.13010235度 180度-90度-tan x度=36.86989765度 tan 36.86989765度=4/ED ED=5.33333333333 哈哈 !!YEAH!!|||||AED唔係90度