Maths question 好深 希望各位幫手

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

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~~
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∠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