超簡單立幾(急)!!!要求詳解

a)A 到 BC 垂足為 F
則 AF = √3/2 ,
FD = √(FC² + CD²) = √[(1/2)² +√2²] = 3/2設 AD 與 BCDE 所成角為 θ ,則 AF / FD = tan θ√3/3 = tan θθ = 30°
bi)
AE² = AB² + BE²
AE² = 1 + BE² .......(1)
AE = √(1 + BE²);
ED² = BC² + (CD - BE)²
ED² = 1 + (√2 - BE)² ......(2)
ED = √[1 + (√2 - BE)²]又 AD = AF / sin 30° = 2AF = √3
AD² = 3 ........(3)(1) + (2) = (3) :1 + BE² + 1 + (√2 - BE)² = 32BE² - 2√2BE + 1 = 0正根 BE = √2/2ii)ACDE 體積 = (1/3) 梯形 BCDE * AF
= (1/3) (BE + CD)BC/2 * AF
= (1/3) (√2/2 + √2)1/2 * √3/2
= (1/3) 3√6/8
= √6/8設所求距離 = d ,
(1/3) d * △AED = √6/8
(1/3) d * (1/2)ED * AE = √6/8
(1/3) d * (1/2)√[1 + (√2 - BE)²] * √(1 + BE²) = √6/8
(d/6) * √[1 + (√2 - √2/2)²] * √[1 + (√2/2)²] = √6/8
(d/6) * √(3/2) * √(3/2) = √6/8
(d/6)(3/2) = √6/8
d = √6/2

2010-12-29 00:27:16 補充：
更正 bii) ,

ACDE 體積
= (1/2) △CDE * AF
= (1/2) CD * BC * AF
= (1/2) √2 * 1 * √3/2
= √6/4

2010-12-29 00:27:22 補充：
設所求距離 = d ,
(1/3) d * △AED = √6/4
(1/3) d * (1/2)ED * AE = √6/4
(1/3) d * (1/2)√[1 + (√2 - BE)²] * √(1 + BE²) = √6/4
(d/6) * √[1 + (√2 - √2/2)²] * √[1 + (√2/2)²] = √6/4
(d/6) * √(3/2) * √(3/2) = √6/4
(d/6)(3/2) = √6/4
d = √6

2010-12-29 00:31:36 補充：
再更正 :

ACDE 體積
= (1/3) △CDE * AF
= (1/3) (1/2)CD * BC * AF
= (1/3) (1/2)√2 * 1 * √3/2
= √6/12

2010-12-29 00:31:51 補充：
設所求距離 = d ,
(1/3) d * △AED = √6/12
(1/3) d * (1/2)ED * AE = √6/12
(1/3) d * (1/2)√[1 + (√2 - BE)²] * √(1 + BE²) = √6/12
(d/6) * √[1 + (√2 - √2/2)²] * √[1 + (√2/2)²] = √6/12
(d/6) * √(3/2) * √(3/2) = √6/12
(d/6)(3/2) = √6/12
d = √6/3

2011-01-01 02:57:31 補充：
點解 AE會垂直於ED?

因為題目 b) 已事先聲明 : 「設ㄥAED = π/2」

係wo係wo , 不好意思