一題想了挺久才完成的圖形題

幻藍

Originally posted by xiirr at 2005-5-2 01:15 AM:

我同你溝通唔到,
我話oscar咁快就學微積分.
這是合比例.

                               
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[ Last edited by xiirr on 2005-5-2 a ...

合比例的义包.....?

lpgexe

答案弄到手!
因為太興奮所以x個月來首度開腔!

let CB be y cm
CB = AC = y(Given)
AB = 開方 (y^2 + y^2) = 開方 (2 x y^2) = 開方(2) x y
AB = DB = 開方(2) x y
The area of tri. ACB = y^2 / 2
The area of tri. ACB = The area of tri. ADB
The height of tri. ADB( 部份步驟省略 )
= y^2 / 開方(2) x y
= y / 開方(2)
sin ang.ABD = The height of tri. ADB / DB
                 = [y / 開方(2)] x [1 / 開方(2) x y]
                 = 1 / 2
     ang.ABD = 30

xiirr

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xiirr

Originally posted by lpgexe at 2005-5-2 11:52 AM:
答案弄到手!
因為太興奮所以x個月來首度開腔!

let CB be y cm
CB = AC = y(Given)
AB = 開方 (y^2 + y^2) = 開方 (2 x y^2) = 開方(2) x y
AB = DB = 開方(2) x y
The area of tri. ACB = y^2 / 2
The a ...

好!
可以修改

兩個tri.係一樣
tri. ACB = y^2 / 2
tri. ADB =AB^2 x 0.5 x sin∠ABD
           =2 x y^2 x 0.5 x sin∠ABD

since tri. ADB =tri. ACB
y^2 / 2 =2 x y^2 x 0.5 x sin∠ABD
y^2 / 2 =y^2 x sin∠ABD
sin∠ABD =1/2
∠ABD =30

JOHNLAM17

Originally posted by lpgexe at 2005-5-2 11:52 AM:
答案弄到手!
因為太興奮所以x個月來首度開腔!

let CB be y cm
CB = AC = y(Given)
AB = 開方 (y^2 + y^2) = 開方 (2 x y^2) = 開方(2) x y
AB = DB = 開方(2) x y
The area of tri. ACB = y^2 / 2
The a ...

原來係依課...
未教ar...唔掛得之我唔識...
果個廢miss教極都係coordinate果課=.=

碧咸

Originally posted by lpgexe at 2005-5-2 11:52 AM:
答案弄到手!
因為太興奮所以x個月來首度開腔!

let CB be y cm
CB = AC = y(Given)
AB = 開方 (y^2 + y^2) = 開方 (2 x y^2) = 開方(2) x y
AB = DB = 開方(2) x y
The area of tri. ACB = y^2 / 2
The a ...

有問題
個height唔係     y     咩
                   2(√2)