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標題: [Core] 請教中四數學題 [打印本頁]

作者: rchan626 時間: 2020-11-3 03:35 PM 標題: 請教中四數學題

中四數學題 part (b) 計來計去都計唔到, 求高人指點, 謝謝.

[ 本帖最後由 rchan626 於 2020-11-3 05:22 PM 編輯 ]

作者: ehbb 時間: 2020-11-3 06:28 PM

known:
1. DE = FG = k
2. AC = 5 (pyth thm)
3. shortest distance from B to AC
thus height of triangle BDG = (a) - k

let DG = EF = y, Area of DEFG = ky
let AE = x, then FC = 5-x-y

AND, Area DEFG = Area of triangle ABC - (Area of triangle BDG + Area of triangle ADE + Area of triangle FCG)
= 3x4/2 - {y[(a)-k]/2 + kx/2 + k(5-x-y)/2}
= 6 - {y[(a)-k]/2 + kx/2 + (5k-kx-ky)/2}
= 6 - {y[(a)-k]/2 + kx/2 + 5k/2 - kx/2 - ky/2}
= 6 - {y[(a)-k]/2 + 5k/2 - ky/2}

Solve for y
如果佢等於5/12(12-5k)可以嗎?
p.s. Area of DEFG = ky = k [5/12(12-5k)]

作者: 風之男 時間: 2020-11-3 07:38 PM

My first instinct with these "rectangle inscribed in a polygon" questions is to look out for similar triangles.

Note that triangles ABC, AED, DBG and GFC are all similar [Why? That's your homework].
Now, by considering AE and FC, can you express EF in terms of k? This hint basically spoils the outline of the solution.

Surprisingly, you don't need to use (a) at all to complete (b) or (c) [Ugh], although it does raise the interesting question "what does it mean by k=12/5 [hmm... where did you see this number before?] or, as odd as it sounds, k=0?".

Btw, your answer might end up looking similar but not quite the same. Don't worry, they are equivalent when you expand out the "to show" expression.

[ 本帖最後由風之男於 2020-11-3 07:57 PM 編輯 ]

作者: rchan626 時間: 2020-11-3 09:59 PM

引用:

原帖由 ehbb 於 2020-11-3 06:28 PM 發表
known:
1. DE = FG = k
2. AC = 5 (pyth thm)
3. shortest distance from B to AC
thus height of triangle BDG = (a) - k

let DG = EF = y, Area of DEFG = ky
let AE = x, then FC = 5-x-y

AND, Area D ...

我既計法都係用相同扣減方法, 計到大約淨下 [12-5k+y(2k-2.4)]/2 就停左, 無辦法變到問題所問既 (5k/12)(12-5k).

作者: 風之男 時間: 2020-11-3 10:54 PM

引用:

原帖由 rchan626 於 2020-11-3 09:59 PM 發表

我既計法都係用相同扣減方法, 計到大約淨下 [12-5k+y(2k-2.4)]/2 就停左, 無辦法變到問題所問既 (5k/12)(12-5k).

I'm not sure how exactly did you get to that expression, maybe there are some algebra errors or something.
Now let me rephrase ehbb's method a little bit, so it "looks" nicer.

Suppose AE = x, EF = y, FC = 5-x-y
How about instead of thinking the area of the rectangle, think about the big triangle ABC?
[It doesn't make a difference, but adding things up instead of subtracting things away "feels" nicer]

One way of expressing the area of ABC is, well, the obviously way - it's just 3*4/2 = 6.
The other way is adding up all the 4 parts - 3 triangles and a rectangle. ehbb did a great job as to what the 4 small parts' areas are.
The trickiest one is DBG, where the base is DG, and the height is what you get from (a) minus k [Why? Pause and ponder]
Therefore, the long chain of adding up areas should equal to 6, which then you have to simplify...
Now don't forget our goal, we want to know what EF (i.e. y) is in terms of k.

I still prefer utilizing similar triangles more, but this approach works.

[ 本帖最後由風之男於 2020-11-3 10:58 PM 編輯 ]

作者: 風之男 時間: 2020-11-3 11:19 PM

Here's a bonus desmos for you.
https://www.desmos.com/calculator/xspfchzosk?lang=zh-TW

For now just assume c as any number that dictates how the rectangle looks.
You can see how the k value (length of FG) as the rectangle varies, which answers the question "what does it mean by k=12/5".

[ 本帖最後由風之男於 2020-11-3 11:21 PM 編輯 ]

作者: rchan626 時間: 2020-11-4 11:25 AM

計到了, 原本自己計既方向都係正確, 只係去到埋尾突然個腦轉唔到, 唔該晒各位.

Area of DEFG=[12-5k+y(2k-2.4)]/2
yk=[12-5k+y(2k-2.4)]/2
2yk=12-5k+y(2k-2.4)
2yk=12-5k+2yk-2.4y
2.4y=12-5k
y=(12-5k)/2.4
y=5(12-5k)/12

therefore yk=5k(12-5k)/12

作者: xmanhk 時間: 2024-3-31 02:36 PM

where is the question?

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