A wire PQ is bent at 3 points X, Y and Z to form 3 right angles as shown in the figure. PX=(x-1)cm, XY=(3x)/2 cm , YZ=(2x+1), ZQ=(5x)/2 cm and the shortest distance between P and Q is 40 cm.
(a)What is the value of x ?
(b)What is the length of the wire PQ?
Plz
PS: I don't know how to put the figure/picture on here,anyone know how to put plz answer me.
Challenging problem F.2 Math
2013-11-24 6:44 am
回答 (6)
2013-11-24 5:02 pm
✔ 最佳答案
Kwok sir, 我覺得幅圖應該是像樓梯級的。我的作法如下：
(a) Extend PX and QZ to meet at R,
so triangle PQR is a right-angled triangle with right angle at R,
PR = (x - 1) + (2x + 1) = 3x
QR = 3x/2 + 5x/2 = 4x
As PQ = 40, using Pythagoras Theorem,
(3x)^2 + (4x)^2 = 40^2
==> x = 8 or -8 (rej)
(b) Length of PQ :
PX + XY + YZ + ZQ
= 7x
= 56 (cm)
2013-11-24 6:17 pm
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候儺咥労
候儺咥労
2013-11-24 5:24 pm
2013-11-24 8:49 am
Alvin:
把圖upload到以下的網站，再把link貼出：
http://imgur.com/
http://postimage.org/
http://upload.lsforum.net/index.php?user=upload
把圖upload到以下的網站，再把link貼出：
http://imgur.com/
http://postimage.org/
http://upload.lsforum.net/index.php?user=upload
2013-11-24 8:45 am
我大至明白個圖會是怎樣的。
從 P 劃線至 QZ 並且垂直於 QZ，交點為 K。
QK = 5x/2 - 3x/2 = 2x/2 = x
PK = (2x+1) - (x-1) = 2x + 1 - x + 1 = x + 2
用畢氏定理可得 40² = QK² + PK² = x² + (x+2)²
1600 = x² + x² + 2x + 4
2x² + 2x + 4 = 1600
x² + x = 798
(x² + x + 1/4) = 798 + (1/4)
[x + (1/2)]² = 798 + (1/4)
x + (1/2) = √[798 + (1/4)] = 28.253318
x = 27.7533 cm
2013-11-24 00:49:14 補充：
PQ 長度 = (x-1) + (3x/2) + (2x+1) + (5x/2) = 7x+2 = 196.2731 cm
從 P 劃線至 QZ 並且垂直於 QZ，交點為 K。
QK = 5x/2 - 3x/2 = 2x/2 = x
PK = (2x+1) - (x-1) = 2x + 1 - x + 1 = x + 2
用畢氏定理可得 40² = QK² + PK² = x² + (x+2)²
1600 = x² + x² + 2x + 4
2x² + 2x + 4 = 1600
x² + x = 798
(x² + x + 1/4) = 798 + (1/4)
[x + (1/2)]² = 798 + (1/4)
x + (1/2) = √[798 + (1/4)] = 28.253318
x = 27.7533 cm
2013-11-24 00:49:14 補充：
PQ 長度 = (x-1) + (3x/2) + (2x+1) + (5x/2) = 7x+2 = 196.2731 cm
2013-11-24 7:53 am
比個fb link尼貼圖~
收錄日期: 2021-04-13 19:50:40
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20131123000051KK00245