1. In the figure,ABCD is a square of side 10cm.
Find the area of the shaded regionPQRS .
(Hints: Draw a line from S perpendicular to DC
and a line from R perpendicular to BC.)
圖如下:
圖片參考：http://imgcld.yimg.com/8/n/HA00805478/o/701108140012013873442360.jpg
ps. 圖可能有d比例不正確
唔該詳細解釋...Plz [中文]
要step
2.30x(1-r%) = 20x(1+r%)
唔該詳細解釋...Plz [中文]
要step
Thank you !
唔該哂!!
F1 超難題目!!Plz!!
2011-08-14 10:16 am
回答 (2)
2011-08-14 10:57 pm
✔ 最佳答案
1. Let AP = a and AQ = b.So PD = 10 - a
DS = AQ - 4 = b - 4.
SC = 10 - DS = 10 - (b - 4) = 14 - b.
QB = 10 - b.
BR = AP + 3 = a + 3
RC = 10 - BR = 10 - (a + 3) = 7 - a.
Area of triangle AQP = (AP)(AQ)/2 = ab/2.
" " " PDS = (PD)(DS)/2 = (10 - a)(b - 4)/2 = (10b - 40 - ab + 4a)/2.
" " " SRC = (SC)(RC)/2 = (14 - b)(7 - a)/2 = (98 - 14a - 7b + ab)/2
" " " QBR = (QB)(BR)/2 = (10 - b)(a + 3)/2 = (10a + 30 - ab - 3b)/2
So sum of area of 4 triangles
= (ab + 10b - 40 - ab + 4a + 98 - 14a - 7b + ab + 10a + 30 - ab - 3b)/2
= 88/2 = 44
So area of shaded region = 100 - 44 = 56 cm^2.
2.
Let r% = x
30(1 - x) = 20(1 + x)
30 - 30x = 20 + 20x
10 = 50x
x = 1/5 = 0.2 = 20%
so r = 20.
2011-08-14 10:49 pm
2.
30*(1-r%)=20*(1+r%)
30*(1-r/100)=20*(1+r/100)
30-3r/10=20+2r/10
10=5r/10
100=5r
r=20
30*(1-r%)=20*(1+r%)
30*(1-r/100)=20*(1+r/100)
30-3r/10=20+2r/10
10=5r/10
100=5r
r=20
收錄日期: 2021-04-25 22:40:29
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20110814000051KK00120