Please do no.12 and no.15 showed on the following webpage.
http://cid-632ec8cb72303e4e.skydrive.live.com/self.aspx/ex/%e6%8e%83%e6%8f%8f0012.jpg
F.3 Area and Volume
2009-05-27 2:50 am
回答 (2)
2009-05-27 3:47 am
12. 160cm^3
15. 8400π
2009-05-26 19:47:53 補充：
12.
Let h be height of the pyramid.
h=√(132-52)
=√144
=12cm
The volume of the pyramid=1/3(1/2x10x8)x12
=160
=165cm3
圖片參考：http://g.imagehost.org/0232/ScreenHunter_02_May_26_19_47.gif
15. 8400π
2009-05-26 19:47:53 補充：
12.
Let h be height of the pyramid.
h=√(132-52)
=√144
=12cm
The volume of the pyramid=1/3(1/2x10x8)x12
=160
=165cm3
圖片參考：http://g.imagehost.org/0232/ScreenHunter_02_May_26_19_47.gif
2009-05-27 3:08 am
12.
let h be height
h^2+(10/2)^2=13^2
h=12
base area={[(10/2)X8]/2} X2 =40
volume=1/3 X40X12=160(cm3)
15.
VQ= l = 29
A= 兀rl
580兀=兀(29)r
r=20
h^2+20^2=29^2
h=21
V=1/3兀(20)^2(21)
V=2800兀(cm3)
let h be height
h^2+(10/2)^2=13^2
h=12
base area={[(10/2)X8]/2} X2 =40
volume=1/3 X40X12=160(cm3)
15.
VQ= l = 29
A= 兀rl
580兀=兀(29)r
r=20
h^2+20^2=29^2
h=21
V=1/3兀(20)^2(21)
V=2800兀(cm3)
參考: me
收錄日期: 2021-04-23 20:43:19
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20090526000051KK01329