~阿maths阿~

Question

A right pyramid with a square base just fits insie A hemisphere. if a slant edge of the
pyramid is 4 cm, find the curved surface area of the hemisphere.


http://hk.geocities.com/leemingki21/6m.bmp
(圖)有一個2分之一的球體,,和一個埃及三角x
畫得不像對不起

Answer 1

參閱圖片(MATH相簿):
http://photos.yahoo.com/gensecardo
金字塔尖垂直於底部的直線長度為r(即球體的半徑),而底部的對角線被這垂直線分成兩半,長度各為r(因為正方形的對角線是球體的直徑)
根據pythagora's theorem, r^2+r^2 = 4^2
i.e. r^2 = 8

curved surface area=(4π r^2) /2 = 2π(8)=50.24cm^2

Answer 2

Cut a cross section of the pyramid across the opposite slant edges, you'll see:
1. a semi-circle which represents the hemisphere
2. a triangle with the diameter of the hemisphere as the base (=2r), and the two slant edges as the other sides. The top corner of the triangle is the top of the hemisphere, call it A, and the other two corners B and C.

Let D be the mid-point of the diameter BC, so D is also the centre of the semi-circle.
AB = 4 cm
BC=2r
AD=r
By Pythagoras theorem,
4^2=AB^2=BD^2+AD^2=2r^2
2r^2=16
r=2sqrt(2)
The curved surface of a hemisphere is half that of a sphere
S=(4 pi r^2)/2=2 pi r^2 = 2*pi*8 = 16 pi

Answer 3

Let r be the radius of the hemisphere

Inside the pyramid,
r^2 + r^2 = 4^2
r^2 = 8
r = 2(sqrt2)

the curved surface = 2(pi)r^2 = 16pi or 50.265 (cm^2)

2007-03-12 00:44:00 補充：
The diagram you drew is very good!!