數學大難題！！！！(共9題)

用戶190083

2007-07-19 12:32 am

1.圖中的立體是由一個半球體和一個圓柱組成的，若圓柱體高度與底半徑的比是2:1，而此立體的總表面面積是252πcm2次，求
a.高度h
b.底半徑r
c.立體的體積（答案以π表示）
圖：(&b=2&(這個位係冇;的)f=190113)
http://www.wretch.cc/album/show.php?i=Puchikoyiu&b=2&f=1901133627&p=0
2.一個正方形體外接一個半徑是√8cm的球體，求
a.正方體的邊長
b.球體與正方體之間的體積（取π＝3.14)
圖：(&b=2&(這個位係冇;的)f=190113)
http://www.wretch.cc/album/show.php?i=Puchikoyiu&b=2&f=1901133628&p=1
3.在圖中，ABCD是一個平行四邊珍，PA及PD分別是∠A及∠D的角平分線，證明∠APD＝90度
圖：(&b=2&(這個位係冇;的)f=190113)
http://www.wretch.cc/album/show.php?i=Puchikoyiu&b=2&f=1901133629&p=2

回答 (1)

用戶82723

2007-07-19 2:33 am

✔ 最佳答案

1.According to the question, h=2r
a)總表面積
= 4Pi(r^2) / 2 + 2 Pi rh + Pi (r^2)
= 3Pi (r^2) + 2 Pi rh
= 3Pi (h/2)^2 + 2 Pi (h^2)/2
= 3/4 Pi h^2 + Pi h^2
= 7/4 Pi h^2

Hence,
7/4 Pi h^2 = 252 Pi
h^2 = 144
h = 12 (because h > 0)

b) h = 2r
12 = 2r
r = 6

c) 體積
= [1/2 x 4/3 Pi (6)^3 + Pi (6^2) x 12] cm^3
= 576 Pi cm^3

2a)Let a cm be the length of the side of the cube.
a^2 + (a^2 + a^2) = (2√8)^2
3a^2 = 32
a^2 = 32/3
a = √(32/3) (because a>0)
So the length of each side of the cube is √(32/3) cm.

b) Volume between the sphere and the cube
= {4/3 Pi (√8) ^ 3 - [√(32/3)]^3 } cm^3
= 59.90 cm^3 (corr. to 2 d.p.)

3.Let ∠PDC = x
Then ∠PDA = x
∠DAB = 180 degree - 2x (int. ∠s, AB//CD)\
∠DAP = ∠DAB/2 = 90 degree - x

In triangle DAP,
∠DAP + ∠PDA + ∠APD = 180 degree (∠ sum of triangle)
90 degree - x + x + ∠APD = 180 degree
Hence, ∠APD = 90 degree

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