maths figures 問題

1.When the height is 10 cm, Let the base area be A1.
When the height is 18 cm, Let the base area be A2.

(1/3)(A1)(10) = 1500
A1 = 450 cm²

(A2)/(A1) = (18/10)²Since they are similar figures

A2 = (18/10)²(A1) = 1458 cm²

Total volume : (1/3)(1458)(18) = 8748 cm³

Volume of water that should be added in order to fill the pyramid completely :

8748 cm³ - 1500 cm³ = 7248 cm³

2014-03-13 20:03:56 補充：
2.　　　Let the radius of the circle containing point J be r1 and radius of the circle containing point M be r2.

　　　　Let height of cone OJK be h1 and height of cone OMN be h2

　　　　Volume of cone OJK = (1/3)(π)(r1)²(h1)

　　　　Volume of cone OMN = (1/3)(π)(r2)²(h2)

2014-03-13 20:08:27 補充：
　　　　r1 : r2 = OJ : OM = 2 : 5

　　　　h1 : h2 = OJ : OM = 2 : 5

　　　　 Volume of cone OMN : Volume of cone OJK = (1/3)(π)(r2)²(h2) : (1/3)(π)(r1)²(h1)

　　　　= (r2/r1)²(h2/h1) = (5/2)²(5/2) = 125/8

　　　　Ratio of volume of cone OJK to that of the frustum

2014-03-13 20:14:21 補充：
Let V1 = volume of cone OJK ; V2 = volume of cone OMN。

　　　　Ratio of volume of cone OJK to that of the frustum

　　　　= volume of cone OJK / (volume of cone OMN - volume of cone OJK)

　　　　= V1 / (V2 - V1)

　　　　= (V1/V1) / [(V2 - V1)/V1] = 1 / (V2/V1 - 1) = 1 / (125/8 - 1) = 1 / (117/8)

　　　　

2014-03-13 20:14:43 補充：
　　　　= 8 : 117