F3 maths 急~

2009-05-10 2:46 am

1) The slant height of a right circular cone is 6cm. If the total surface area of the cone is 51.84pie cm^2 , find the radius of the cone.

2) The figure shows a metal soild right circular cone with base radius 5cm and height 10cm. It is melted and recast into a cube of side x cm.
a) Find X
b) Find the percentage change in the total surface area when the right circular cone is recast into the cube.

3) The command module of a spacecraft is in the shape of a frustum of a right circular cone. The radius of the upper base is 0.5m and the radius of the lower base is 2m. The height of the module is 4m.
a) Find the volume of the module in terms of pie.
b) To protect the module, the curved surface of the module is coated
with heat shield. If the weight of 1m^2 of the heat shield is 38kg, find the total weight of the heat shield.

thz~~

回答 (1)

老爺子

2009-05-10 5:02 am

✔ 最佳答案

1)
Let r cm be the radius of the cone.

(1/2)(6)(2πr) + (πr2) = 51.84π
r2 + 6r - 51.84 = 0
(r - 4.8)(r + 10.8) = 0
r = 4.8 or r = -10.8 (rejected)

The radius of the cone = 4.8 cm

2)
a)
The volume is unchanged after recast into a cube:
x3 = (1/3)π(5)2(10)
x = 6.397

b)
Total surface area of the cone
= (1/2)(2π x 5)√[(5)2 + (10)2] cm2
= 175.62 cm2

Total surface area of the cube
= 6(6.397)2
= 245.53

% change in total surface area
= [(245.53 - 175.62)/175.62] x 100%
= +39.8%

3)
(a)
The frustum is the result of a small right circular cone of base radius 0.5 m from a large right circular cone of base radius of 2 m.

Let h cm be the height of the small cone.
h/0.5 = (h + 4)/2
2h = 0.5h + 2
h = 4/3

Volume of the frustum
= [(1/3)π(2)2(4 + 4/3) - (1/3)π(0.5)2(4/3)] m3
= 7π m3

(b)
Weight of the heat shield
= {(1/3)π(2)2√[(2)2 + (4+ 4/3)2] - (1/3)π(0.5)2√[(0.5)2 + (4/3)2]} x 38 kg
= 892.5 kg
=

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