Math questions about cones~~

[匿名]

2010-05-22 1:43 am

Don't know how to solve these problems......can anyone help?

1) A piece of paper in the shape of a circular sector is bent to form a right circular co ne. The radius of the sector is 5 cm and the area of the sector is 20π cm3.

a) Find the slant height of the cone.
b) Find the base radius of the cone.
c) Find the volume of the cone.

2) A right circular cone of base radius 8 cm and height 20 cm is cut into two portions by a plane parallel to its base. The upper part is removed and the height of the remaining frustum is 15 cm.

a) Find the radius of the upper part of the frustum.
b) Find the volume of the frustum.
c) Find the total surface area of the frustum.

3) A cylindrical hole with base radius 4 cm is drilled through the centre of the base of a metallic solid cone with base radius 12 cm and height 30 cm.

a) Find the height of the cylindrical hole.
b) Find the volume of the remaining portion.
c) If the density of the metal is 6.5g/cm3, find the weight of the remaining portion

Thankssss :D

更新1:

sorry.....still have one more.... 4) The surface area of a spherical balloon is increased by 21%. Find a) the percentage change in radius. b) the percentage change in volume.

更新2:

多謝你呀:DD 不過我唔係好明2c....你可唔可以解釋下? Thankssss sooooo much!

回答 (1)

STEVIE-G™

2010-05-22 2:18 am

✔ 最佳答案

1a)
the slant height=radius of the sector=5cm

b)
Let the angle at the center be @
(@/360)*π*(5)^2=20π
@=288

arc length of the sector=(288/360)*2π(5)=8π
2*π*(base radius)=8π
base radius=4cm

c)
volume of the cone
=(1/3)πr^2(h)
=(1/3)π(4^2)(√(5^2-4^2))
=16π cm^3

2a)
Using property of similar triangle,
let the radius of the upper part be r

20/8=(20-15)/r
5/2=5/r
r=2

b)
volume of the frustum
=(1/3)π(8^2)*20-(1/3)π(2^2)(20-15)
=420π cm^3

c)
total surface area
=π(2^2)+π(8^2)+π(2)(√(5^2+2^2))+π(8)(√(8^2+20^2))
=(68+34√29)π cm^2

3a)
Let h be the height of the hole

Using property of similar triangle
(30-h)/4=30/12
12(30-h)=30*4
h=20

b)
volume of the remaining part
=(1/3)π(12^2)(30)-π(4^2)(20)
=1120π cm^3

c)
the weight
=1120π*6.5
=7280π g

2010-05-21 18:33:16 補充：
4a)
Surface area of a sphere=4πr²
Let the radius be r
4π(r')²=(1.21)(4πr²)
r'=1.1r

The percentage increase in radius=(1.1r-r)/r*100%=10%

b)
New volume V'=(4/3)π(r')³=1.331*(4/3)πr³=1.331V (V=original volume)
percentage increased in volume
=(1.331V-V)/V*100%
=33.1%

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