40 marks - f3 maths

2009-05-22 10:01 pm

1. The volume of a cone is 108兀cm ^3 and the raio of its height to base radius is 3: 2 . Find a) the base radius and the slant edge of the cone. b) the curved surface area of the cone.
2.A metallic solid cone with base radius 12cm and height 30cm. Acylindricaql hole with base radius 4cm is drilled through the center of the base of the cone.find a) the height of the cylindrical hole, b) the volume of the remaining portion c) if the density of the metal is 6.5 g/ cm^3, find the weight of the remaining portion.
3. The surface area of a spherical balloon is increased by 21%. find a) the percentage change in radius b) the percentage change in volume.
4.If the volume of a sphere is increased by 33.1%, what will be the percentage increase in its radius?

回答 (1)

老爺子

2009-05-22 11:08 pm

✔ 最佳答案

1.
a)
Let 3y cm be the height of the cone.
Then, base radius of the cone = 2y cm

Volume of the cone:
(1/3)π(2y)2(3y) = 108π
4y3 = 108
y = 3

base radius
= 2y cm
= 6 cm

height
= 3y cm
= 9 cm

slant edge
= √[(6)2 + (9)2] cm
= 3√13 cm
(or 10.82 cm)

b)
The curve surface area of the cone
= (1/2)[2π(6)](3√13) cm2
= 18π√13 cm2
(or 203.9 cm2)

2.
(a)
Let h cm be the height of the cylindrical hole.
30/12 = (30 - h)/4
h = 20
Height of the cylindrical hole = 20 cm

(b)
Volume of the remaining portion
= [(1/3)π(12)2(30) - π(4)2(20)] cm3
= 1120π cm3
(or 3518.6 cm3)

(c)
Weight of the remaining portion
= 1120π x 6.5 g
= 7280π g
(or 22871 g)

3.
a)
Old surface area : New surface area
= 1 : (1 + 21%)
= 1 : 1.21

Old radius : New radius
= √1 : √1.21
= 1 : 1.1

% change in radius
= [(1.1 - 1)/1] x 100%
= +10%

b)
Old volume : New volume
= (1)3 : (1.1)3
= 1 : 1.331

% change in volume
= [(1.331 - 1)/1] x 100%
= +33.1%

4.
Old volume : New volume
= 1 : (1 + 33.1%)
= 1 : 1.331

Old radius : New radius
= (1)1/3 : (1.331)1/3
= 1 : 1.1

% increase in radius
= [(1.1 - 1)/1] x 100%
= 10%
=

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