Mathematics--mensuration

[匿名]

2011-02-03 9:04 pm

1.

圖片參考：http://imgcld.yimg.com/8/n/HA05298198/o/701102030025613873376680.jpg

Find the curved surface area of the right circular cone as shown in the figure

2. A metallic sphere with a radius of R is melted. How many identical right circular cones with radii and height R/8 can be recast?

3.

圖片參考：http://imgcld.yimg.com/8/n/HA05298198/o/701102030025613873376691.jpg

The figure shows a solid which is composed of a right circular cone and a hemisphere with the same base. If the ratio of the volume of the cone to that of the hemisphere is 5:3, find the ratio of r to h.

4.

圖片參考：http://imgcld.yimg.com/8/n/HA05298198/o/701102030025613873376692.jpg

In the figure, the height of the pyramid VPQRS is 8h. ABCD adn PQRS are squares with sides a and 8a respectively. Find the ratio of the volume of the frustum ABCDSPQR to that of the pyramid VPQRS.

5. If the surface area of a sphere decreases by 51%, find the percentage change in its volume.

6. The base radius of a right circular cone is 18cm. If the total surface area of the cone is 675pi sq.cm, find the slant height of the cone.

7. If a solid sphere is fitted exactly into a cubical container with sides 7 cm, find the volume of the empty space in the container after the sphere is put inside.
(Correct your answer to 2 decimal places.)

8.

圖片參考：http://imgcld.yimg.com/8/n/HA05298198/o/701102030025613873376703.jpg

Find the volume of the right rectangular pyramid as shown in the figure.
(Correct your answer to 3 sig.fig.)

Just try your best to answer! You don't have to answer all the questions if you don't know some of them! I need steps for all of them. Try Q1-4 first as I don't know how to do them. I just want to check my answer for Q5-8. THX!!!

回答 (1)

Godfrey

2011-02-04 7:07 pm

✔ 最佳答案

Question 1:
Cos 60º = r/d where r = radius and s = slant height
s = r/cos Cos 60º = 22 cm/0.5 = 22 cm
curved surface area = (pi) x (r) x (s)
curved surface area = (pi) x (11 cm) x (22 cm) = 243 pi sq. cm = 760.3 cm^2

Question 2:
Volume of sphere = 4/3 (pi) (R^3)
Volume of cone = 1/3 (pi) (r^2)(h) = 1/3 (pi) ((R/8)^2)(R/8) = 1/3 (pi) ((R/8)^3)
Number of cones = Volume of sphere/ Volume of cone
Number of cones = 4/3 (pi) (R^3) / 1/3 (pi) ((R/8)^3)
Number of cones = 4/3 (pi) (R^3) / 1/3 (pi) (0.125^3)(R^3) = 4/(0.125^3) = 2048

Question 3:
Volume of cone = 1/3 (pi)(r^2)(h)
Volume of hemisphere = ½ volume of sphere = ½ (4/3) (pi)(r^3)
Volume of cone / volume of hemisphere = 5/3
1/3 (pi)(r^2)(h)/ ½ (4/3) (pi)(r^3)= 5/3
1/3 (pi)(r^2)(h)/ 2/3 (pi)(r^3)= 5/3
h/ 2r = 5/3
2r/h = 3/5
r /h = 3/10
The ratio of r to h 3:10

Question 4:
Volume of frustum = Volume of pyramid VPQRS – Volume of pyramid VABCD
Volume of a square base pyramid = 1/3(s^2)(h)
Volume of pyramid VPQRS = 1/3[(8a)^2](8h)
Ratio of height to half side of square base is a constant
Height of pyramid VABCD/ a = 8h/8a
Height of pyramid VABCD = h
Volume of pyramid VABCD = 1/3[(a)^2](h)
Volume of frustum = 1/3[(8a)^2](8h) – 1/3[(a)^2](h)
Volume of frustum = 1/3[512(a)^2](h) – 1/3[(a)^2](h)
Volume of frustum = 1/3[511(a)^2](h)
The ratio of Volume of frustum to Volume of pyramid VPQRS
= 1/3[511(a)^2](h)/ 1/3[(8a)^2](8h)
= 1/3[511(a)^2](h)/ 1/3[512(a)^2](h)
= 511/512
The ratio of Volume of frustum to Volume of pyramid VPQRS 511 : 512

Question 5:
Surface of a sphere = 4 (pi) (r^2)
0.49 = 4 (pi) ((kr)^2)/ 4 (pi) (r^2)
0.49 = k^2
k = 0.7
radius decreases by 30%
Volume of a sphere = 4/3 (pi) (r^3)
Change in volume of a sphere = 4/3 (pi) [(0.7r)^3]/ 4/3 (pi) (r^3)
= 0.34.3
Volume decreases by (1 – 0.343)100 = 65.7%

Volume decreases by 65.7%

Question 6:
A = pir^2 + pi r s
675 pi cm^2 = pi(18^2) +pi 18s
675 = 324 + 18s
slant height, s = (675 – 324)/18 = 19.5 cm

Run out of space... Let other reader do the rest.

2011-02-04 11:26:51 補充：
Ques 7:
V = 7^3 - 4/3 (pi)(3.5)^3 = 163.41 cm^3

Ques 8:
V = 1/3(10)(6)(15.198) = 304 cm^3

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