標題:
area and volume (III)
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最佳解答:
1. Height of the cylinder=10cm-1cm=9cm Volume of the test tube =TT(1^2)(9)+[(4/3)TT(1^3)]/2 =9TT+(2/3)TT =(29/3)TT cm^3 2. Let x cm^2 be the curved surface area of the small circular cone. (2/5)^2=x/125 4/25=x/125 x=20 Therefore the curved surface area of the small circular cone=20cm^2 3a. Height of the equil. triangle=√(8^2-4^2)=√48 --------- (Pyth. thereom) Area = (8*√48)/2 = 4√48 = 16√3 cm^2 3b. Volume = (1/3)*(16√3)*9 = 48√3 cm^3
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area and volume (III)
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唔識呀!!! help me pls!! thx!!1. A test tube consisting of a cylindrical upper part of radius 1cm and a hemispherical lower part of the same radius. If the height of the test tube is 10cm, find the volume of the space in the test tube. (Give the answer in the terms of π.)2. There are two similar circular... 顯示更多 唔識呀!!! help me pls!! thx!! 1. A test tube consisting of a cylindrical upper part of radius 1cm and a hemispherical lower part of the same radius. If the height of the test tube is 10cm, find the volume of the space in the test tube. (Give the answer in the terms of π.) 2. There are two similar circular cones. The ratio of their base diameters is 2 : 5. If the curved surface area of the large circular cone is 125cm^2, find the curved surface area of the small circular cone. 3. (a) There is an equilateral triangel of side 8cm. Find its area. (b) The base of a pyramid is the equilateral triangle shown in part (a), and the height of the pyamid is 9cm. Find the volume of the pyramid. (Leave the answers in surd form.)最佳解答:
1. Height of the cylinder=10cm-1cm=9cm Volume of the test tube =TT(1^2)(9)+[(4/3)TT(1^3)]/2 =9TT+(2/3)TT =(29/3)TT cm^3 2. Let x cm^2 be the curved surface area of the small circular cone. (2/5)^2=x/125 4/25=x/125 x=20 Therefore the curved surface area of the small circular cone=20cm^2 3a. Height of the equil. triangle=√(8^2-4^2)=√48 --------- (Pyth. thereom) Area = (8*√48)/2 = 4√48 = 16√3 cm^2 3b. Volume = (1/3)*(16√3)*9 = 48√3 cm^3
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