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maths AL and CE question ,help
Help me for the following questions: http://blog.yahoo.com/_QKI2XC34JEFFNUWEBTTXBZYU4A/articles/1238059/index 更新: re:： LUNG. ( 小學級 4 級 ) how about the other
最佳解答:
1.(a) In Figure 7(a),let h cm be the height of the smaller cone. Then, we have h/(h+8)=3/6 h=8 The volume of the frustum =1/3pi (6^2)(16)-1/3pi(3^2)(8) =168pi cm^3 The volume of X =2/3pi(6^3)+168pi =312pi cm^3 the volume of Y =(開方9/4)^3(312pi) =(27/8)(312pi) =1053pi cm^3 1(b) The volume of X' =312pi +4/3pi(1^3) =940pi/3 cm^3 The volume of Y' =1053pi +4/3pi(2^3) =3191/3 pi cm^3 The volume of X'/The volume of Y' =940/3191 NOT equal to 8/27 Thus X' AND Y' ARE NOT SIMILAR. 2013-08-24 19:07:06 補充： 2(a) Let r cm and h cm be the base radius and the height of X respectively. 2πr=2π(OA)(216/360) 2πr=2π(20)(216/360) r=12 h=開方(OA^2-r^2) h=開方(20^2-12^2) h=16 (b)The volume of X =π/3 x (12^2)(16) =768π cm^3 2013-08-24 19:07:23 補充： c) Let S cm be the base radius of Y. 2πs=2π(10)(108/360) s=3 The slant height of Y/The slant height of X=10/20=1/2 The base radius of Y/The base radius of X=3/12=1/4 The slant height of Y/The slant height of X not equal to The base radius of Y/The base radius of X Thus, X and Y are not similar. 2013-08-24 19:15:31 補充： 3.(a)(i) The capacity of the container =π/3(12^2)(18) =864π cm^3 (ii) Let r cm be the base radius of the lower part of the container. Then, we have (18-6)/18=r/12 Solving, we have r=8. The volume of frustum =864π-π/3(8^2)(18-6) =864π-256π =608π cm^3 2013-08-24 19:19:34 補充： b(i) The capacity of the lower part of the vessel =π(8^2)(10) =640π cm^3 Let h cm be the depth of water in the vessel. [(h-10+(18-6))/18]^3=[(884π-640π)+(864π-608π)]/864π [(h+2)/18]^3=125/216 (h+2)/18=5/6 h=13 Thus, the depth of water in the vessel is 13cm. 2013-08-24 19:19:42 補充： (ii) The volume not occupied by water in the vessel =640π+640π-884π =364π cm^3 Note that the volume of metal is 1000cm^3. So,the volume of metal is less than 364π cm^3. Thus, the water wil not overflow 2013-08-24 19:27:00 補充： 4.(c)(i) The volume of the wooden block APQRES =2400(4/16)^2 =150CM^3 (ii)Note that PQ/BC=1/4 and QR/CD=1 Also note that the two ratios are not equal. Thus, the two blocks are not similar.
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maths AL and CE question ,help
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發問:Help me for the following questions: http://blog.yahoo.com/_QKI2XC34JEFFNUWEBTTXBZYU4A/articles/1238059/index 更新: re:： LUNG. ( 小學級 4 級 ) how about the other
最佳解答:
1.(a) In Figure 7(a),let h cm be the height of the smaller cone. Then, we have h/(h+8)=3/6 h=8 The volume of the frustum =1/3pi (6^2)(16)-1/3pi(3^2)(8) =168pi cm^3 The volume of X =2/3pi(6^3)+168pi =312pi cm^3 the volume of Y =(開方9/4)^3(312pi) =(27/8)(312pi) =1053pi cm^3 1(b) The volume of X' =312pi +4/3pi(1^3) =940pi/3 cm^3 The volume of Y' =1053pi +4/3pi(2^3) =3191/3 pi cm^3 The volume of X'/The volume of Y' =940/3191 NOT equal to 8/27 Thus X' AND Y' ARE NOT SIMILAR. 2013-08-24 19:07:06 補充： 2(a) Let r cm and h cm be the base radius and the height of X respectively. 2πr=2π(OA)(216/360) 2πr=2π(20)(216/360) r=12 h=開方(OA^2-r^2) h=開方(20^2-12^2) h=16 (b)The volume of X =π/3 x (12^2)(16) =768π cm^3 2013-08-24 19:07:23 補充： c) Let S cm be the base radius of Y. 2πs=2π(10)(108/360) s=3 The slant height of Y/The slant height of X=10/20=1/2 The base radius of Y/The base radius of X=3/12=1/4 The slant height of Y/The slant height of X not equal to The base radius of Y/The base radius of X Thus, X and Y are not similar. 2013-08-24 19:15:31 補充： 3.(a)(i) The capacity of the container =π/3(12^2)(18) =864π cm^3 (ii) Let r cm be the base radius of the lower part of the container. Then, we have (18-6)/18=r/12 Solving, we have r=8. The volume of frustum =864π-π/3(8^2)(18-6) =864π-256π =608π cm^3 2013-08-24 19:19:34 補充： b(i) The capacity of the lower part of the vessel =π(8^2)(10) =640π cm^3 Let h cm be the depth of water in the vessel. [(h-10+(18-6))/18]^3=[(884π-640π)+(864π-608π)]/864π [(h+2)/18]^3=125/216 (h+2)/18=5/6 h=13 Thus, the depth of water in the vessel is 13cm. 2013-08-24 19:19:42 補充： (ii) The volume not occupied by water in the vessel =640π+640π-884π =364π cm^3 Note that the volume of metal is 1000cm^3. So,the volume of metal is less than 364π cm^3. Thus, the water wil not overflow 2013-08-24 19:27:00 補充： 4.(c)(i) The volume of the wooden block APQRES =2400(4/16)^2 =150CM^3 (ii)Note that PQ/BC=1/4 and QR/CD=1 Also note that the two ratios are not equal. Thus, the two blocks are not similar.
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