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Q8: http://postimg.org/image/ezugr032t/Q9(a): http://postimg.org/image/wr635gihh/Q11: http://postimg.org/image/jbj2g09zp/Q12(b)(ii): http://postimg.org/image/9fhzgd47p/http://postimg.org/image/uey37v5w5/http://postimg.org/image/ymsvgm7bp/Q13:... 顯示更多 Q8: http://postimg.org/image/ezugr032t/ Q9(a): http://postimg.org/image/wr635gihh/ Q11: http://postimg.org/image/jbj2g09zp/ Q12(b)(ii): http://postimg.org/image/9fhzgd47p/ http://postimg.org/image/uey37v5w5/ http://postimg.org/image/ymsvgm7bp/ Q13: http://postimg.org/image/wkse2d9cl/ http://postimg.org/image/wkse2d9cl/ http://postimg.org/image/49bfj2etx/
最佳解答:
Q8 (a) angle OPQ=292-202=90度 所以OPQ是直角三角形 (b) k^2=13^2-12^2 k=5 the perimeter of OPQ =5+13+12 =30 units Q9a coordinates of A' : (4-10,-2)=(-6,-2) coordinates of B' : (-7,3) Q11 (a) AC=DE,AD=CG(given) angle ACF=angle ADE(corr. angles, CG//DE) ADE全等GCA(SAS) (B)(i) angle HBF=angle HGA(alt. angles, BF//AG) angle FBH=angle AGH(alt. angles, BF//AG) BFH~GAH(AA) (b)(ii) ADE,GCA(FROM A) Q12bii 將prqs分做pqr同pqs兩個三角形 prqs的面積為(PQ*4)/2 =4 sq units 2014-04-01 19:11:08 補充： Q13a (i) the capacity of the circular conical container =10^2 pi * 15 * 1/3 =500 pi cm^3 (ii) the volume of the frustum =500pi * [1-(9/15)^3] =392pi cm^3 2014-04-01 19:18:57 補充： Q13b (i) 留意器皿內的水大於392pi 所以 FRUSTUM 一定係裝滿水 先減去FRUSTUM既水,然後再計圓柱體既水深 裝在圓柱的水=737.6pi - 392 pi=345.6 pi cm^3 Let x be the depth of the right circular cylinder Let y be the radius of the right circular cylinder y/10=9/15 y=6 6^2 pi * x =345.6 pi x=9.6 so that the depth of water in the vessel =6+9.6=15.6 2014-04-01 19:19:04 補充： (ii) total capacity of the vessel =392 pi +6^2 pi *16 =968 pi =3041 cm^3(cor. to nearest integer) the total volume of metal and water =737.6 pi +600 =2917 cm^3 少於3041 cm^3 所以,不會溢出
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help!! f.5 maths LQX4!!(20點)發問:
Q8: http://postimg.org/image/ezugr032t/Q9(a): http://postimg.org/image/wr635gihh/Q11: http://postimg.org/image/jbj2g09zp/Q12(b)(ii): http://postimg.org/image/9fhzgd47p/http://postimg.org/image/uey37v5w5/http://postimg.org/image/ymsvgm7bp/Q13:... 顯示更多 Q8: http://postimg.org/image/ezugr032t/ Q9(a): http://postimg.org/image/wr635gihh/ Q11: http://postimg.org/image/jbj2g09zp/ Q12(b)(ii): http://postimg.org/image/9fhzgd47p/ http://postimg.org/image/uey37v5w5/ http://postimg.org/image/ymsvgm7bp/ Q13: http://postimg.org/image/wkse2d9cl/ http://postimg.org/image/wkse2d9cl/ http://postimg.org/image/49bfj2etx/
最佳解答:
Q8 (a) angle OPQ=292-202=90度 所以OPQ是直角三角形 (b) k^2=13^2-12^2 k=5 the perimeter of OPQ =5+13+12 =30 units Q9a coordinates of A' : (4-10,-2)=(-6,-2) coordinates of B' : (-7,3) Q11 (a) AC=DE,AD=CG(given) angle ACF=angle ADE(corr. angles, CG//DE) ADE全等GCA(SAS) (B)(i) angle HBF=angle HGA(alt. angles, BF//AG) angle FBH=angle AGH(alt. angles, BF//AG) BFH~GAH(AA) (b)(ii) ADE,GCA(FROM A) Q12bii 將prqs分做pqr同pqs兩個三角形 prqs的面積為(PQ*4)/2 =4 sq units 2014-04-01 19:11:08 補充： Q13a (i) the capacity of the circular conical container =10^2 pi * 15 * 1/3 =500 pi cm^3 (ii) the volume of the frustum =500pi * [1-(9/15)^3] =392pi cm^3 2014-04-01 19:18:57 補充： Q13b (i) 留意器皿內的水大於392pi 所以 FRUSTUM 一定係裝滿水 先減去FRUSTUM既水,然後再計圓柱體既水深 裝在圓柱的水=737.6pi - 392 pi=345.6 pi cm^3 Let x be the depth of the right circular cylinder Let y be the radius of the right circular cylinder y/10=9/15 y=6 6^2 pi * x =345.6 pi x=9.6 so that the depth of water in the vessel =6+9.6=15.6 2014-04-01 19:19:04 補充： (ii) total capacity of the vessel =392 pi +6^2 pi *16 =968 pi =3041 cm^3(cor. to nearest integer) the total volume of metal and water =737.6 pi +600 =2917 cm^3 少於3041 cm^3 所以,不會溢出
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