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1.At present, a father and his daugther are x years old and y years old respectively.The sum of the squares of their ages is 936. Six years later, the age of the father is three times the age of his daughter.Find the present age of the father and his daugther.2.ABCD and DEFG are two squares.If EC =4cm and the... 顯示更多 1.At present, a father and his daugther are x years old and y years old respectively.The sum of the squares of their ages is 936. Six years later, the age of the father is three times the age of his daughter.Find the present age of the father and his daugther. 2.ABCD and DEFG are two squares.If EC =4cm and the sum of their areas is 400 cm^2,find the lengths of the sidesvof the two squares.

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1.At present, a father and his daugther are x years old and y years old respectively.The sum of the squares of their ages is 936. Six years later, the age of the father is three times the age of his daughter.Find the present age of the father and his daugther. x^2 + y^2 = 936 --------(1) x+6 = 3y+18 x = 3y + 12 -------------(2) Sub (2) into (1), (3y+12)^2 + y^2 = 936 9y^2 + 72y + 144 + y^2 = 936 10y^2 + 72y - 792 = 0 5y^2 + 36y - 396 = 0 (y-6)(5y+66) = 0 y = 6 or -66/5 (rej) Sub y = 6 into (2) x = 3(6) + 12 x = 30 The present age of his daughter is 6 years old . The present age of the father is 30 years old. 2.ABCD and DEFG are two squares.If EC =4cm and the sum of their areas is 400 cm^2,find the lengths of the sidesvof the two squares. Let S1 cm and S2 cm be the side of ABCD and DEFG repectively, A1 cm^2 and A2 cm^2 be the area of ABCD and DEFG respectively, A1 + A2 = 400 (S1)^2 + (S2)^2 = 400 ----------(1) S1 + S2 = 4 S1 = 4 - S2 -------------(2) Sub (2) into (1), (4-S2)^2 + (S2)^2 = 400 16 - 8(S2) + (S2)^2 + (S2)^2 = 400 2(S2)^2 - 8(S2) - 384 = 0 (S2)^2 - (S2) - 192 = 0 S2 = 14.4 or -13.4 (rej) Sub S2 into (2), S1 = 4 - 14.4 S1 = -10.4 Therefore, Length of side of ABCD is 10.4cm and length of side of DEFG is 14.4cm 圖片參考:http://imgcld.yimg.com/8/n/HA00526499/o/701003100161113873425410.jpg

 

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