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1. http://img262.imageshack.us/i/maths7002.jpg/ 第29題點做呀..我唔識做..thx~
最佳解答:
29) a.) i) Let x be the length of Q1R1 PH - Q1A1 = t - n QR - A1B1 = s -n The area of the biggest triangle PQR = st / 2 n^2 + n(t - n)/2 + n(s - n)/2 = st / 2 2n^2 + n(t - n) + n(s - n) = st n( t - n + s - n + 2n ) = st n(s + t) = st n = st / (s+t) Therefore, Q1R1 = st / (s+t) ii)Similar to a(i) , Let y be the length of Q2R2 the area of triangle PQ1R1 = ( st / (s+t) )( s - ( st / (s+t) ) ) /2 y = ( t / (s+t) )(1 - t / (s+t) ) y = st / (s+t)^2 Let z be the length of Q3R3 z = st / (s+t)^3 b)i) T1 = ( st / (s+t) )^2 T1 = (st)^2 / (s+t)^2 T2 = ( st / (s+t)^2 )^2 T2 = (st)^2 / (s+t)^4 ii) Tn = (st)^2 / (s+t)^2n c)唔識做==
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數學一題等比問題發問:
1. http://img262.imageshack.us/i/maths7002.jpg/ 第29題點做呀..我唔識做..thx~
最佳解答:
29) a.) i) Let x be the length of Q1R1 PH - Q1A1 = t - n QR - A1B1 = s -n The area of the biggest triangle PQR = st / 2 n^2 + n(t - n)/2 + n(s - n)/2 = st / 2 2n^2 + n(t - n) + n(s - n) = st n( t - n + s - n + 2n ) = st n(s + t) = st n = st / (s+t) Therefore, Q1R1 = st / (s+t) ii)Similar to a(i) , Let y be the length of Q2R2 the area of triangle PQ1R1 = ( st / (s+t) )( s - ( st / (s+t) ) ) /2 y = ( t / (s+t) )(1 - t / (s+t) ) y = st / (s+t)^2 Let z be the length of Q3R3 z = st / (s+t)^3 b)i) T1 = ( st / (s+t) )^2 T1 = (st)^2 / (s+t)^2 T2 = ( st / (s+t)^2 )^2 T2 = (st)^2 / (s+t)^4 ii) Tn = (st)^2 / (s+t)^2n c)唔識做==
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