更新1: TO LUCY:都唔啱嘅
更新2: To ee*** :都唔系好啱
a) Let the vertical stick be x and the horizontal one will be 76-x. Area of the kite: 2x〔( 76-x )/ 2 〕 =76x-x^2 ------------------------------------------------------------------------------------------------------------ b) 76x - x^2 ≥ 720 x^2 - 76x + 720 ≤ 0 11.1 ≤ x ≤ 64.9 (corr. to 3 sig. fig.) 2010-10-11 22:57:48 补充： Sorry for texting wrong wer I think this might be right: 2010-10-11 22:57:53 补充： Let the vertical stick be x and the horizontal one will be 76-x. Area of the kite: 2(x〔( 76-x )/ 2 〕)/2 =38x-1/2x^2 ------------------------------------------------------------------------------------------------------------ b) 38x-1/2x^2 ≥ 720 x^2-76x+1440 ≤ 0 36 ≤ x ≤ 40
Let the horizontal stick = x cm. ( In fact it does not matter which stick is assumed to be x cm.) so the vertical stick = (76 - x) cm. long. Area of kite = 2 x Area of triangle with height (x/2) and base (76 - x) = 2[(x/2)(76 - x)]/2 = x(76 - x)/2 Now x(76 - x)/2 > 720 76x - x^2 > 1440 x^2 - 76x + 1440 < 0 (x - 40)(x - 36) < 0 so 36 < x < 40.