题目矩形的两邻边长为2和5,经过它的短边上的点作直线,使得所截得的直角三角形的周长为8,求矩形留下部分面积的最小值.图1这是2004年俄罗斯全国统一高考题,文[1]给出的解法太繁琐,本文将给出两种简捷解法,供读者参考.如图1,在矩形ABCD中,AB=5,BC=2,P、Q两点分别在边BC、CD上,为 The two adjacent sides of the title rectangle are 2 and 5, passing through the points on its short side as a straight line, so that the perimeter of the right-angled triangle is 8 and the minimum area of ​​the rectangle is left. Figure 1 This is In 2004, the unified national college entrance examination in Russia, the solution given in [1] is too tedious. This article will give two kinds of simple solutions for readers’ reference. As shown in figure 1, in the rectangle ABCD, AB=5, BC=2,P The two points Q and Q are on the side BC and CD respectively.