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当一个多边形的周长一定且有一边长固定时,其面积的最大值在何时取到? 对于三角形,利用海伦公式,易得定理1 设△ABC中,BC=m(定长)且AB+AC=n(定长),(n>m)则当且仅当AB=AC=n/2时,△ABC的面积最大,且最大值为S_(max)=(m/4)(n~2-m~2)~(1/2)证明S=(m+n/2(m+n/2-m)(m+n/2-AC)(m+n/2-AB)~)(1/2)≤(1/2)(n~2-m~2)~(1/2)·((m+n)-(AB+AC))/2=(m/4)(n~2-m~2)~(1/2)显然其中等式成立的充要条件是AB=AC=n/2 对于四边形,也有类似的结果; 定理2 设四边形ABCD中,AB=a(定长),并且有BC+CD+DA=3r(定长)(3r>a),则当且仅当
When a polygon has a constant circumference and one side has a fixed length, when is the maximum value of its area taken? For triangles, using Helen’s formula, the easy theorem 1 is set in △ABC, BC=m (fixed length) and AB +AC=n (fixed length), (n>m), if and only if AB=AC=n/2, △ABC has the largest area, and the maximum value is S_(max)=(m/4)(n ~2-m~2)~(1/2) proved that S=(m+n/2(m+n/2-m)(m+n/2-AC)(m+n/2-AB)~ (1/2)≤(1/2)(n~2-m~2)~(1/2)((m+n)-(AB+AC))/2=(m/4)( n~2-m~2)~(1/2) Obviously, the necessary and sufficient condition for the equality to be established is AB=AC=n/2. For the quadrilateral, there are similar results. Theorem 2 In the quadrilateral ABCD, AB=a ( Fixed length), and there is BC+CD+DA=3r (fixed length) (3r>a), if and only if