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\(\sqrt{2^2-H^2}+\sqrt{3^2-H^2}=4-1,\frac{S}{4+1}=\frac{H}{2},H=\frac{4\sqrt{2}}{3},S=\frac{10\sqrt{2}}{3}\)
这样也行。\(梯形A(4)B(3)C(1)D,延长AD,延长BC交于E,BE=4,AE=\frac{8}{3}\)
\(\frac{等腰三角形ABE面积}{梯形ABCD面积}=\frac{4^2}{4^2-1^2},等腰三角形ABE面积=\sqrt{4^2-(\frac{4}{3})^2}*\frac{8}{3}*\frac{1}{2}\) |
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