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HMMT 二月 2000 · 几何 · 第 5 题

HMMT February 2000 — Geometry — Problem 5

专题
Discrete Math / 离散数学
难度
L3
来源
HMMT

题目详情

  1. Side AB = 3. 4 AB F is an equilateral triangle. Side D E = AB = AF = GE . 6 Æ F E D = 60 . F G = 1. Cal ulate the area of AB C D E . A G F C B E D p
解析
  1. The area of 4 AB F = bh = 3 h = (3 sin 60 ) = sin 60 . The area of 4 F C D E = 2 2 2 2 9 1 Æ Æ Æ area of 4 AB F - area of 4 F GC = sin 60 sin 60 = 4 sin 60 . Therefore, area of 2 2 p 17 3 9 17 Æ Æ Æ AB C D E = area of 4 AB F + area of F C D E = sin 60 + 4 sin 60 = sin 60 = . 2 2 4 p 3 2