HMMT 二月 2000 · 几何 · 第 5 题
HMMT February 2000 — Geometry — Problem 5
题目详情
英文原题
- 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 .
C BFGA
E Dp
解析
英文解析
- 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 Æarea of 4 AB F - area of 4 F GC = sin 60 sin 60 = 4 sin 60 . Therefore, area of
2 2
17 3 p
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
3 p 2