PUMaC 2010 · 几何（B 组） · 第 3 题
PUMaC 2010 — Geometry (Division B) — Problem 3
题目详情
- As in the following diagram, square ABCD and square CEF G are placed side by side (i.e. C is between B and E and G is between C and D ). Now CE = 14, AB > 14, compute the minimal area of 4 AEG . 1
解析
- As in the following diagram, square ABCD and square CEF G are placed side by side (i.e. C is between B and E and G is between C and D ). Now CE = 14, AB > 14, compute minimal area of 4 AEG . [Answer] 98 [Solution] [Solution] Connect AC, Note that two triangles AEG and CEG share same base and have equal height. So the area of 4 AEG is equal to area of 4 CEG = 14 × 14 / 2 = 98.