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HMMT 二月 2003 · 几何 · 第 1 题

HMMT February 2003 — Geometry — Problem 1

专题
Discrete Math / 离散数学
难度
L3
来源
HMMT

题目详情

  1. AD and BC are both perpendicular to AB , and CD is perpendicular to AC . If AB = 4 and BC = 3, find CD . D C A B
解析
  1. AD and BC are both perpendicular to AB , and CD is perpendicular to AC . If AB = 4 and BC = 3, find CD . D C A B Solution: 20 / 3 ◦ 6 6 6 By Pythagoras in 4 ABC , AC = 5. But CAD = 90 − BAC = ACB , so right triangles CAD, BCA are similar, and CD/AC = BA/CB = 4 / 3 ⇒ CD = 20 / 3.