AIME 2006 I · 第 1 题

AIME 2006 I — Problem 1

专题: Contest Math
难度: L4
来源: AIME

题目详情

Problem

In quadrilateral $ABCD$ , $\angle B$ is a right angle, diagonal $\overline{AC}$ is perpendicular to $\overline{CD}$ , $AB=18$ , $BC=21$ , and $CD=14$ . Find the perimeter of $ABCD$ .

解析

Solution 1

We construct the following diagram:

$AIME diagram$

Using the Pythagorean Theorem, we get the following two equations:

$AD^2 = AC^2 + CD^2$ $AC^2 = AB^2 + BC^2$ Substituting $AB^2 + BC^2$ for $AC^2$ gives us $AD^2 = AB^2 + BC^2 + CD^2$ . Plugging in the given information, we get $AD^2 = 18^2 + 21^2 + 14^2 = 961 \implies AD = 31$ , so the perimeter is $AB+BC+CD+AD = 18+21+14+31 = \boxed{084}$ .