2021 AMC 10A Spring 第 13 题

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所有题目均经美国数学协会(MAA)官方合法授权使用。

13.

四面体 ABCDABCD 的边长为 AB=2AB = 2AC=3AC = 3AD=4AD = 4BC=13BC = \sqrt{13}BD=25BD = 2\sqrt{5}CD=5CD = 5。它的体积是多少?

What is the volume of tetrahedron ABCDABCD with edge lengths AB=2,AB = 2, AC=3,AC = 3, AD=4,AD = 4, BC=13,BC = \sqrt{13}, BD=25,BD = 2\sqrt{5}, and CD=5?CD = 5?

33

232\sqrt{3}

44

333\sqrt{3}

66

答案:C
知识点:体积立体几何勾股定理
难度评级:1370
视频讲解:
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文字解答:

A=(0,0,0)A=(0,0,0)B=(2,0,0)B=(2,0,0)C=(0,3,0)C=(0,3,0)D=(0,0,4)D=(0,0,4)。则

BC=22+32=13,BD=22+42=25,CD=32+42=5, \begin{aligned} BC &= \sqrt{2^2+3^2}=\sqrt{13}, \\ BD &= \sqrt{2^2+4^2}=2\sqrt5, \\ CD &= \sqrt{3^2+4^2}=5, \end{aligned}

所以这个坐标模型与题目给出的所有棱长相符。三条从 AA 出发的棱互相垂直,长度分别为 2,3,42,3,4,因此体积为

16(2)(3)(4)=4.\frac16(2)(3)(4)=4.

所以正确答案是 C

Place A=(0,0,0),A=(0,0,0), B=(2,0,0),B=(2,0,0), C=(0,3,0),C=(0,3,0), and D=(0,0,4).D=(0,0,4). Then

BC=22+32=13,BD=22+42=25,CD=32+42=5, \begin{aligned} BC &= \sqrt{2^2+3^2}=\sqrt{13}, \\ BD &= \sqrt{2^2+4^2}=2\sqrt5, \\ CD &= \sqrt{3^2+4^2}=5, \end{aligned}

so this coordinate model matches all the given edge lengths. The tetrahedron is a rectangular-corner tetrahedron with perpendicular edge lengths 2,3,42,3,4 from A,A, so its volume is

16(2)(3)(4)=4.\frac16(2)(3)(4)=4.

Thus, C is the correct answer.

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