AMC12 2015 A

AMC12 2015 A · Q16

AMC12 2015 A · Q16. It mainly tests Triangles (properties), Pythagorean theorem.

Tetrahedron $ABCD$ has $AB=5$, $AC=3$, $BC=4$, $BD=4$, $AD=3$, and $CD=\frac{12}{5}\sqrt{2}$. What is the volume of the tetrahedron?

四面体 $ABCD$ 满足 $AB=5$，$AC=3$，$BC=4$，$BD=4$，$AD=3$，且 $CD=\frac{12}{5}\sqrt{2}$。求该四面体的体积。

(A) $3\sqrt{2}$ $3\sqrt{2}$

(B) $2\sqrt{5}$ $2\sqrt{5}$

(D) $3\sqrt{3}$ $3\sqrt{3}$

(E) $\frac{24}{5}\sqrt{2}$ $\frac{24}{5}\sqrt{2}$

Answer

Correct choice: (C)

正确答案：（C）

Solution

Answer (C): Triangles $ABC$ and $ABD$ are $3-4-5$ right triangles with area $6$. Let $CE$ be the altitude of $\triangle ABC$. Then $CE=\frac{12}{5}$. Likewise in $\triangle ABD$, $DE=\frac{12}{5}$. Triangle $CDE$ has sides $\frac{12}{5}$, $\frac{12}{5}$, and $\frac{12}{5}\sqrt{2}$, so it is an isosceles right triangle with right angle $CED$. Therefore $DE$ is the altitude of the tetrahedron to base $ABC$. The tetrahedron’s volume is $\frac{1}{3}\cdot 6\cdot \frac{12}{5}=\frac{24}{5}$.

答案（C）：三角形 $ABC$ 和 $ABD$ 是面积为 $6$ 的 $3-4-5$ 直角三角形。设 $CE$ 为 $\triangle ABC$ 的高，则 $CE=\frac{12}{5}$。同理在 $\triangle ABD$ 中，$DE=\frac{12}{5}$。三角形 $CDE$ 的三边分别为 $\frac{12}{5}$、$\frac{12}{5}$ 和 $\frac{12}{5}\sqrt{2}$，因此它是一个等腰直角三角形，且直角在 $\angle CED$。因此 $DE$ 是该四面体到底面 $ABC$ 的高。四面体体积为 $\frac{1}{3}\cdot 6\cdot \frac{12}{5}=\frac{24}{5}$。

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