AMC10 2021 A

AMC10 2021 A · Q13

AMC10 2021 A · Q13. It mainly tests Pythagorean theorem, 3D geometry (volume).

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

四面体 $ABCD$ 的边长 $AB = 2$，$AC = 3$，$AD = 4$，$BC = \sqrt{13}$，$BD = 2\sqrt{5}$，$CD = 5$，其体积是多少？

(A) 3 3

(B) 2\sqrt{3} 2\sqrt{3}

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

(E) 6 6

Answer

Correct choice: (C)

正确答案：（C）

Solution

Drawing the tetrahedron out and testing side lengths, we realize that the $\triangle ACD, \triangle ABC,$ and $\triangle ABD$ are right triangles by the Converse of the Pythagorean Theorem. It is now easy to calculate the volume of the tetrahedron using the formula for the volume of a pyramid. If we take $\triangle ADC$ as the base, then $\overline{AB}$ must be the altitude. The volume of tetrahedron $ABCD$ is $\dfrac{1}{3} \cdot \dfrac{3 \cdot 4}{2} \cdot 2=\boxed{\textbf{(C)} ~4}.$

画出四面体并检验边长，我们发现 $\triangle ACD$、$\triangle ABC$ 和 $\triangle ABD$ 由勾股定理的逆命题是直角三角形。现在使用金字塔体积公式很容易计算四面体的体积。如果取 $\triangle ADC$ 作为底面，则 $\overline{AB}$ 必须是高度。四面体 $ABCD$ 的体积为 $\dfrac{1}{3} \cdot \dfrac{3 \cdot 4}{2} \cdot 2=\boxed{\textbf{(C)} ~4}$。

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