Difference between revisions of "2021 AMC 10A Problems/Problem 13"

Revision as of 11:39, 24 April 2021

Problem

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

$\textbf{(A)} ~3 \qquad\textbf{(B)} ~2\sqrt{3} \qquad\textbf{(C)} ~4\qquad\textbf{(D)} ~3\sqrt{3}\qquad\textbf{(E)} ~6$

Solution 1 (Observations)

Drawing the tetrahedron out and testing side lengths, we realize that the $\triangle ABD, \triangle ABC,$ 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 $AB$ must be the height. $\dfrac{1}{3} \cdot \dfrac{3 \cdot 4}{2} \cdot 2$ , so we have an answer of $\boxed{\textbf{(C) } 4}$ .

Solution 2 (Bash: Coordinate Geometry)

We will place tetrahedron $ABCD$ in the $xyz$ -plane. By the Converse of the Pythagorean Theorem, we know that $\triangle ACD$ is a right triangle. Without the loss of generality, let $A=(0,0,0), C=(3,0,0), D=(0,4,0),$ and $B=(x,y,z).$ Applying the Distance Formula to $\overline{BA},\overline{BC},$ and $\overline{BD},$ respectively, we have the following system: $\begin{align*} x^2+y^2+z^2&=2^2, \\ (x-3)^2+y^2+z^2&=\sqrt{13}^2, \\ x^2+(y-4)^2+z^2&=\left(2\sqrt5\right)^2. \end{align*}$

~MRENTHUSIASM

Video Solution (Simple & Quick)

https://youtu.be/bRrchiDCrKE

~ Education, the Study of Everything

Video Solution (Using Pythagorean Theorem, 3D Geometry - Tetrahedron)

https://youtu.be/i4yUaXVUWKE

~ pi_is_3.14

Video Solution by TheBeautyofMath

https://youtu.be/t-EEP2V4nAE?t=813

~IceMatrix

@@ Line 8: / Line 8: @@
 ==Solution 2 (Bash: Coordinate Geometry)==
-We will place tetrahedron <math>ABCD</math> in the <math>xyz</math>-plane. By the Converse of the Pythagorean Theorem, we know that <math>\triangle ACD</math> is a right triangle. Without the loss of generality, let <math>A=(0,0,0), C=(3,0,0), D=(0,4,0),</math> and <math>B=(x,y,z).</math>
+We will place tetrahedron <math>ABCD</math> in the <math>xyz</math>-plane. By the Converse of the Pythagorean Theorem, we know that <math>\triangle ACD</math> is a right triangle. Without the loss of generality, let <math>A=(0,0,0), C=(3,0,0), D=(0,4,0),</math> and <math>B=(x,y,z).</math> Applying the Distance Formula to <math>\overline{BA},\overline{BC},</math> and <math>\overline{BD},</math> respectively, we have the following system:
+<cmath>\begin{align*}
+x^2+y^2+z^2&=2^2, \\
+(x-3)^2+y^2+z^2&=\sqrt{13}^2, \\
+x^2+(y-4)^2+z^2&=\left(2\sqrt5\right)^2.
+\end{align*}</cmath>
 ~MRENTHUSIASM

2021 AMC 10A (Problems • Answer Key • Resources)
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