2021 AMC 10A Problems/Problem 13

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: $\begin{align*} x^2+y^2+z^2&=2^2, &(1) \\ (x-3)^2+y^2+z^2&=\sqrt{13}^2, &(2) \\ x^2+(y-4)^2+z^2&=\left(2\sqrt5\right)^2. &\hspace{1mm} (3) \end{align*}$ Subtracting $(1)$ from $(2)$ gives $-6x+9=9,$ from which $x=0.$

Subtracting $(1)$ from $(3)$ gives $-8y+16=16,$ from which $y=0.$

Substituting $(x,y)=(0,0)$ into $(1)$ produces $z^2=4,$ or $|z|=2.$

Finally, we find the volume of tetrahedron $ABCD$ using $\triangle ACD$ as the base: $\begin{align*} \mathrm{Volume}_{ABCD}&=\frac13\cdot\left(\mathrm{Base \ Area}\right)\cdot\left(\mathrm{Height}\right) \\ &=\frac13\cdot\underbrace{[ACD]}_{\tfrac12\cdot AC \cdot AD}\cdot\underbrace{\left(\mathrm{Height \ from \ }B\right)}_{|z|} \\ &=\frac13\cdot\left(\frac12\cdot 3 \cdot 4\right)\cdot(2) \\ &=\boxed{\textbf{(C)} ~4}. \end{align*}$

~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

2021 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 12	Followed by Problem 14
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

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