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

Revision as of 20:53, 9 August 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 (Three Right Triangles)

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: One Right Triangle)

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 $B=(x,y,z).$

We apply the Distance Formula to $\overline{BA},\overline{BC},$ and $\overline{BD},$ respectively: $\begin{align*} x^2+y^2+z^2&=2^2, &(1) \\ (x-3)^2+y^2+z^2&=\sqrt{13}^2, &(2) \\ x^2+(y-4)^2+z^2&=\left(2\sqrt5\right)^2. &\hspace{1mm} (3) \end{align*}$ 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 $|z|=2.$

Let the brackets denote areas. Finally, we find the volume of tetrahedron $ABCD$ using $\triangle ACD$ as the base: $\begin{align*} V_{ABCD}&=\frac13\cdot[ACD]\cdot h_B \\ &=\frac13\cdot\left(\frac12\cdot AC \cdot AD\right)\cdot |z| \\ &=\boxed{\textbf{(C)} ~4}. \end{align*}$ ~MRENTHUSIASM

Trirectangular Tetrahedron Solution

https://mathworld.wolfram.com/TrirectangularTetrahedron.html

Given the observations from Solution 1, where $\triangle ABD, \triangle ABC,$ and $\triangle ABD$ are right triangles, the base is $\triangle ABD$ . We can apply the information about a trirectangular tetrahedron (all of the face angles are right angles), which states that the volume is $V=1/6 ABC$ , where $A B C$ are the side lengths.

$AB = 2$ , $AD = 4$ , $BD = 3$ : The volume is $1/6 * 2 * 4 * 3 = 4$ .

The answer is $\boxed{\textbf{(C)} ~4}$ .

-AMC60

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 38: / Line 38: @@
 Given the observations from Solution 1, where <math>\triangle ABD, \triangle ABC,</math> and <math>\triangle ABD</math> are right triangles, the base is <math>\triangle ABD</math>. We can apply the information about a trirectangular tetrahedron (all of the face angles are right angles), which states that the volume is <math>V=1/6 ABC</math>, where <math>A B C</math> are the side lengths.
-<math>AB = 2</math>, <math>AD = 4</math>, <math>BD = 3</math>: The volume is <math>1/6 * 2 * 3 * 4 = 4</math>.
+<math>AB = 2</math>, <math>AD = 4</math>, <math>BD = 3</math>: The volume is <math>1/6 * 2 * 4 * 3 = 4</math>.
 The answer is <math>\boxed{\textbf{(C)} ~4}</math>.

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

Page

Toolbox

Search