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

Revision as of 10:10, 4 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

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 $\boxed{\textbf{(C) } 4}$ .

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 5: / Line 5: @@
 ==Solution==
-Drawing the tetrahedron out and testing side lengths, we realize that the <math>\triangle ABD, ABC, and ABD</math> 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 <math>\triangle ADC</math> as the base, then <math>AB</math> must be the height. <math>\dfrac{1}{3} \cdot \dfrac{3 \cdot 4}{2} \cdot 2</math>, so we have an answer of <math>\boxed{\textbf{(C) } 4}</math>.
+Drawing the tetrahedron out and testing side lengths, we realize that the <math>\triangle ABD, \triangle ABC,</math> and <math>\triangle ABD</math> 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 <math>\triangle ADC</math> as the base, then <math>AB</math> must be the height. <math>\dfrac{1}{3} \cdot \dfrac{3 \cdot 4}{2} \cdot 2</math>, so we have an answer of <math>\boxed{\textbf{(C) } 4}</math>.
 ==Similar Problem==

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