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

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==Problem==
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What is the volume of tetrahedron <math>ABCD</math> with edge lengths <math>AB = 2</math>, <math>AC = 3</math>, <math>AD = 4</math>, <math>BC = \sqrt{13}</math>, <math>BD = 2\sqrt{5}</math>, and <math>CD = 5</math> ?
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<math>\textbf{(A)} ~3 \qquad\textbf{(B)} ~2\sqrt{3} \qquad\textbf{(C)} ~4\qquad\textbf{(D)} ~3\sqrt{3}\qquad\textbf{(E)} ~6</math>
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==Solution==
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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>.
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==Similar Problem==
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https://artofproblemsolving.com/wiki/index.php/2015_AMC_10A_Problems/Problem_21
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==Video Solution (Simple & Quick)==
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https://youtu.be/bRrchiDCrKE
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~ Education, the Study of Everything
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== Video Solution (Using Pythagorean Theorem, 3D Geometry - Tetrahedron) ==
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https://youtu.be/i4yUaXVUWKE
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~ pi_is_3.14
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==Video Solution by TheBeautyofMath==
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https://youtu.be/t-EEP2V4nAE?t=813
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~IceMatrix
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==See also==
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{{AMC10 box|year=2021|ab=A|num-b=12|num-a=14}}
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{{MAA Notice}}

Revision as of 11:10, 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 $\boxed{\textbf{(C) } 4}$.

Similar Problem

https://artofproblemsolving.com/wiki/index.php/2015_AMC_10A_Problems/Problem_21

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

See also

2021 AMC 10A (ProblemsAnswer KeyResources)
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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