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

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==Solution==
 
==Solution==
 
Drawing the tetrahedron out and testing side lengths, we realize that the triangles ABD and ABC are right triangles. It is now easy to calculate the volume of the tetrahedron using the formula for the volume of a pyramid: <math>\frac{3\cdot4\cdot2}{3\cdot2}=4</math>, so we have an answer of <math>\boxed{C}</math>. ~IceWolf10
 
Drawing the tetrahedron out and testing side lengths, we realize that the triangles ABD and ABC are right triangles. It is now easy to calculate the volume of the tetrahedron using the formula for the volume of a pyramid: <math>\frac{3\cdot4\cdot2}{3\cdot2}=4</math>, so we have an answer of <math>\boxed{C}</math>. ~IceWolf10
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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
  
 
==See also==
 
==See also==
 
{{AMC10 box|year=2021|ab=A|num-b=12|num-a=14}}
 
{{AMC10 box|year=2021|ab=A|num-b=12|num-a=14}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 16:20, 11 February 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 triangles ABD and ABC are right triangles. It is now easy to calculate the volume of the tetrahedron using the formula for the volume of a pyramid: $\frac{3\cdot4\cdot2}{3\cdot2}=4$, so we have an answer of $\boxed{C}$. ~IceWolf10


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

https://youtu.be/i4yUaXVUWKE

~ pi_is_3.14

See also

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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