Difference between revisions of "2015 AMC 10B Problems/Problem 17"

(Solution 2)
(6 intermediate revisions by 2 users not shown)
Line 26: Line 26:
 
The octahedron is just two congruent pyramids glued together by their base. The base of one pyramid is a rhombus with diagonals <math>4</math> and <math>5</math>, for an area <math>A = 10</math>. The height <math>h</math>, of one pyramid, is <math>\dfrac{3}{2}</math>, so the volume of one pyramid is <math>\dfrac{Ah}{3}=5</math>. Thus, the octahedron has volume <math>2\cdot5=\boxed{{(B)}\\10}</math>
 
The octahedron is just two congruent pyramids glued together by their base. The base of one pyramid is a rhombus with diagonals <math>4</math> and <math>5</math>, for an area <math>A = 10</math>. The height <math>h</math>, of one pyramid, is <math>\dfrac{3}{2}</math>, so the volume of one pyramid is <math>\dfrac{Ah}{3}=5</math>. Thus, the octahedron has volume <math>2\cdot5=\boxed{{(B)}\\10}</math>
  
==Solution 2==
+
==Solution 2 (Scaling)==
Scale the prism down by two, but let one vertice coincide with the big prism (The small prism is still inside the big one). Three of the vertices are the centers of faces. Since the vertices of the octahedron are made by the centers of the faces of the prism, connect those three vertices. Then draw a plane using those 3 points, and "slice" the prism. Notice how they are congruent. You do this for all of the eight prisms in the big prism. Since all of them are half the area, the area of the octahedron is 3 ⋅4 ⋅5=\boxed{{(B)}\\10}$
+
Because it is connected by the midpoints, the "base" of the octahedron is half the base of the rectangular prism. In addition, the volume of an octahedron is <math>\dfrac{1}{3}</math> of its respective prism. Thus, the octahedron's volume is <math>\dfrac{1}{2} \cdot \dfrac{1}{3} = \dfrac{1}{6}</math> of the rectangular prism's volume, meaning that the answer is <math>3 \cdot 4 \cdot 5 \cdot \dfrac{1}{6} = \boxed{\\10}</math>
  
 
==See Also==
 
==See Also==
Line 34: Line 34:
  
 
[[Category: Introductory Geometry Problems]]
 
[[Category: Introductory Geometry Problems]]
 +
[[Category: 3D Geometry Problems]]

Revision as of 19:13, 28 March 2021

Problem

The centers of the faces of the right rectangular prism shown below are joined to create an octahedron. What is the volume of this octahedron?

[asy] import three; size(2inch); currentprojection=orthographic(4,2,2); draw((0,0,0)--(0,0,3),dashed); draw((0,0,0)--(0,4,0),dashed); draw((0,0,0)--(5,0,0),dashed); draw((5,4,3)--(5,0,3)--(5,0,0)--(5,4,0)--(0,4,0)--(0,4,3)--(0,0,3)--(5,0,3)); draw((0,4,3)--(5,4,3)--(5,4,0)); label("3",(5,0,3)--(5,0,0),W); label("4",(5,0,0)--(5,4,0),S); label("5",(5,4,0)--(0,4,0),SE); [/asy]

$\textbf{(A) } \dfrac{75}{12} \qquad\textbf{(B) } 10 \qquad\textbf{(C) } 12 \qquad\textbf{(D) } 10\sqrt2 \qquad\textbf{(E) } 15$

Solution 1

The octahedron is just two congruent pyramids glued together by their base. The base of one pyramid is a rhombus with diagonals $4$ and $5$, for an area $A = 10$. The height $h$, of one pyramid, is $\dfrac{3}{2}$, so the volume of one pyramid is $\dfrac{Ah}{3}=5$. Thus, the octahedron has volume $2\cdot5=\boxed{{(B)}\\10}$

Solution 2 (Scaling)

Because it is connected by the midpoints, the "base" of the octahedron is half the base of the rectangular prism. In addition, the volume of an octahedron is $\dfrac{1}{3}$ of its respective prism. Thus, the octahedron's volume is $\dfrac{1}{2} \cdot \dfrac{1}{3} = \dfrac{1}{6}$ of the rectangular prism's volume, meaning that the answer is $3 \cdot 4 \cdot 5 \cdot \dfrac{1}{6} = \boxed{\\10}$

See Also

2015 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 16
Followed by
Problem 18
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. AMC logo.png