2005 AMC 12A Problems/Problem 17

Revision as of 13:56, 19 January 2021 by Anmol04 (talk | contribs) (moved answer choices after diagram)
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

Problem

A unit cube is cut twice to form three triangular prisms, two of which are congruent, as shown in Figure 1. The cube is then cut in the same manner along the dashed lines shown in Figure 2. This creates nine pieces. What is the volume of the piece that contains vertex $W$?

[asy] path a=(0,0)--(10,0)--(10,10)--(0,10)--cycle; path b = (0,10)--(6,16)--(16,16)--(16,6)--(10,0); path c= (10,10)--(16,16); path d= (0,0)--(3,13)--(13,13)--(10,0); path e= (13,13)--(16,6); draw(a,linewidth(0.7)); draw(b,linewidth(0.7)); draw(c,linewidth(0.7)); draw(d,linewidth(0.7)); draw(e,linewidth(0.7)); draw(shift((20,0))*a,linewidth(0.7)); draw(shift((20,0))*b,linewidth(0.7)); draw(shift((20,0))*c,linewidth(0.7)); draw(shift((20,0))*d,linewidth(0.7)); draw(shift((20,0))*e,linewidth(0.7)); draw((20,0)--(25,10)--(30,0),dashed); draw((25,10)--(31,16)--(36,6),dashed); draw((15,0)--(10,10),Arrow); draw((15.5,0)--(30,10),Arrow); label("$W$",(15.2,0),S); label("Figure 1",(5,0),S); label("Figure 2",(25,0),S); [/asy]

$(\mathrm {A}) \ \frac{1}{12} \qquad (\mathrm {B}) \ \frac{1}{9} \qquad (\mathrm {C})\ \frac{1}{8} \qquad (\mathrm {D}) \ \frac{1}{6} \qquad (\mathrm {E})\ \frac{1}{4}$

Solution

It is a pyramid with height $1$ and base area $\frac{1}{4}$, so using the formula for the volume of a pyramid, $\frac{1}{3} \cdot \left(\frac{1}{4}\right) \cdot (1) = \frac {1}{12} \Rightarrow \boxed{(\mathrm {A})}$.

See also

2005 AMC 12A (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 12 Problems and Solutions

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