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Difference between revisions of "2012 AMC 8 Problems/Problem 5"
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[[File:2012amc85.png]]
 
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The figure is the same height on both sides, so the sum of the lengths contributing to the height on the left side will equal the sum of the lengths contributing to the height on the left side.
 
  
 
<math>1 + 1 + 1 + 2 + X = 1 + 2 + 1 + 6\\
 
<math>1 + 1 + 1 + 2 + X = 1 + 2 + 1 + 6\\

Revision as of 17:20, 8 October 2022

Problem

In the diagram, all angles are right angles and the lengths of the sides are given in centimeters. Note the diagram is not drawn to scale. What is , $X$ in centimeters?

[asy] pair A,B,C,D,E,F,G,H,I,J,K,L,M,N,O,P,Q,R; A=(4,0); B=(7,0); C=(7,4); D=(8,4); E=(8,5); F=(10,5); G=(10,7); H=(7,7); I=(7,8); J=(5,8); K=(5,7); L=(4,7); M=(4,6); N=(0,6); O=(0,5); P=(2,5); Q=(2,3); R=(4,3); draw(A--B--C--D--E--F--G--H--I--J--K--L--M--N--O--P--Q--R--cycle); label("$X$",(3.4,1.5)); label("6",(7.6,1.5)); label("1",(7.6,3.5)); label("1",(8.4,4.6)); label("2",(9.4,4.6)); label("2",(10.4,6)); label("3",(8.4,7.4)); label("1",(7.5,7.8)); label("2",(6,8.5)); label("1",(4.7,7.8)); label("1",(4.3,7.5)); label("1",(3.5,6.5)); label("4",(1.8,6.5)); label("1",(-0.5,5.5)); label("2",(0.8,4.5)); label("2",(1.5,3.8)); label("2",(2.8,2.6));[/asy]

$\textbf{(A)}\hspace{.05in}1\qquad\textbf{(B)}\hspace{.05in}2\qquad\textbf{(C)}\hspace{.05in}3\qquad\textbf{(D)}\hspace{.05in}4\qquad\textbf{(E)}\hspace{.05in}5$

Solution

2012amc85.png


$1 + 1 + 1 + 2 + X = 1 + 2 + 1 + 6\\ 5 + X = 10\\ X = 5$

Thus, the answer is $\boxed{\textbf{(E)}\ 5}$.

Video Solution

https://youtu.be/m4g-Nmot-c8 ~savannahsolver

See Also

2012 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 4 		Followed by
Problem 6
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 AJHSME/AMC 8 Problems and Solutions

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