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Difference between revisions of "2012 AMC 8 Problems/Problem 5"

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<math> \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 </math>
 
<math> \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 </math>
  
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==Solution==
+
==Solution 1==
 
[[File:2012amc85.png]]
 
[[File:2012amc85.png]]
  
−
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\\
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Thus, the answer is <math> \boxed{\textbf{(E)}\ 5} </math>.
 
Thus, the answer is <math> \boxed{\textbf{(E)}\ 5} </math>.
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 +
==Solution 2==
 +
[[File:bcd57a9d78159bce4d3873f81f5d879beaed1d5a.png|300px|center]]
 +
 +
Note that we only need to consider the value below the marked red line, so we have the equation:
 +
<cmath> X + 2 = 6 + 1 </cmath>
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<cmath> X = 5 </cmath>
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 +
Hence, the answer is <math> \boxed{\textbf{(E)}\ 5} </math>.
 +
 +
~[[User:Bloggish|Bloggish]]
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 +
==Video Solution==
 +
https://youtu.be/m4g-Nmot-c8 ~savannahsolver
  
 
==See Also==
 
==See Also==
 
{{AMC8 box|year=2012|num-b=4|num-a=6}}
 
{{AMC8 box|year=2012|num-b=4|num-a=6}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 20:16, 3 January 2024

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 1

2012amc85.png


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

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

Solution 2

Bcd57a9d78159bce4d3873f81f5d879beaed1d5a.png

Note that we only need to consider the value below the marked red line, so we have the equation: \[X + 2 = 6 + 1\] \[X = 5\]

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

~Bloggish

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

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

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