Difference between revisions of "2012 AMC 8 Problems/Problem 5"
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Note that we only need to consider the value below the marked red line, so we have equation: | Note that we only need to consider the value below the marked red line, so we have equation: | ||
<cmath> X + 2 = 6 + 1 </cmath> | <cmath> X + 2 = 6 + 1 </cmath> | ||
| − | <cmath> X = </cmath> | + | <cmath> X = 5 </cmath> |
Hence, the answer is <math> \boxed{\textbf{(E)}\ 5} </math>. | Hence, the answer is <math> \boxed{\textbf{(E)}\ 5} </math>. | ||
Revision as of 07:27, 1 March 2023
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 , 
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]](http://latex.artofproblemsolving.com/b/c/d/bcd57a9d78159bce4d3873f81f5d879beaed1d5a.png)
Solution 1
Thus, the answer is 
.
Solution 2
Note that we only need to consider the value below the marked red line, so we have equation:
![\[X + 2 = 6 + 1\]](http://latex.artofproblemsolving.com/4/d/f/4df9ced0f2c10dfa5f6b6590a125d641804f37fc.png)
![\[X = 5\]](http://latex.artofproblemsolving.com/d/4/2/d428366384771ecc1c4a712b9898e2c680a889a8.png)
Hence, the answer is .
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. 
