Difference between revisions of "2012 AMC 8 Problems/Problem 5"
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+ | ==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 , <math> X </math> in centimeters? | 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 , <math> X </math> in centimeters? | ||
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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> | ||
+ | |||
+ | ==Solution 1== | ||
+ | [[File:2012amc85.png]] | ||
+ | |||
+ | |||
+ | <math>1 + 1 + 1 + 2 + X = 1 + 2 + 1 + 6\ | ||
+ | 5 + X = 10\ | ||
+ | X = 5</math> | ||
+ | |||
+ | Thus, the answer is <math> \boxed{\textbf{(E)}\ 5} </math>. | ||
+ | |||
+ | ==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> | ||
+ | <cmath> X = 5 </cmath> | ||
+ | |||
+ | Hence, the answer is <math> \boxed{\textbf{(E)}\ 5} </math>. | ||
+ | |||
+ | ~[[User:Bloggish|Bloggish]] | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://youtu.be/m4g-Nmot-c8 ~savannahsolver | ||
+ | |||
+ | ==See Also== | ||
+ | {{AMC8 box|year=2012|num-b=4|num-a=6}} | ||
+ | {{MAA Notice}} |
Latest revision as of 19:16, 3 January 2024
Contents
[hide]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?
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 the equation:
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.