2011 AMC 10B Problems/Problem 18: Difference between revisions
No edit summary |
|||
| Line 4: | Line 4: | ||
<math> \textbf{(A)}\ 15 \qquad\textbf{(B)}\ 30 \qquad\textbf{(C)}\ 45 \qquad\textbf{(D)}\ 60 \qquad\textbf{(E)}\ 75</math> | <math> \textbf{(A)}\ 15 \qquad\textbf{(B)}\ 30 \qquad\textbf{(C)}\ 45 \qquad\textbf{(D)}\ 60 \qquad\textbf{(E)}\ 75</math> | ||
[[Category: Introductory Geometry Problems]] | |||
==Solution== | ==Solution== | ||
Revision as of 10:43, 13 August 2014
Problem
Rectangle 
has 
and 
. Point 
is chosen on side 
so that 
. What is the degree measure of 
?
Solution
![[asy] unitsize(10mm); defaultpen(linewidth(.5pt)+fontsize(10pt)); dotfactor=3; pair A=(0,3), B=(6,3), C=(6,0), D=(0,0); pair M=(0.80385,3); draw(A--B--C--D--cycle); draw(M--C); draw(M--D); draw(anglemark(A,M,D)); draw(anglemark(D,M,C)); draw(anglemark(C,D,M)); pair[] ps={A,B,C,D,M}; dot(ps); label("$A$",A,NW); label("$B$",B,NE); label("$C$",C,SE); label("$D$",D,SW); label("$M$",M,N); label("$6$",midpoint(C--M),SW); label("$6$",midpoint(A--B),N); label("$3$",midpoint(B--C),E); [/asy]](http://latex.artofproblemsolving.com/6/e/5/6e52470a180b9bf804714a44c042322ee1ffdadd.png)
It is given that 
. Since and 
are alternate interior angles and 
, 
. Use the Base Angle Theorem to show 
. We know that is a rectangle, so it follows that 
. We notice that 
is a 
triangle, and 
. If we let 
be the measure of 
then
See Also
| 2011 AMC 10B (Problems • Answer Key • Resources) | ||
| Preceded by Problem 17 |
Followed by Problem 19 | |
| 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 10 Problems and Solutions | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: File missing
