Difference between revisions of "2014 AMC 12B Problems/Problem 9"
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<math> \textbf{(A)}\ 30\qquad\textbf{(B)}\ 36\qquad\textbf{(C)}\ 40\qquad\textbf{(D)}\ 48\qquad\textbf{(E)}\ 58.5 </math> | <math> \textbf{(A)}\ 30\qquad\textbf{(B)}\ 36\qquad\textbf{(C)}\ 40\qquad\textbf{(D)}\ 48\qquad\textbf{(E)}\ 58.5 </math> | ||
| + | [[Category: Introductory Geometry Problems]] | ||
==Solution== | ==Solution== | ||
Revision as of 10:56, 13 August 2014
Problem
Convex quadrilateral 
has 
, 
, 
, 
, and 
, as shown. What is the area of the quadrilateral?
![[asy] pair A=(0,0), B=(-3,0), C=(-3,-4), D=(48/5,-36/5); draw(A--B--C--D--A); label("$A$",A,N); label("$B$",B,NW); label("$C$",C,SW); label("$D$",D,E); draw(rightanglemark(A,B,C,25)); [/asy]](http://latex.artofproblemsolving.com/e/d/6/ed685cbdb58e7beeff868e0123c51cd79633def7.png)
Solution
Note that by the pythagorean theorem, 
. Also note that 
is a right angle because 
is a right triangle. The area of the quadrilateral is the sum of the areas of 
and which is equal to
![\[\frac{3\times4}{2} + \frac{5\times12}{2} = 6 + 30 = \boxed{\textbf{(B)}\ 36}\]](http://latex.artofproblemsolving.com/d/6/0/d60d5dac5538495750c4b3cbf16b14c243bd6d22.png)
See also
| 2014 AMC 12B (Problems • Answer Key • Resources) | |
| Preceded by Problem 8 |
Followed by Problem 10 |
| 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 12 Problems and Solutions | |
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
