Difference between revisions of "2014 AMC 12B Problems/Problem 9"
Kevin38017 (talk | contribs) |
Kevin38017 (talk | contribs) (→Solution) |
||
| Line 14: | Line 14: | ||
==Solution== | ==Solution== | ||
| − | Note that <math>AC=5</math>. Also note that <math>\angle CAD</math> is a right angle because <math>\triangle CAD</math> is a right triangle. The area of the quadrilateral is the sum of the areas of <math>\triangle ABC</math> and <math>\triangle CAD</math> which is equal to | + | Note that by the pythagorean theorem, <math>AC=5</math>. Also note that <math>\angle CAD</math> is a right angle because <math>\triangle CAD</math> is a right triangle. The area of the quadrilateral is the sum of the areas of <math>\triangle ABC</math> and <math>\triangle CAD</math> which is equal to |
<cmath>\frac{3*4}{2} + \frac{5*12}{2} = 6 + 30 = \boxed{\textbf{(B)}\ 36}</cmath> | <cmath>\frac{3*4}{2} + \frac{5*12}{2} = 6 + 30 = \boxed{\textbf{(B)}\ 36}</cmath> | ||
Revision as of 23:07, 20 February 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
![\[\frac{3*4}{2} + \frac{5*12}{2} = 6 + 30 = \boxed{\textbf{(B)}\ 36}\]](http://latex.artofproblemsolving.com/f/1/9/f1942c94940cdd5ad8c2d230f4d970ec20f810de.png)
