Difference between revisions of "2013 AMC 10A Problems/Problem 18"
Countingkg (talk | contribs) |
(→Solution) |
||
| Line 6: | Line 6: | ||
<math> \textbf{(A)}\ 54\qquad\textbf{(B)}\ 58\qquad\textbf{(C)}\ 62\qquad\textbf{(D)}\ 70\qquad\textbf{(E)}\ 75 </math> | <math> \textbf{(A)}\ 54\qquad\textbf{(B)}\ 58\qquad\textbf{(C)}\ 62\qquad\textbf{(D)}\ 70\qquad\textbf{(E)}\ 75 </math> | ||
| − | == | + | The area of the quadrilateral ABCD is 7.5, therefore the area of each of the equal area pieces is 3.75. |
| − | + | Since the dividing line passes through point A, the lower piece must be a triangle with base AD. | |
| + | Since the formula for the area of a triangle is b*h/2 and b=4, h must equal 15/8. | ||
| + | The point on line CD with y-coordinate 15/8 is (27/8, 15/8). | ||
| + | 27+8+15+8=58, therefore the answer is (B). | ||
==See Also== | ==See Also== | ||
Revision as of 01:05, 8 February 2013
Problem
Let points 
, 
, 
, and 
. Quadrilateral 
is cut into equal area pieces by a line passing through 
. This line intersects 
at point 
, where these fractions are in lowest terms. What is 
?
The area of the quadrilateral ABCD is 7.5, therefore the area of each of the equal area pieces is 3.75. Since the dividing line passes through point A, the lower piece must be a triangle with base AD. Since the formula for the area of a triangle is b*h/2 and b=4, h must equal 15/8. The point on line CD with y-coordinate 15/8 is (27/8, 15/8). 27+8+15+8=58, therefore the answer is (B).
See Also
| 2013 AMC 10A (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 | ||
| 2013 AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by Problem 12 |
Followed by Problem 14 |
| 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 | |
