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Difference between revisions of "2013 AMC 10A Problems/Problem 18"

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<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>
  
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==Solution==
 
The area of the quadrilateral ABCD is 7.5, therefore the area of each of the equal area pieces is 3.75.
 
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 dividing line passes through point A, the lower piece must be a triangle with base AD.

Revision as of 07:31, 8 February 2013

Problem

Let points $A = (0, 0)$, $B = (1, 2)$, $C=(3, 3)$, and $D = (4, 0)$. Quadrilateral $ABCD$ is cut into equal area pieces by a line passing through $A$. This line intersects $\overline{CD}$ at point $(\frac{p}{q}, \frac{r}{s})$, where these fractions are in lowest terms. What is $p+q+r+s$?


$\textbf{(A)}\ 54\qquad\textbf{(B)}\ 58\qquad\textbf{(C)}\ 62\qquad\textbf{(D)}\ 70\qquad\textbf{(E)}\ 75$

Solution

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 (ProblemsAnswer KeyResources)
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 (ProblemsAnswer KeyResources)
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
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