Art of Problem Solving

2013 AMC 10A Problems/Problem 16: Difference between revisions

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==Problem==
==Problem==


A triangle with vertices <math>(6, 5)</math>, <math>(8, -3)</math>, and <math>(9, 1) is reflected about the line </math>x=8<math> to create a second triangle.  What is the area of the union of the two triangles
A triangle with vertices <math>(6, 5)</math>, <math>(8, -3)</math>, and <math>(9, 1)</math> is reflected about the line <math>x=8</math> to create a second triangle.  What is the area of the union of the two triangles?


</math> \textbf{(A)}\ 9 \qquad\textbf{(B)}\ \frac{28}{3} \qquad\textbf{(C)}\ 10 \qquad\textbf{(D)}\ \frac{31}{3} \qquad\textbf{(E)}\ \frac{32}{3} $
<math> \textbf{(A)}\ 9 \qquad\textbf{(B)}\ \frac{28}{3} \qquad\textbf{(C)}\ 10 \qquad\textbf{(D)}\ \frac{31}{3} \qquad\textbf{(E)}\ \frac{32}{3} </math>


==Solution==
==Solution==

Revision as of 19:51, 7 February 2013

Problem

A triangle with vertices $(6, 5)$, $(8, -3)$, and $(9, 1)$ is reflected about the line $x=8$ to create a second triangle. What is the area of the union of the two triangles?

$\textbf{(A)}\ 9 \qquad\textbf{(B)}\ \frac{28}{3} \qquad\textbf{(C)}\ 10 \qquad\textbf{(D)}\ \frac{31}{3} \qquad\textbf{(E)}\ \frac{32}{3}$

Solution