2013 AMC 10A Problems/Problem 16
Problem
A triangle with vertices , , and is reflected about the line to create a second triangle. What is the area of the union of the two triangles?
Solution 1
Let be at , B be at , and be at . Reflecting over the line , we see that , (as the x-coordinate of B is 8), and . Line can be represented as , so we see that is on line .
We see that if we connect to , we get a line of length (between and ). The area of is equal to .
Now, let the point of intersection between and be . If we can just find the area of and subtract it from , we are done.
We realize that because the diagram is symmetric over , the intersection of lines and should intersect at an x-coordinate of . We know that the slope of is . Thus, we can represent the line going through and as . Plugging in , we find that the y-coordinate of F is . Thus, the height of is . Using the formula for the area of a triangle, the area of is .
To get our final answer, we must subtract this from .
Solution 2
First, realize that is the midpoint of and is the midpoint of . Connect to to form . Let the midpoint of be . Connect to . is a median of .
Because is isosceles, is also an altitude of . We know the length of and from the given coordinates. The area of is .
Let the intesection of , and be . is the centroid of . Therefore, it splits into and . The area of quadrilateral
~Zeric Hang
Solution 3
Since we have to find the area on a coordinate plane, we can use the Shoelace Theorem to find the area of the intersection. When you reflect it, it makes a quadrilateral.
Since it is reflected around , the point remains the same on both. The top right corners are , and its reflection . Now to find the 4th point, point F, we can use the equation of the line DE(, and substitute , to get . Now we can use the theorem:
~idk12345678
See Also
2013 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 15 |
Followed by Problem 17 | |
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