Difference between revisions of "2013 AMC 10A Problems/Problem 16"
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To get our final answer, we must subtract this from <math>16</math>. <math>[ABD] - [ADF] = 16 - \frac{16}{3} = \boxed{\textbf{(E) }\frac{32}{3}}</math> | To get our final answer, we must subtract this from <math>16</math>. <math>[ABD] - [ADF] = 16 - \frac{16}{3} = \boxed{\textbf{(E) }\frac{32}{3}}</math> | ||
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+ | ==Solution 2== | ||
+ | |||
+ | First, realize that <math>E</math> is the midpoint of <math>AB</math> and <math>C</math> is the midpoint of <math>BD</math>. Connect <math>A</math> to <math>D</math> to form <math>\triangle ABD</math>. Let the midpoint of <math>AD</math> be <math>G</math>. Connect <math>B</math> to <math>G</math>. <math>BG</math> is a median of <math>\triangle ABD</math>. | ||
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+ | Because <math>\triangle ABD</math> is isosceles, <math>BG</math> is also an altitude of <math>\triangle ABD</math>. We know the length of <math>AD</math> and <math>BG</math> from the given coordinates. The area of <math>\triangle ABD</math> is <math>\frac{bh}{2} = \frac{4(8)}{2} = 16</math>. | ||
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+ | Let the intesection of <math>AC</math>, <math>DE</math> and <math>BG</math> be <math>F</math>. <math>F</math> is the centroid of <math>\triangle ABD</math>. Therefore, it splits <math>BG</math> into <math>BF={2 \over 3}(BG)</math> and <math>FG={1\over 3}(BG)</math>. The area of quadrilateral <math>ABDF = 16\cdot {2 \over 3} = \boxed{\textbf{(E) }\frac{32}{3}}</math> | ||
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+ | ~Zeric Hang | ||
==See Also== | ==See Also== |
Revision as of 14:52, 15 July 2018
Contents
[hide]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
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 16, 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
See Also
2013 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 15 |
Followed by Problem 17 | |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.