Difference between revisions of "2014 AMC 10A Problems/Problem 16"
(→Solution 1) |
(→Solution 2) |
||
Line 51: | Line 51: | ||
===Solution 2=== | ===Solution 2=== | ||
− | Let the area of the shaded region be <math>x</math>. Let the other two vertices of the kite be <math>I</math> and <math>J</math> with <math>I</math> closer to <math>AD</math> than <math>J</math>. Note that <math> [ABCD] = [ABF] + [DCH] - x + [ADI] + [BCJ]</math>. The area of <math>ABF</math> is <math>1</math> and the area of <math>DCH</math> is <math>\dfrac{1}{2}</math>. We will solve for the areas of <math>ADI</math> and <math>BCJ</math> in terms of x by noting that the area of each triangle is the length of the perpendicular from <math>I</math> to <math>AD</math> and <math>J</math> to <math>BC</math> respectively. Because the area of <math>x</math> = <math>\dfrac{1}{2}* IJ</math> based on the area of a kite formula, <math>\dfrac{ab}{2}</math> for diagonals of length <math>a</math> and <math>b</math>, <math>IJ = 2x</math>. So each perpendicular is length <math>\dfrac{1-2x}{2}</math>. So taking our numbers and plugging them into <math> [ABCD] =[ABF] + [DCH] - x + [ADI] + [BCJ]</math> gives us <math>2 = \dfrac{5}{2} - 3x</math> Solving this equation for <math>x</math> gives us | + | Let the area of the shaded region be <math>x</math>. Let the other two vertices of the kite be <math>I</math> and <math>J</math> with <math>I</math> closer to <math>AD</math> than <math>J</math>. Note that <math> [ABCD] = [ABF] + [DCH] - x + [ADI] + [BCJ]</math>. The area of <math>ABF</math> is <math>1</math> and the area of <math>DCH</math> is <math>\dfrac{1}{2}</math>. We will solve for the areas of <math>ADI</math> and <math>BCJ</math> in terms of x by noting that the area of each triangle is the length of the perpendicular from <math>I</math> to <math>AD</math> and <math>J</math> to <math>BC</math> respectively. Because the area of <math>x</math> = <math>\dfrac{1}{2}* IJ</math> based on the area of a kite formula, <math>\dfrac{ab}{2}</math> for diagonals of length <math>a</math> and <math>b</math>, <math>IJ = 2x</math>. So each perpendicular is length <math>\dfrac{1-2x}{2}</math>. So taking our numbers and plugging them into <math> [ABCD] =[ABF] + [DCH] - x + [ADI] + [BCJ]</math> gives us <math>2 = \dfrac{5}{2} - 3x</math> Solving this equation for <math>x</math> gives us <math> x = {\textbf{(C)}\ \frac{1}{6}}</math> |
==See Also== | ==See Also== |
Revision as of 20:47, 7 February 2014
Contents
[hide]Problem
In rectangle ,
,
, and points
,
, and
are midpoints of
,
, and
, respectively. Point
is the midpoint of
. What is the area of the shaded region?
Solution
Solution 1
Note that the region is a kite; hence its diagonals are perpendicular and it has area for diagonals of length
and
. Since
as both
and
are midpoints of parallel sides of rectangle
and
, we let
. Now all we need to do is to find
.
Let the other two vertices of the kite be and
with
closer to
than
. This gives us
. Now let
. We thus find that the equation of
is
and that of
is
. Solving this system gives us
, so the
-coordinate of
is
; in other words,
is
from
. By symmetry,
is also the same distance from
, so as
we have
. Hence the area of the kite is
.
Solution 2
Let the area of the shaded region be . Let the other two vertices of the kite be
and
with
closer to
than
. Note that
. The area of
is
and the area of
is
. We will solve for the areas of
and
in terms of x by noting that the area of each triangle is the length of the perpendicular from
to
and
to
respectively. Because the area of
=
based on the area of a kite formula,
for diagonals of length
and
,
. So each perpendicular is length
. So taking our numbers and plugging them into
gives us
Solving this equation for
gives us
See Also
2014 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.