Difference between revisions of "2018 AMC 8 Problems/Problem 22"
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Solving, we get <math>x = 9</math>. The area of square <math>ABCD</math> is <math>90+2x = 90 + 2 \cdot 9 = \boxed{\textbf{(B) } 108}</math>. | Solving, we get <math>x = 9</math>. The area of square <math>ABCD</math> is <math>90+2x = 90 + 2 \cdot 9 = \boxed{\textbf{(B) } 108}</math>. | ||
− | Sidenote: <math>\triangle CEF \sim \triangle ABF</math> is a 1 | + | Sidenote: <math>\triangle CEF \sim \triangle ABF</math> is a 1:2 ratio because it was given. |
==Solution 2== | ==Solution 2== |
Revision as of 16:14, 22 December 2023
Contents
Problem 22
Point is the midpoint of side
in square
and
meets diagonal
at
The area of quadrilateral
is
What is the area of
Solution 1
Let the area of be
. Thus, the area of triangle
is
and the area of the square is
.
By AA similarity, with a 1:2 ratio, so the area of triangle
is
. Now, consider trapezoid
. Its area is
, which is three-fourths the area of the square. We set up an equation in
:
Solving, we get
. The area of square
is
.
Sidenote: is a 1:2 ratio because it was given.
Solution 2
We can use analytic geometry for this problem.
Let us start by giving the coordinate
,
the coordinate
, and so forth.
and
can be represented by the equations
and
, respectively. Solving for their intersection gives point
coordinates
.
Now, ’s area is simply
or
. This means that pentagon
’s area is
of the entire square, and it follows that quadrilateral
’s area is
of the square.
The area of the square is then .
Solution 3
has half the area of the square.
has base equal to half the square side length, and by AA Similarity with
, it has 1/(1+2)= 1/3 the height, so has
th area of square. Thus, the area of the quadrilateral is
th the area of the square. The area of the square is then
.
Solution 4
Extend and
to meet at
. Drop an altitude from
to
and call it
. Also, call
. As stated before, we have
, so the ratio of their heights is in a
ratio, making the altitude from
to
. Note that this means that the side of the square is
. In addition,
by AA Similarity in a
ratio. This means that the side length of the square is
, making
.
Now, note that . We have
and
Subtracting makes
We are given that
so
Therefore,
so our answer is
- moony_eyed
Solution 5
Solution with Cartesian and Barycentric Coordinates:
We start with the following:
Claim: Given a square , let
be the midpoint of
and let
. Then
.
Proof: We use Cartesian coordinates. Let be the origin,
. We have that
and
are governed by the equations
and
, respectively. Solving,
. The result follows.
Now, we apply Barycentric Coordinates w.r.t. . We let
. Then
.
In the barycentric coordinate system, the area formula is where
is a random triangle and
is the reference triangle. Using this, we find that
Let
so that
. Then, we have
, so the answer is
.
Video Solution by OmegaLearn
https://youtu.be/FDgcLW4frg8?t=4038
- pi_is_3.14
Video Solutions
- Happytwin
~savannahsolver
See Also
2018 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 21 |
Followed by Problem 23 | |
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 AJHSME/AMC 8 Problems and Solutions |
Set s to be the bottom left triangle.
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.