2018 AMC 8 Problems/Problem 22: Difference between revisions
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<math>\textbf{(A) } 100 \qquad \textbf{(B) } 108 \qquad \textbf{(C) } 120 \qquad \textbf{(D) } 135 \qquad \textbf{(E) } 144</math> | <math>\textbf{(A) } 100 \qquad \textbf{(B) } 108 \qquad \textbf{(C) } 120 \qquad \textbf{(D) } 135 \qquad \textbf{(E) } 144</math> | ||
=Solution 1= | |||
Let the area of <math>\triangle CEF</math> be <math>x</math>. Thus, the area of triangle <math>\triangle ACD</math> is <math>45+x</math> and the area of the square is <math>2(45+x) = 90+2x</math>. | Let the area of <math>\triangle CEF</math> be <math>x</math>. Thus, the area of triangle <math>\triangle ACD</math> is <math>45+x</math> and the area of the square is <math>2(45+x) = 90+2x</math>. | ||
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The area of the square is then <math>\frac{45}{\frac{5}{12}}=9\cdot12=\boxed{\textbf{(B)}108}</math>. | The area of the square is then <math>\frac{45}{\frac{5}{12}}=9\cdot12=\boxed{\textbf{(B)}108}</math>. | ||
=See Also= | =See Also= | ||
{{AMC8 box|year=2018|num-b=21|num-a=23}} | {{AMC8 box|year=2018|num-b=21|num-a=23}} | ||
Set s to be the bottom left triangle. | Set s to be the bottom left triangle. | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 14:27, 20 August 2019
Problem 22
Point 
is the midpoint of side 
in square 
and 
meets diagonal 
at 
The area of quadrilateral 
is 
What is the area of 
![[asy] size(5cm); draw((0,0)--(6,0)--(6,6)--(0,6)--cycle); draw((0,6)--(6,0)); draw((3,0)--(6,6)); label("$A$",(0,6),NW); label("$B$",(6,6),NE); label("$C$",(6,0),SE); label("$D$",(0,0),SW); label("$E$",(3,0),S); label("$F$",(4,2),E); [/asy]](http://latex.artofproblemsolving.com/7/c/1/7c13c3d233bfe4e1b9a7ecd2e6343fe5b8aa2e3a.png)
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 :
![\[45+4x = \frac{3}{4}\left(90+2x\right)\]](http://latex.artofproblemsolving.com/2/5/e/25ee5c5006bebcb296565d98e6deac0ebf21e1e9.png)
Solving, we get 
. The area of square 
is 
.
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 
.
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. These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: Unable to save thumbnail to destination
