![\[ \begin{tabular}{c c c c}
{} & \textbf{Adams} & \textbf{Baker} & \textbf{Adams and Baker} \\
\textbf{Boys:} & 71 & 81 & 79 \\
\textbf{Girls:} & 76 & 90 & ? \\
\textbf{Boys and Girls:} & 74 & 84 & \\
\end{tabular}
\]](http://latex.artofproblemsolving.com/8/5/c/85c75770c3f90af49df0db600dead2a29e6674b3.png)
Y by ETS1331, Adventure10
Quadrilateral 
satisfies 
, and 
. Diagonals 
and 
intersect at point 
, and 
. What is the area of quadrilateral ?
satisfies 
, and 
. Diagonals 
and 
intersect at point 
, and 
. What is the area of quadrilateral ?
This post has been edited 1 time. Last edited by LauraZed, Jan 31, 2020, 6:34 PM
Reason: added AMC 12 # to source
Reason: added AMC 12 # to source
and 
for simplicity. Note that 
,
.
and 
.
and 
,
since 
.
,
,
.![$$[ABCD] = \frac{1}{2} \cdot 2 \sqrt{10} \cdot 6 \sqrt{10} + \frac{1}{2} \cdot 20 \cdot 30 = \boxed{360}.$$](http://latex.artofproblemsolving.com/b/b/8/bb82e735bb9e98c59633e94281668baf28535f3a.png)
to 
and similar triangles finishes
.
.
and 
which gives 
and the answer is $360#
cuz area of 
is 300, and when u draw the circumcircle of 
and so the length of the altitude from 
by symmetry with 
so area of 
) w/ graph paper and compass, just draw lines with slopes 
and 
for 
and then draw the circle w/ diameter 
![[asy]usepackage("tikz");label("\begin{tikzpicture}
\coordinate[label=above:$A$] (A) at (0,4);
\coordinate[label=below:$C$] (C) at (0,0);
\coordinate[label=left:$B$] (B) at (-1.24,3.57);
\coordinate[label=below:$D$] (D) at (6,0);
\coordinate[label=left:$M$] (M) at (0,2);
\coordinate[label=below:$O$] (O) at (1.94,2);
\coordinate[label=above:$N$] (N) at (3,2);
\draw[thick] (A)--(B)--(C)--(D)--cycle;
\draw[thick] (A)--(C);
\draw[thick] (B)--(D);
\coordinate[label=below left:$E$] (E) at (0,2.95);
\draw[thick,red] (B)--(M)--(N);
\foreach \s in {A,B,C,D,M,O,N,E} \filldraw (\s) circle (2pt);
\path (C)--(D) node[midway,below]{30};
\path (A)--(E) node[right,midway]{5};
\path (E)--(M) node[right,midway]{5};
\path (C)--(M) node[right,midway]{10};
\end{tikzpicture}");[/asy]](http://latex.artofproblemsolving.com/7/5/3/7539933764d9bc616aecd50dadabb72b103459cc.png)
and 
be a midpoint of 
,
and 
.![\[\tan \theta = \frac{CE}{CD} = \frac{CE}{CD} = \frac{1}{2}\quad \text{and}\quad \tan (\theta + \alpha) = \frac{AC}{AD} = \frac{2}{3}\]](http://latex.artofproblemsolving.com/9/e/8/9e86f6dbf1b1d156c273e33286ea0df05a5162ab.png)
Then, we have 
,![\[\sin \alpha = \frac{1}{\sqrt{65}} \quad \text{and}\quad \sin \theta = \frac{1}{\sqrt{5}}\quad \text{and}\quad \cos \theta = \frac{2}{\sqrt{5}}\]](http://latex.artofproblemsolving.com/1/c/a/1caa344f2b303ad9413de07be93d228e66d537da.png)
Because 
,![\[\angle MOB =\angle CDO = \theta\]](http://latex.artofproblemsolving.com/7/4/6/74682669a09e847d244f29d5068fcb66dfccb214.png)
We have 
which give us 
.
.![\[\frac{BM}{\sin \theta} = \frac{BO}{\sin (180^\circ- 2\theta)} = \frac{BO}{2\sin \theta \cos \theta}\Longrightarrow BO = 8\sqrt{5}\]](http://latex.artofproblemsolving.com/9/8/7/98794056865ca8143ec1bbf417383ab8e47a9333.png)
From 
we get 
.
.
.![\begin{align*}
[ABCD] &= [BCD] + [ADB] \\
&= \frac{1}{2} \cdot 30 \cdot 18\sqrt{5} \cdot \frac{1}{\sqrt{5}} + \frac{1}{2} \cdot 18\sqrt{5} \cdot 10\sqrt{13} \cdot \frac{1}{\sqrt{65}}\\
&= 270 + 90\\
&= \boxed{360}
\end{align*}](http://latex.artofproblemsolving.com/2/2/b/22b635a3f8859aebd926d81038c086203c6173aa.png)
![[asy]
size(300);
draw(circle((0,0), 10));
draw((0,10)--(30,10));
draw((30,10)--(-6,-8));
draw((-6,-8)--(0,10));
draw((0,10)--(0,-10));
draw((0,-10)--(-6,-8));
draw((0,0)--(10,0));
dot((0,0));
label("$O$", (0.29,0.6725), NE * 0.5);
dot((0,10));
label("$A$", (0.29,10.6625), N);
dot((30,10));
label("$D$", (30.26,10.6625), N);
dot((0,-10));
label("$C$", (0.29,-9.3175), S * 2);
dot((10,0));
label("$P$", (10.28,0.6725), NE);
dot((0,-5));
label("$E$", (0.29,-4.3225), S + E * 0.5);
dot((-6,-8));
label("$B$", (-7.0675,-8.9125), SW * 0.25);
[/asy]](http://latex.artofproblemsolving.com/9/c/9/9c9950c9b71a4545f5cdbdddc9414d20dffbd27c.png)
,
.
is the circumcenter 
be the point on 
such that 
.
,
and the Pythagorean Theorem gives 
,
and 
.![$[ABCD]=\frac{20\cdot 30}{2}+\frac{2\sqrt{10}\cdot 6\sqrt{10}}{2}=360$](http://latex.artofproblemsolving.com/0/8/2/082fba762ebfbcc3c57039d57aac09a000ee5c76.png)
.
,
is similar to triangle 
,
,
as 
by our ratio. Then we also know that triangle 
is similar to triangle 
by 
.![$[ABCD]=[ABC]+[ACD]=\frac{1}{2}(6)(20)+\frac{1}{2}(20)(30)=360 \rightarrow \fbox{D}$](http://latex.artofproblemsolving.com/d/f/3/df39bae95562104428de8cf6e94faa744ddeb644.png)
as the origin, 
,
.
has equation 
.
,
with 
,![\[ (x -10)^2 + (2x - 10)^2 = 100\implies 5x^2 - 60x + 200 = 100\implies x^2 - 12x + 20 = 0\implies x = 10, 2.\]](http://latex.artofproblemsolving.com/0/3/3/0334b2cae12f1c717bf5b1990aaf00a5284faba6.png)
Hence, the intersections are 
and 
.
.
.![\[ [ABC] =\frac{1}{2} \text{dist}(B, AC)\cdot AC = \frac{1}{2}\cdot 6\cdot 20 = 60.\]](http://latex.artofproblemsolving.com/b/7/1/b7190060d1718456c187c0f85daf4fd363eb99d5.png)
FInally, our answer is ![$[ABC] + [ACD] = 60 + 300 = \boxed{360}$](http://latex.artofproblemsolving.com/c/7/7/c779844c1559c4a5af1865fc3f9bcf2532b5dfbc.png)
.
and 
(and 
)
.
,
,
,
.![\[ (5-a)(15+a)=4a^2.\]](http://latex.artofproblemsolving.com/b/c/e/bcefa2a213bdc82a48f3b14c06c546024021a9f3.png)
Solving yields 
.![\[ [ABCD]=[ABC]+[ACD]=60+300=\boxed{360}.\]](http://latex.artofproblemsolving.com/0/8/7/0870854ad8aabc3d6e49f3a5d13ceb7e9f523c8c.png)
