2002 AMC 10P Problems/Problem 14
Problem 14
The vertex 
of a square 
is at the center of square 
The length of a side of 
is 
and the length of a side of is 
Side 
intersects 
at 
and 
intersects 
at 
If angle 
the area of quadrilateral 
is
Solution 1
Draw a diagram. Split quadrilateral into 
and 
Let the perpendicular from point intersect at 
, and let the perpendicular from point intersect at 
We know 
because 
since is a square, 
as given, and 
so \angle EJD = 360^{\circ}-120^{\circ}-90^\{circ}-90^\{circ}-60^\{circ}=120\{circ}.
E
ABCD
EX=EY=\frac{1}{2}.
30^{\circ}-60^{\circ}-90^{\circ}ED=\frac{EX}{\sqrt{3}}=EC=\frac{EY}{\sqrt{3}}=\frac{1}{3}. Additionally, we know 
so 
and we know 
so 
From here, we can sum the areas of and to get the area of quadrilateral 
Therefore,
\begin{align*} [EIDJ]&=[EIJ]+[JDI] // &=\frac{1}{2}\frac{1}{\sqrt{3}}\frac{1}{\sqrt{3}} + \frac{1}{2} (1-\frac{1}{2}-\frac{1}{2 \sqrt{3}}) (\frac{1}{2}+\frac{1}{2 \sqrt{3}}) // &=\frac{1}{2}\frac{1}{3}+\frac{\frac{1}{4}-\frac{1}{12}}{2} // &=\frac{1}{6}+\frac{1}{12} // &=\frac{1}{4} // \end{align*}
See also
| 2002 AMC 10P (Problems • Answer Key • Resources) | ||
| Preceded by Problem 13 |
Followed by Problem 15 | |
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