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 
Since is at the center of square , 
By 
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} (\frac{1}{2}-\frac{1}{2 \sqrt{3}}) (\frac{1}{2}+\frac{1}{2 \sqrt{3}}) \\ &=\frac{1}{2}(\frac{1}{3})+\frac{1}{2}(\frac{1}{4}-\frac{1}{12}) \\ &=\frac{1}{6}+\frac{1}{12} \\ &=\frac{1}{4} \\ \end{align*}
Thus, our answer is 
See also
| 2002 AMC 10P (Problems • Answer Key • Resources) | ||
| Preceded by Problem 13 |
Followed by Problem 15 | |
| 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 | ||
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