2002 AMC 10P Problems/Problem 14
Contents
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 
Solution 2
If we draw a diagram as explained by the prompt, 
as 
and 
are both 
because they are angles of the squares, and 
given by the question. Similar to solution 1, if we draw 
and 
perpendicular to and respectively, square 
with a side length of 
will be formed. Looking at 
, we will notice 
. Therefore 
and 
are congruent by 
as they both have a and a 
angle and 
. Thus, the area of quadrilateral is the same as the area of square , which equals 
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
