Difference between revisions of "2002 AMC 10P Problems/Problem 14"
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[EIDJ]&=[EIJ]+[JDI] \\ | [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}{\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}{2}(\frac{1}{3})+\frac{1}{2}(\frac{1}{4}-\frac{1}{12}) \\ |
&=\frac{1}{6}+\frac{1}{12} \\ | &=\frac{1}{6}+\frac{1}{12} \\ | ||
&=\frac{1}{4} \\ | &=\frac{1}{4} \\ |
Revision as of 17:14, 15 July 2024
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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