Difference between revisions of "2002 AMC 10P Problems/Problem 14"
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== Solution 1== | == Solution 1== | ||
− | Draw a diagram. Split quadrilateral <math>EIDJ</math> into <math>\triangle EIJ</math> and <math>\triangle JDI.</math> Let the perpendicular from point <math>E</math> intersect <math>AD</math> at <math>X</math>, and let the perpendicular from point <math>E</math> intersect <math>CD</math> at <math>Y.</math> We know <math>\angle EJD=120^{\circ}</math> because <math>\angle JDC=90^{\circ}</math> since <math>ABCD</math> is a square, <math>\angle DCE=60^{\circ}</math> as given, and <math>\angle CEJ = 90^{\circ},</math> so <math>\angle EJD = 360^{\circ}-120^{\circ}-90^\ | + | Draw a diagram. Split quadrilateral <math>EIDJ</math> into <math>\triangle EIJ</math> and <math>\triangle JDI.</math> Let the perpendicular from point <math>E</math> intersect <math>AD</math> at <math>X</math>, and let the perpendicular from point <math>E</math> intersect <math>CD</math> at <math>Y.</math> We know <math>\angle EJD=120^{\circ}</math> because <math>\angle JDC=90^{\circ}</math> since <math>ABCD</math> is a square, <math>\angle DCE=60^{\circ}</math> as given, and <math>\angle CEJ = 90^{\circ},</math> so <math>\angle EJD = 360^{\circ}-120^{\circ}-90^{\circ}-90^{\circ}-60^{\circ}=120{\circ}.</math> Since <math>E</math> is at the center of square <math>ABCD</math>, <math>EX=EY=\frac{1}{2}.</math> By the <math>30^{\circ}-60^{\circ}-90^{\circ},</math> <math>ED=\frac{EX}{\sqrt{3}}=EC=\frac{EY}{\sqrt{3}}=\frac{1}{3}.</math> Additionally, we know <math>JD=AD-AX-XJ,</math> so <math>JD=1-\frac{1}{2}-\frac{1}{2 \sqrt{3}}</math> and we know <math>ID=DY+IY,</math> so <math>ID=\frac{1}{2}+\frac{1}{2 \sqrt{3}}.</math> From here, we can sum the areas of <math>\triangle EIJ</math> and <math>\triangle JDI.</math> to get the area of quadrilateral <math>EIDJ.</math> Therefore, |
\begin{align*} | \begin{align*} | ||
[EIDJ]&=[EIJ]+[JDI] \ | [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}{\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}{2}(\frac{1}{3})+\frac{\frac{1}{4}-\frac{1}{12}}{2} \ |
&=\frac{1}{6}+\frac{1}{12} \ | &=\frac{1}{6}+\frac{1}{12} \ | ||
&=\frac{1}{4} \ | &=\frac{1}{4} \ | ||
\end{align*} | \end{align*} | ||
− | Thus, our answer is \boxed{\textbf{(A) } \frac{1}{4}}. | + | Thus, our answer is <math>\boxed{\textbf{(A) } \frac{1}{4}}.</math> |
== See also == | == See also == | ||
{{AMC10 box|year=2002|ab=P|num-b=13|num-a=15}} | {{AMC10 box|year=2002|ab=P|num-b=13|num-a=15}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 07:52, 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 the Additionally, we know so and we know so From here, we can sum the areas of and to get the area of quadrilateral Therefore,
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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