Difference between revisions of "2002 AMC 10P Problems/Problem 14"
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− | 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 | + | 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 <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}}=\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*} |
Revision as of 09:07, 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{\frac{1}{4}-\frac{1}{12}}{2} \\ &=\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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