Difference between revisions of "2002 AMC 10P Problems/Problem 14"
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Thus, our answer is <math>\boxed{\textbf{(A) } \frac{1}{4}}.</math> | Thus, our answer is <math>\boxed{\textbf{(A) } \frac{1}{4}}.</math> | ||
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+ | ==Solution 2== | ||
+ | If we draw a diagram as explained by the prompt, <math>\angle EJD = 360^{\circ}-90^{\circ}-90^{\circ}-60^{\circ} = 120^{\circ}</math> as <math>\angle IEJ</math> and <math>\angle JDI</math> are both <math>90^{\circ}</math> because they are angles of the squares, and <math>\angle EID=60^{\circ}</math> given by the question. Similar to solution 1, if we draw <math>EX</math> and <math>EY</math> perpendicular to <math>AD</math> and <math>CD</math> respectively, square <math>EXDY</math> with a side length of <math>\frac{1}{2}</math> will be formed. Looking at <math>\triangle JXE</math>, we will notice <math>\angle EJX = 180^{\circ}-120^{\circ}=60^{\circ}</math>. Therefore <math>\triangle EYI</math> and <math>\triangle EXJ</math> are congruent by <math>A.A.S.</math> as they both have a <math>90^{\circ}</math> and a <math>60^{\circ}</math> angle and <math>EX=EY</math>. Thus, the area of quadrilateral <math>EIDJ</math> is the same as the area of square <math>EXDY</math>, which equals <math>\frac{1}{2} \times \frac{1}{2} = \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}} |
Latest revision as of 18:16, 18 July 2024
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 |
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