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| Let <math>ABCD</math> be a square with side length 10. Points <math>P</math> and <math>Q</math> lie on sides <math>\overline{AB}</math> and <math>\overline{BC}</math>, respectively, such that <math>AP = BQ = 2</math>. Let <math>R</math> be the intersection of segments <math>\overline{AQ}</math> and <math>\overline{DP}</math>.
| | #redirect [[2025 AMC 12A Problems/Problem 15]] |
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| What is the area of triangle <math>\triangle APR</math>?
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| <asy>
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| pair A,B,C,D,P,Q,R;
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| A=(0,0);
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| B=(20,0);
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| C=(20,20);
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| D=(0,20);
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| P=(4,0);
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| Q=(20,4);
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| R=intersectionpoint(A--Q, D--P);
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| draw(A--B--C--D--cycle);
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| draw(A--Q--C);
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| draw(D--P--B);
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| dot(A);
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| dot(B);
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| dot(C);
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| dot(D);
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| dot(P);
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| dot(Q);
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| dot(R);
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| label("$A$",A,SW);
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| label("$B$",B,SE);
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| label("$C$",C,NE);
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| label("$D$",D,NW);
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| label("$P$",P,NW);
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| label("$Q$",Q,SE);
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| label("$R$",R,N);
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| label("$10$", (A+B)/2, S);
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| label("$2$", (A+P)/2, S);
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| label("$2$", (B+Q)/2, E);
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| </asy>
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| <math>\textbf{(A)}\ 2 \qquad\textbf{(B)}\ 5/3 \qquad\textbf{(C)}\ 10/3 \qquad\textbf{(D)}\ 5 \qquad\textbf{(E)}\ 6</math>
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| ==Solution==
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| We can find the area of <math>\triangle APR</math> by finding its base and height.
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| * Base: <math>AP = 2</math>
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| * Height: To find the height, we can use similar triangles.
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| * <math>\triangle APR \sim \triangle AQD</math>
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| * So, <math>\frac{AP}{AQ} = \frac{PR}{QD}</math>
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| * Substituting the values, we get: <math>\frac{2}{12} = \frac{PR}{10}</math>
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| * Solving for <math>PR</math>, we get <math>PR = \frac{5}{3}</math>.
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| Therefore, the area of <math>\triangle APR = \frac{1}{2} \cdot base \cdot height = \frac{1}{2} \cdot 2 \cdot \frac{5}{3} = \textbf{(B)} \frac{5}{3} \qquad\square</math>
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| ==See also==
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| {{AMC10 box|year=2025|ab=A|before=First Problem|num-a=2}}
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| {{MAA Notice}}
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