2025 AMC 10A Problems/Problem 21

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Let $ABCD$ be a square with side length 10. Points $P$ and $Q$ lie on sides $\overline{AB}$ and $\overline{BC}$, respectively, such that $AP = BQ = 2$. Let $R$ be the intersection of segments $\overline{AQ}$ and $\overline{DP}$.


What is the area of triangle $\triangle APR$?


[asy]  pair A,B,C,D,P,Q,R;  A=(0,0);  B=(20,0);  C=(20,20);  D=(0,20);  P=(4,0);  Q=(20,4);  R=intersectionpoint(A--Q, D--P);    draw(A--B--C--D--cycle);  draw(A--Q--C);  draw(D--P--B);  dot(A);  dot(B);  dot(C);  dot(D);  dot(P);  dot(Q);  dot(R);  label("$A$",A,SW);  label("$B$",B,SE);  label("$C$",C,NE);  label("$D$",D,NW);  label("$P$",P,NW);  label("$Q$",Q,SE);  label("$R$",R,N);  label("$10$", (A+B)/2, S);  label("$2$", (A+P)/2, S);  label("$2$", (B+Q)/2, E);  [/asy]


$\textbf{(A)}\ 2 \qquad\textbf{(B)}\ 5/3 \qquad\textbf{(C)}\ 10/3 \qquad\textbf{(D)}\ 5 \qquad\textbf{(E)}\ 6$


Solution

We can find the area of $\triangle APR$ by finding its base and height. 


  • Base: $AP = 2$
  • Height: To find the height, we can use similar triangles. 
  • $\triangle APR \sim \triangle AQD$
  • So, $\frac{AP}{AQ} = \frac{PR}{QD}$
  • Substituting the values, we get: $\frac{2}{12} = \frac{PR}{10}$
  • Solving for $PR$, we get $PR = \frac{5}{3}$


Therefore, the area of $\triangle APR = \frac{1}{2} \cdot base \cdot height = \frac{1}{2} \cdot 2 \cdot \frac{5}{3} = \textbf{(B)}\ 5/3 \qquad\