2025 AMC 10A Problems/Problem 21

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 $\triangle APR = \frac{1}{2} \cdot base \cdot height = \frac{1}{2} \cdot 2 \cdot \frac{5}{3} = \textbf{(B)} \frac{5}{3} \qquad\square$

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2025 AMC 10A Problems/Problem 21

Solution

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