Difference between revisions of "2025 AMC 10A Problems/Problem 21"

Latest revision as of 17:20, 11 December 2024

2025 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 20	Followed by Problem 22
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

2025 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 17	Followed by Problem 19
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 12 Problems and Solutions

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions.

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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>.
+==See also==
+{{AMC10 box|year=2025|ab=A|num-b=20|num-a=22}}
-What is the area of triangle <math>\triangle APR</math>?
+{{AMC12 box|year=2025|ab=A|num-b=17|num-a=19}}
+{{MAA Notice}}
-<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>
-<math>\textbf{(A)}\ 2 \qquad\textbf{(B)}\ 5/3 \qquad\textbf{(C)}\ 10/3 \qquad\textbf{(D)}\ 5 \qquad\textbf{(E)}\ 6</math>
-==Solution==
-We can find the area of <math>\triangle APR</math> by finding its base and height.
-* Base: <math>AP = 2</math>
-* Height: To find the height, we can use similar triangles.
-* <math>\triangle APR \sim \triangle AQD</math>
-* So, <math>\frac{AP}{AQ} = \frac{PR}{QD}</math>
-* Substituting the values, we get: <math>\frac{2}{12} = \frac{PR}{10}</math>
-* Solving for <math>PR</math>, we get <math>PR = \frac{5}{3}</math>.
-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\

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Difference between revisions of "2025 AMC 10A Problems/Problem 21"

Latest revision as of 17:20, 11 December 2024

See also