Difference between revisions of "2025 AMC 12A Problems/Problem 18"

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{{duplicate|[[2025 AMC 10A Problems/Problem 21|2025 AMC 10A #21]] and [[2025 AMC 12A Problems/Problem 18|2024 AMC 12A #18]]}}
 
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>.
 
 
 
 
What is the area of triangle <math>\triangle APR</math>?
 
 
 
 
<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 <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>
 
  
 
==See also==
 
==See also==
 
 
{{AMC10 box|year=2025|ab=A|num-b=20|num-a=22}}
 
{{AMC10 box|year=2025|ab=A|num-b=20|num-a=22}}
 
{{AMC12 box|year=2025|ab=A|num-b=17|num-a=19}}
 
{{AMC12 box|year=2025|ab=A|num-b=17|num-a=19}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 16:20, 11 December 2024

See also

2025 AMC 10A (ProblemsAnswer KeyResources)
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 (ProblemsAnswer KeyResources)
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

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