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

(Created page with "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>, respect...")
 
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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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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|num-b=20|num-a=22}}
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{{AMC12 box|year=2025|ab=A|num-b=17|num-a=19}}
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{{MAA Notice}}

Latest revision as of 23:57, 23 November 2024

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

See also

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

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