2025 AMC 10A Problems/Problem 21
Let 
be a square with side length 10. Points 
and 
lie on sides 
and 
, respectively, such that 
. Let 
be the intersection of segments 
and 
.
What is the area of triangle 
?
![[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]](http://latex.artofproblemsolving.com/c/0/9/c09bebc96c0340e3959fef1c5715b08f9ace4693.png)
Solution
We can find the area of by finding its base and height.
- Base:
- Height: To find the height, we can use similar triangles.
- So,
- Substituting the values, we get:
- Solving for
, we get 
.
, we get 
.
Therefore, the area of 
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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
