2021 Fall AMC 10B Problems/Problem 15
Problem
In square 
, points 
and 
lie on 
and 
, respectively. Segments 
and 
intersect at right angles at 
, with 
and 
. What is the area of the square?
![[asy] size(170); defaultpen(linewidth(0.6)+fontsize(10)); real r = 3.5; pair A = origin, B = (5,0), C = (5,5), D = (0,5), P = (0,r), Q = (5-r,0), R = intersectionpoint(B--P,C--Q); draw(A--B--C--D--A^^B--P^^C--Q^^rightanglemark(P,R,C,7)); dot("$A$",A,S); dot("$B$",B,S); dot("$C$",C,N); dot("$D$",D,N); dot("$Q$",Q,S); dot("$P$",P,W); dot("$R$",R,1.3*S); label("$7$",(P+R)/2,NE); label("$6$",(R+B)/2,NE); [/asy]](http://latex.artofproblemsolving.com/9/3/0/9303b2a7f1e03de801c7915e1a61d7b755a6a43d.png)
Solution
Note that 
Then, it follows that 
Thus, 
Define 
to be the length of side 
then 
Because 
is the altitude of the triangle, we can use the property that 
Substituting the given lengths, we have ![\[(13-x) \cdot x = 36.\]](http://latex.artofproblemsolving.com/c/a/6/ca645dc62b5b9083e880a83a0e861099eeec54a2.png)
Solving, gives 
and 
We eliminate the possibilty of 
because 
Thus, the side lengnth of the square, by Pythagorean Theorem, is ![\[\sqrt{9^2 +6^2} = \sqrt{81+36} = \sqrt{117}.\]](http://latex.artofproblemsolving.com/b/3/a/b3ab8ad3129aca4e4a429de69029760e4aaa7cf0.png)
Thus, the area of the sqaure is 
Thus, the answer is 
~NH14
See Also
| 2021 Fall AMC 10B (Problems • Answer Key • Resources) | ||
| Preceded by Problem 14 |
Followed by Problem 16 | |
| 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 | ||
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