Difference between revisions of "2009 AMC 10A Problems/Problem 10"
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Revision as of 10:40, 13 August 2014
Problem
Triangle 
has a right angle at 
. Point 
is the foot of the altitude from , 
, and 
. What is the area of 
?
![[asy] unitsize(5mm); defaultpen(linewidth(.8pt)+fontsize(8pt)); dotfactor=4; pair B=(0,0), C=(sqrt(28),0), A=(0,sqrt(21)); pair D=foot(B,A,C); pair[] ps={B,C,A,D}; draw(A--B--C--cycle); draw(B--D); draw(rightanglemark(B,D,C)); dot(ps); label("$A$",A,NW); label("$B$",B,SW); label("$C$",C,SE); label("$D$",D,NE); label("$3$",midpoint(A--D),NE); label("$4$",midpoint(D--C),NE); [/asy]](http://latex.artofproblemsolving.com/4/b/e/4bea2faf81f4ba6dca40fdb9ce720c9ed82b168c.png)
Solution
It is a well-known fact that in any right triangle with the right angle at and the foot of the altitude from onto 
we have 
. (See below for a proof.) Then 
, and the area of the triangle is 
.
Proof: Consider the Pythagorean theorem for each of the triangles , 
, and 
. We get:
.
.
Substituting equations 2 and 3 into the left hand side of equation 1, we get 
.
Alternatively, note that 
. 
See Also
| 2009 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 9 |
Followed by Problem 11 | |
| 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 | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
