Difference between revisions of "2009 AMC 10B Problems/Problem 18"
(New page: == Problem == Rectangle <math>ABCD</math> has <math>AB=8</math> and <math>BC=6</math>. Point <math>M</math> is the midpoint of diagonal <math>\overline{AC}</math>, and <math>E</math> is o...) |
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Revision as of 11:54, 4 July 2013
Problem
Rectangle 
has 
and 
. Point 
is the midpoint of diagonal 
, and 
is on 
with 
. What is the area of 
?
Solution
![[asy] unitsize(0.75cm); defaultpen(0.8); pair A=(0,0), B=(8,0), C=(8,6), D=(0,6), M=(A+C)/2; path ortho = shift(M)*rotate(-90)*(A--C); pair Ep = intersectionpoint(ortho, A--B); draw( A--B--C--D--cycle ); draw( A--C ); draw( M--Ep ); filldraw( A--M--Ep--cycle, lightgray, black ); draw( rightanglemark(A,M,Ep) ); label("$A$",A,SW); label("$B$",B,SE); label("$C$",C,NE); label("$D$",D,NW); label("$E$",Ep,S); label("$M$",M,NW); [/asy]](http://latex.artofproblemsolving.com/0/6/c/06c316058781219dd8057b6f50468c97fa6199f7.png)
By the Pythagorean theorem we have 
, hence 
.
The triangles 
and 
have the same angle at 
and a right angle, thus all their angles are equal, and therefore these two triangles are similar.
The ratio of their sides is 
, hence the ratio of their areas is 
.
And as the area of triangle is 
, the area of triangle is 
.
See Also
| 2009 AMC 10B (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 10 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
