2017 AMC 10B Problems/Problem 15
Contents
Problem
Rectangle 
has 
and 
. Point 
is the foot of the perpendicular from 
to diagonal 
. What is the area of 
?
Solution
![[asy] pair A,B,C,D,E; A=(0,4); B=(3,4); C=(3,0); D=(0,0); draw(A--B--C--D--cycle); label("$A$",A,N); label("$B$",B,N); label("$C$",C,S); label("$D$",D,S); E=foot(B,A,C); draw(E--B); draw(A--C); draw(rightanglemark(B,E,C)); label("$E$",E,N); draw(D--E); label("$3$",A--B,N); label("$4$",B--C,E); [/asy]](http://latex.artofproblemsolving.com/8/1/e/81ebb7263d49644ae842d3b0e2a3cfce7c5885ce.png)
First, note that 
because 
is a right triangle. In addition, we have ![$AB\cdot BC=2[ABC]=AC\cdot BE$](http://latex.artofproblemsolving.com/9/8/7/98758b43351b7737fc5a633ef12749dedd2574ca.png)
, so 
. Using similar triangles within , we get that 
and 
.
Let 
be the foot of the perpendicular from to 
. Since 
and 
are parallel, 
is similar to 
. Therefore, we have 
. Since , 
. Note that 
is an altitude of 
from 
, which has length 
. Therefore, the area of is 
Solution 2
Alternatively, we can use coordinates. Denote 
as the origin. We find the equation for 
as 
, and 
as 
. Solving for 
yields 
. Our final answer then becomes
See Also
| 2017 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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