Difference between revisions of "2009 AMC 10B Problems/Problem 18"
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path ortho = shift(M)*rotate(-90)*(A--C); | path ortho = shift(M)*rotate(-90)*(A--C); | ||
pair Ep = intersectionpoint(ortho, A--B); | pair Ep = intersectionpoint(ortho, A--B); | ||
− | draw( A--B--C-- | + | draw( A--B--C--Dsxklskl--cycle ); |
draw( A--C ); | draw( A--C ); | ||
draw( M--Ep ); | draw( M--Ep ); |
Revision as of 14:45, 18 May 2021
Contents
Problem
Rectangle has
and
. Point
is the midpoint of diagonal
, and
is on
with
. What is the area of
?
Solution 1 (Coordinate Geo)
Set to
. Since
is the midpoint of the diagonal, it would be
. The diagonal
would be the line
. Since
is perpendicular to
, its line would be in the form
. Plugging in
and
for
and
would give
. To find the x-intercept of
we plug in
for
and get
. Then, using the Shoelace Formula for
,
, and
, we find the area is
.
Solution 2
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--Dsxklskl--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); (Error making remote request. Unknown error_msg)
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
.
Solution 3 (Only Pythagorean Theorem)
Draw as shown from the diagram. Since
is of length
, we have that
is of length
, because of the midpoint
. Through the Pythagorean theorem, we know that
, which means
. Define
to be
for the sake of clarity. We know that
. From here, we know that
. From here, we can write the expression
. Now, remember
.
, since we set
in the start of the solution. Now to find the area
Video Solution
https://www.youtube.com/watch?v=VNJrZ-ABtS4
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 |
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