Difference between revisions of "2013 AMC 8 Problems/Problem 24"
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J = (2,0); | J = (2,0); | ||
X= extension(I,J,A,B); | X= extension(I,J,A,B); | ||
− | dot(X); | + | dot(X,red); |
draw(I--X--B,red); | draw(I--X--B,red); | ||
draw(A--B); | draw(A--B); | ||
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label("$H$", H, N); | label("$H$", H, N); | ||
label("$I$", I, NE); | label("$I$", I, NE); | ||
+ | label("$X$", X,SE); | ||
label("$J$", J, SE);</asy> | label("$J$", J, SE);</asy> | ||
Revision as of 19:33, 27 November 2013
Problem
Squares ,
, and
are equal in area. Points
and
are the midpoints of sides
and
, respectively. What is the ratio of the area of the shaded pentagon
to the sum of the areas of the three squares?
Solution
First let (where
is the side length of the squares) for simplicity. We can extend
until it hits the extension of
. Call this point
. The area of triangle
then is
The area of rectangle
is
. Thus, our desired area is
. Now, the ratio of the shaded area to the combined area of the three squares is
.
See Also
2013 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 23 |
Followed by Problem 25 | |
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 AJHSME/AMC 8 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.