Difference between revisions of "2013 AMC 10B Problems/Problem 16"
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===Solution 3=== | ===Solution 3=== | ||
− | From the solution above, we can find the length of the diagonals to be 6 and 4.5. Now, since AEDC is a | + | From the solution above, we can find the length of the diagonals to be 6 and 4.5. Now, since AEDC is a trapezoid, we use the area formula to find that the total area is <math>\frac{6\times4.5}{2} = \frac{27}{2} = 13.5 = \boxed{B}</math> |
== See also == | == See also == |
Revision as of 17:30, 2 January 2016
Problem
In triangle , medians
and
intersect at
,
,
, and
. What is the area of
?
Solution
Solution 1
Let us use mass points:
Assign mass
. Thus, because
is the midpoint of
,
also has a mass of
. Similarly,
has a mass of
.
and
each have a mass of
because they are between
and
and
and
respectively. Note that the mass of
is twice the mass of
, so AP must be twice as long as
. PD has length
, so
has length
and
has length
. Similarly,
is twice
and
, so
and
. Now note that triangle
is a
right triangle with the right angle
. This means that the quadrilateral
is a kite. The area of a kite is half the product of the diagonals,
and
. Recall that they are
and
respectively, so the area of
is
Solution 2
Note that triangle is a right triangle, and that the four angles that have point
are all right angles. Using the fact that the centroid (
) divides each median in a
ratio,
and
. Quadrilateral
is now just four right triangles. The area is
Solution 3
From the solution above, we can find the length of the diagonals to be 6 and 4.5. Now, since AEDC is a trapezoid, we use the area formula to find that the total area is
See also
2013 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 15 |
Followed by Problem 17 | |
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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