Difference between revisions of "2013 AMC 10B Problems/Problem 16"
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In triangle <math>ABC</math>, medians <math>AD</math> and <math>CE</math> intersect at <math>P</math>, <math>PE=1.5</math>, <math>PD=2</math>, and <math>DE=2.5</math>. What is the area of <math>AEDC</math>? | In triangle <math>ABC</math>, medians <math>AD</math> and <math>CE</math> intersect at <math>P</math>, <math>PE=1.5</math>, <math>PD=2</math>, and <math>DE=2.5</math>. What is the area of <math>AEDC</math>? | ||
− | \qquad\textbf{(A)}13\qquad\textbf{(B)}13.5\qquad\textbf{(C)}14\qquad\textbf{(D)}14.5\qquad\textbf{(E)}<math> | + | <math>\qquad\textbf{(A)}13\qquad\textbf{(B)}13.5\qquad\textbf{(C)}14\qquad\textbf{(D)}14.5\qquad\textbf{(E)}</math> |
==Solution== | ==Solution== | ||
Let us use mass points: | Let us use mass points: | ||
− | Assign < | + | Assign <math>B</math> mass <math>1</math>. Thus, because <math>E</math> is the midpoint of <math>AB</math>, <math>A</math> also has a mass of <math>1</math>. Similarly, <math>C</math> has a mass of <math>1</math>. <math>D</math> and <math>E</math> each have a mass of <math>2</math> because they are between <math>B</math> and <math>C</math> and <math>A</math> and <math>B</math> respectively. Note that the mass of <math>D</math> is twice the mass of <math>A</math>, so AP must be twice as long as <math>PD</math>. PD has length <math>2</math>, so <math>AP</math> has length <math>4</math> and <math>AD</math> has length <math>6</math>. Similarly, <math>CP</math> is twice <math>PE</math> and <math>PE=1.5</math>, so <math>CP=3</math> and <math>CE=4.5</math>. Now note that triangle <math>PED</math> is a <math>3-4-5</math> right triangle with the right angle <math>DPE</math>. This means that the quadrilateral <math>AEDC</math> is a kite. The area of a kite is half the product of the diagonals, <math>AD</math> and <math>CE</math>. Recall that they are <math>6</math> and <math>4.5</math> respectively, so the area of <math>AEDC</math> is <math>6*4.5/2=\boxed{\textbf{(B)} 13.5}</math> |
==Solution 2== | ==Solution 2== | ||
− | Note that triangle < | + | Note that triangle <math>DPE</math> is a right triangle, and that the four angles that have point <math>P</math> are all right angles. Using the fact that the centroid (<math>P</math>) divides each median in a <math>2:1</math> ratio, <math>AP=4</math> and <math>CP=3</math>. Quadrilateral <math>AEDC</math> is now just four right triangles. The area is <math>\frac{4\cdot 2+4\cdot 3+3\cdot 2+2\cdot 1.5}{2}=\boxed{\textbf{(B)} 13.5}</math> |
Revision as of 23:16, 21 February 2013
Problem
In triangle , medians
and
intersect at
,
,
, and
. What is the area of
?
Solution
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