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>? | ||
− | |||
<asy> | <asy> | ||
+ | unitsize(0.2cm); | ||
pair A,B,C,D,E,P; | pair A,B,C,D,E,P; | ||
A=(0,0); | A=(0,0); | ||
Line 23: | Line 23: | ||
dot(E); | dot(E); | ||
dot(P); | dot(P); | ||
− | label("A",A, | + | label("A",A,SW); |
− | label("B",B, | + | label("B",B,SE); |
− | label("C",C, | + | label("C",C,N); |
− | label("D",D, | + | label("D",D,NE); |
label("E",E,SSE); | label("E",E,SSE); | ||
− | label("P",P, | + | label("P",P,SSW); |
</asy> | </asy> | ||
− | + | <math>\textbf{(A) }13 \qquad \textbf{(B) }13.5 \qquad \textbf{(C) }14 \qquad \textbf{(D) }14.5 \qquad \textbf{(E) }15</math> | |
− | + | ||
+ | ==Solution 1== | ||
Let us use mass points: | Let us use mass points: | ||
− | 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>. | + | 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 <math>AP</math> 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>. Since the diagonals of quadrilaterals <math>AEDC</math>, <math>AD</math> and <math>CE</math>, are perpendicular, the area of <math>AEDC</math> is <math>\frac{6 \times 4.5}{2}=\boxed{\textbf{(B)} 13.5}</math> |
+ | |||
+ | ==Solution 2== | ||
+ | Note that triangle <math>DPE</math> is a right triangle, and that the four angles (angles <math>APC, CPD, DPE,</math> and <math>EPA</math>) 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 1.5+4\cdot 3+3\cdot 2+2\cdot 1.5}{2}=\boxed{\textbf{(B)} 13.5}</math> | ||
+ | |||
+ | ==Solution 3== | ||
+ | |||
+ | From the solution above, we can find that the lengths of the diagonals are <math>6</math> and <math>4.5</math>. Now, since the diagonals of AEDC are perpendicular, we use the area formula to find that the total area is <math>\frac{6\times4.5}{2} = \frac{27}{2} = 13.5 = \boxed{\textbf{(B)} 13.5}</math> | ||
+ | |||
+ | ==Solution 4== | ||
− | + | From the solutions above, we know that the sides CP and AP are 3 and 4 respectively because of the properties of medians that divide cevians into 1:2 ratios. We can then proceed to use the heron's formula on the middle triangle EPD and get the area of EPD as 3/2, (its simple computation really, nothing large). Then we can find the areas of the remaining triangles based on side and ratio length of the bases. | |
− | |||
+ | ==Solution 5== | ||
+ | We know that <math>[AEDC]=\frac{3}{4}[ABC]</math>, and <math>[ABC]=3[PAC]</math> using median properties. So Now we try to find <math>[PAC]</math>. Since <math>\triangle PAC\sim \triangle PDE</math>, then the side lengths of <math>\triangle PAC</math> are twice as long as <math>\triangle PDE</math> since <math>D</math> and <math>E</math> are midpoints. Therefore, <math>\frac{[PAC]}{[PDE]}=2^2=4</math>. It suffices to compute <math>[PDE]</math>. Notice that <math>(1.5, 2, 2.5)</math> is a Pythagorean Triple, so <math>[PDE]=\frac{1.5\times 2}{2}=1.5</math>. This implies <math>[PAC]=1.5\cdot 4=6</math>, and then <math>[ABC]=3\cdot 6=18</math>. Finally, <math>[AEDC]=\frac{3}{4}\times 18=\boxed{13.5}</math>. | ||
− | + | ~CoolJupiter | |
− | + | ==Solution 6== | |
+ | As from Solution 4, we find the area of <math>\triangle DPE</math> to be <math>\frac{3}{2}</math>. Because <math>DE = \frac{5}{2}</math>, the altitude perpendicular to <math>DE = \frac{6}{5}</math>. Also, because <math>DE || AC</math>, <math>\triangle ABC</math> is similar to <math>\triangle{DBE}</math> with side length ratio <math>2:1</math>, so <math>AC=5</math> and the altitude perpendicular to <math>AC = \frac{12}{5}</math>. The altitude of trapezoid <math>ACDE</math> is then <math>\frac{18}{5}</math> and the bases are <math>\frac{5}{2}</math> and <math>5</math>. So, we use the formula for the area of a trapezoid to find the area of <math>ACDE = \boxed{13.5}</math> | ||
== See also == | == See also == |
Latest revision as of 19:33, 11 January 2024
Contents
Problem
In triangle , medians
and
intersect at
,
,
, and
. What is the area of
?
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
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
. Since the diagonals of quadrilaterals
,
and
, are perpendicular, the area of
is
Solution 2
Note that triangle is a right triangle, and that the four angles (angles
and
) 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 that the lengths of the diagonals are and
. Now, since the diagonals of AEDC are perpendicular, we use the area formula to find that the total area is
Solution 4
From the solutions above, we know that the sides CP and AP are 3 and 4 respectively because of the properties of medians that divide cevians into 1:2 ratios. We can then proceed to use the heron's formula on the middle triangle EPD and get the area of EPD as 3/2, (its simple computation really, nothing large). Then we can find the areas of the remaining triangles based on side and ratio length of the bases.
Solution 5
We know that , and
using median properties. So Now we try to find
. Since
, then the side lengths of
are twice as long as
since
and
are midpoints. Therefore,
. It suffices to compute
. Notice that
is a Pythagorean Triple, so
. This implies
, and then
. Finally,
.
~CoolJupiter
Solution 6
As from Solution 4, we find the area of to be
. Because
, the altitude perpendicular to
. Also, because
,
is similar to
with side length ratio
, so
and the altitude perpendicular to
. The altitude of trapezoid
is then
and the bases are
and
. So, we use the formula for the area of a trapezoid to find the area of
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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