Difference between revisions of "2006 AMC 10B Problems/Problem 23"
(extra solution) |
m (→Solution 2 (mass points)) |
||
Line 65: | Line 65: | ||
==Solution 2 (mass points)== | ==Solution 2 (mass points)== | ||
− | We see that <math>EF:FB=3:7</math> and <math>AF=FD</math>. We assign a mass of <math>7</math> to <math>E</math> and <math>3</math> to <math> | + | We see that <math>EF:FB=3:7</math> and <math>AF=FD</math>. We assign a mass of <math>7</math> to <math>E</math> and <math>3</math> to <math>B</math>, making <math>F</math> have mass <math>10</math> and <math>A</math> and <math>D</math> each have mass 5. Now, <math>C</math> has mass <math>2</math>. Therefore, the area of triangle <math>CEB</math> is <math>10 \cdot 2.5=25</math>, so the area of <math>CEFD</math> is <math>\boxed{(D), 18}</math> |
+ | |||
== See also == | == See also == | ||
{{AMC10 box|year=2006|ab=B|num-b=22|num-a=24}} | {{AMC10 box|year=2006|ab=B|num-b=22|num-a=24}} |
Revision as of 14:15, 30 January 2016
Problem
A triangle is partitioned into three triangles and a quadrilateral by drawing two lines from vertices to their opposite sides. The areas of the three triangles are 3, 7, and 7 as shown. What is the area of the shaded quadrilateral?
Solution 1
Label the points in the figure as shown below, and draw the segment . This segment divides the quadrilateral into two triangles, let their areas be
and
.
Since triangles and
share an altitude from
and have equal area, their bases must be equal, hence
.
Since triangles and
share an altitude from
and their respective bases are equal, their areas must be equal, hence
.
Since triangles and
share an altitude from
and their respective areas are in the ratio
, their bases must be in the same ratio, hence
.
Since triangles and
share an altitude from
and their respective bases are in the ratio
, their areas must be in the same ratio, hence
, which gives us
.
Substituting into the second equation we get
, which solves to
. Then
, and the total area of the quadrilateral is
.
Solution 2 (mass points)
We see that and
. We assign a mass of
to
and
to
, making
have mass
and
and
each have mass 5. Now,
has mass
. Therefore, the area of triangle
is
, so the area of
is
See also
2006 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 22 |
Followed by Problem 24 | |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.