Difference between revisions of "2019 AMC 8 Problems/Problem 24"
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==Solution 1== | ==Solution 1== | ||
− | The | + | Draw <math>X</math> so that <math>XD</math> is parallel to <math>BC</math>. That makes triangles <math>BEF</math> and <math>EXD</math> congruent since <math>BE</math>=<math>ED</math>. <math>FC</math>=3<math>XD</math> so <math>BC</math>=4<math>BC</math>. Since <math>AF</math>=3<math>EF</math>( <math>XE</math>=<math>EF</math> and <math>AX</math>=1/3<math>AF</math>, so <math>XE</math>=<math>EF</math>=1/3<math>AF</math>), the altitude of triangle <math>BEF</math> is equal to 1/3 of the altitude of <math>ABC</math>. The area of <math>ABC</math> is 360, so the area of <math>BEF</math>=1/3*1/4*360=<math>\boxed{(B) 30}</math> |
==See Also== | ==See Also== |
Revision as of 19:42, 20 November 2019
Problem 24
In triangle , point
divides side
s that
. Let
be the midpoint of
and left
be the point of intersection of line
and line
. Given that the area of
is
, what is the area of
?
Solution 1
Draw so that
is parallel to
. That makes triangles
and
congruent since
=
.
=3
so
=4
. Since
=3
(
=
and
=1/3
, so
=
=1/3
), the altitude of triangle
is equal to 1/3 of the altitude of
. The area of
is 360, so the area of
=1/3*1/4*360=
See Also
2019 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 |
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