Difference between revisions of "2018 AMC 10A Problems/Problem 24"
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The area of <math>\bigtriangleup ABG</math> to the area of <math>\bigtriangleup ACG</math> is <math>5:1</math> by Law of Sines. So the area of <math>\bigtriangleup ABG</math> is <math>100</math>. Since <math>\overline{DE}</math> is the midsegment of <math>\bigtriangleup ABC</math>, so <math>\overline{DF}</math> is the midsegment of <math>\bigtriangleup ABG</math> . So the area of <math>\bigtriangleup ACG</math> to the area of <math>\bigtriangleup ABG</math> is <math>1:4</math> , so the area of <math>\bigtriangleup ACG</math> is <math>25</math>, by similar triangles. Therefore the area of quad <math>FDBG</math> is <math>100-25=\boxed{75}</math> | The area of <math>\bigtriangleup ABG</math> to the area of <math>\bigtriangleup ACG</math> is <math>5:1</math> by Law of Sines. So the area of <math>\bigtriangleup ABG</math> is <math>100</math>. Since <math>\overline{DE}</math> is the midsegment of <math>\bigtriangleup ABC</math>, so <math>\overline{DF}</math> is the midsegment of <math>\bigtriangleup ABG</math> . So the area of <math>\bigtriangleup ACG</math> to the area of <math>\bigtriangleup ABG</math> is <math>1:4</math> , so the area of <math>\bigtriangleup ACG</math> is <math>25</math>, by similar triangles. Therefore the area of quad <math>FDBG</math> is <math>100-25=\boxed{75}</math> | ||
(steakfails) | (steakfails) | ||
+ | |||
+ | ==Solution 4 == | ||
+ | The area of quadrilateral <math>FDBG</math> is the area of <math>\bigtriangleup ABG</math> minus the area of <math>\bigtriangleup ADF</math>. Notice, <math>\overline{DE} || \overline{BC}</math>, so <math>\bigtriangleup ABG \sim \bigtriangleup ADF</math>, and since <math>\overline{AD}:\overline{AB}=1:2</math>, the area of <math>\bigtriangleup ADF:\bigtriangleup ABG=(1:2)^2=1:4</math>. Given that the area of <math>\bigtriangleup ABC</math> is <math>120</math>, using <math>\frac{bh}{2}</math> on side <math>AB</math> yields <math>\frac{50h}{2}=120\implies h=\frac{240}{50}=\frac{24}{5}</math>. Using angle bisector theorem, <math>\overline{BG}:\overline{BC}=50:(10+50)=5:6</math>, so the height of <math>\bigtriangleup ABG: \bigtriangleup ACB=5:6</math>, and so our answer is <math>\big[ FDBG\big] = \big[ABG\big]-\big[ ADF\big] = \big[ ABG\big]\big(1-\frac{1}{4}\big)=\frac{3}{4}\cdot \frac{bh}{2}=\frac{3}{8}\cdot 50\cdot \frac{5}{6}\cdot \frac{24}{5}=\frac{3}{8}\cdot 200=\boxed{75}</math> | ||
+ | -Solution by ktong | ||
==See Also== | ==See Also== |
Revision as of 12:52, 10 February 2018
Problem
Triangle with
and
has area
. Let
be the midpoint of
, and let
be the midpoint of
. The angle bisector of
intersects
and
at
and
, respectively. What is the area of quadrilateral
?
Solution
By angle bisector theorem, . By similar triangles,
, and the height of this trapezoid is
, where
is the length of the altitude to
. Then
and we wish to compute
.
Solution 2
is midway from
to
, and
. Therefore,
is a quarter of the area of
, which is
. Subsequently, we can compute the area of quadrilateral
to be
. Using the angle bisector theorem in the same fashion as the previous problem, we get that
is
times the length of
. We want the larger piece, as described by the problem. Because the heights are identical, one area is
times the other, and
.
Solution 3
The area of to the area of
is
by Law of Sines. So the area of
is
. Since
is the midsegment of
, so
is the midsegment of
. So the area of
to the area of
is
, so the area of
is
, by similar triangles. Therefore the area of quad
is
(steakfails)
Solution 4
The area of quadrilateral is the area of
minus the area of
. Notice,
, so
, and since
, the area of
. Given that the area of
is
, using
on side
yields
. Using angle bisector theorem,
, so the height of
, and so our answer is
-Solution by ktong
See Also
2018 AMC 10A (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 AMC 10 Problems and Solutions |
2018 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 17 |
Followed by Problem 19 |
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 12 Problems and Solutions |
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