Difference between revisions of "2000 AMC 12 Problems/Problem 19"
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==Solution 2== | ==Solution 2== | ||
− | The area of <math>ADE</math> is <math>\frac{DE\cdot h}{2}=\frac{DE}{BC} \cdot \frac{BC\cdot h}{2}=\frac{DE}{BC}[ABC]</math> where <math>h</math> is the height of triangle <math>ABC</math>. Using Angle Bisector Theorem, we find <math>\frac{13}{BE}=\frac{15}{14-BE}</math>, which we solve to get <math>BE=\frac{13}{2}</math>. <math>D</math> is the midpoint of <math>BC</math> so <math>BD=7</math>. Thus we get the base of triangle <math>ADE, DE</math>, to be <math>\frac{1}{2}</math> units long. We can now use Heron's Formula on <math>ABC</math>. | + | The area of <math>ADE</math> is <math>\frac{DE\cdot h}{2}=\frac{DE}{BC} \cdot \frac{BC\cdot h}{2}=\frac{DE}{BC}[ABC]</math> where <math>h</math> is the height of triangle <math>ABC</math>. Using Angle Bisector Theorem, we find <math>\frac{13}{BE}=\frac{15}{14-BE}</math>, which we solve to get <math>BE=\frac{13}{2}</math>. <math>D</math> is the midpoint of <math>BC</math> so <math>BD=7</math>. Thus we get the base of triangle <math>ADE, DE</math>, to be <math>\frac{1}{2}</math> units long. We can now use [[Heron's Formula]] on <math>ABC</math>. |
<cmath>s=\frac{AB+BC+AC}{2}=21</cmath> | <cmath>s=\frac{AB+BC+AC}{2}=21</cmath> | ||
<cmath>[ABC]=\sqrt{(s)(s-AB)(s-BC)(s-AC)}=\sqrt{(21)(8)(7)(6)}=84</cmath> | <cmath>[ABC]=\sqrt{(s)(s-AB)(s-BC)(s-AC)}=\sqrt{(21)(8)(7)(6)}=84</cmath> | ||
<cmath>\frac{DE}{BC}[ABC]=\frac{\frac{1}{2}}{14}\cdot 84=3</cmath> | <cmath>\frac{DE}{BC}[ABC]=\frac{\frac{1}{2}}{14}\cdot 84=3</cmath> | ||
Therefore, the answer is <math>\mathrm{C}</math>. | Therefore, the answer is <math>\mathrm{C}</math>. | ||
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==Solution 3== | ==Solution 3== |
Revision as of 00:23, 31 December 2020
Problem
In triangle ,
,
,
. Let
denote the midpoint of
and let
denote the intersection of
with the bisector of angle
. Which of the following is closest to the area of the triangle
?
Solution 1
The answer is exactly , choice
.
We can find the area of triangle
by using the simple formula
. Dropping an altitude from
, we see that it has length
(we can split the large triangle into a
and a
triangle). Then we can apply the Angle Bisector Theorem on triangle
to solve for
. Solving
, we get that
.
is the midpoint of
so
. Thus we get the base of triangle
, to be
units long. Applying the formula
, we get
.
Solution 2
The area of is
where
is the height of triangle
. Using Angle Bisector Theorem, we find
, which we solve to get
.
is the midpoint of
so
. Thus we get the base of triangle
, to be
units long. We can now use Heron's Formula on
.
Therefore, the answer is
.
Solution 3
Let's find the area of by Heron,
Then,
Knowing that D is the midpoint of BC, then .
By Angle Bisector Theorem we know that:
Also, we know that:
And, we can easily see that , so,
See also
2000 AMC 12 (Problems • Answer Key • Resources) | |
Preceded by Problem 18 |
Followed by Problem 20 |
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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