Difference between revisions of "2000 AMC 12 Problems/Problem 19"
(→Solution 3) |
(→Solution 3) |
||
Line 37: | Line 37: | ||
label("$15$",A--C,NE); | label("$15$",A--C,NE); | ||
label("$14$",B--C,S); | label("$14$",B--C,S); | ||
+ | label("$6.5$",B--E,N); | ||
+ | label("$7$",C--D,N); | ||
</asy> | </asy> | ||
Line 47: | Line 49: | ||
<math>A^2=s(s-a)(s-b)(s-c)\\\\A^2=21(21-14)(21-15)(21-13)\\\\\boxed{A=84}</math> | <math>A^2=s(s-a)(s-b)(s-c)\\\\A^2=21(21-14)(21-15)(21-13)\\\\\boxed{A=84}</math> | ||
− | Knowing that D is the midpoint of BC, | + | Knowing that D is the midpoint of BC, then <math>BD=CD=7</math>. |
By [[Angle Bisector Theorem]] we know that: | By [[Angle Bisector Theorem]] we know that: |
Revision as of 16:27, 7 August 2017
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.