Difference between revisions of "2015 AMC 8 Problems/Problem 6"
(→Solution 1) |
(→Solution 2) |
||
| Line 7: | Line 7: | ||
===Solution 2=== | ===Solution 2=== | ||
| − | Splitting the isosceles triangle in half, we get a right triangle with hypotenuse <math>29</math> and leg <math>21</math>. Using the Pythagorean Theorem , we get <math>841-441</math>. Now that we know the height, the area is | + | Splitting the isosceles triangle in half, we get a right triangle with hypotenuse <math>29</math> and leg <math>21</math>. Using the Pythagorean Theorem , we get <math>sqrt(841-441)=20</math>. Now that we know the height, the area is |
<math>\dfrac{(20)(42)}{2} = \boxed{\textbf{(B)}~420}</math>. | <math>\dfrac{(20)(42)}{2} = \boxed{\textbf{(B)}~420}</math>. | ||
Revision as of 13:00, 30 November 2015
In 
, 
, and 
. What is the area of ?
Solution 1
We know the semi-perimeter of 
is 
. Next, we use Heron's Formula to find that the area of the triangle is just 
.
Solution 2
Splitting the isosceles triangle in half, we get a right triangle with hypotenuse 
and leg 
. Using the Pythagorean Theorem , we get 
. Now that we know the height, the area is
.
See Also
| 2015 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 5 |
Followed by Problem 7 | |
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
