Difference between revisions of "2015 AMC 8 Problems/Problem 6"
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We know the semi-perimeter of <math>\triangle ABC</math> is <math>\frac{29+29+42}{2}=50</math>. Next, we use Heron's Formula to find that the area of the triangle is just <math>\sqrt{50(50-29)^2(50-42)}=\sqrt{50 \cdot 21^2 \cdot 8}=\boxed{\textbf{(B) }420}</math>. | We know the semi-perimeter of <math>\triangle ABC</math> is <math>\frac{29+29+42}{2}=50</math>. Next, we use Heron's Formula to find that the area of the triangle is just <math>\sqrt{50(50-29)^2(50-42)}=\sqrt{50 \cdot 21^2 \cdot 8}=\boxed{\textbf{(B) }420}</math>. | ||
| − | ===Solution 2=== | + | ===Solution 2 (easier)=== |
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 know the height is <math>\sqrt{29^2-21^2}=20</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 know the height is <math>\sqrt{29^2-21^2}=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 01:31, 24 December 2022
Contents
Problem
In 
, 
, and 
. What is the area of ?
Video Solution
https://www.youtube.com/watch?v=Bl3_W2i5zwc
Solutions
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 (easier)
Splitting the isosceles triangle in half, we get a right triangle with hypotenuse 
and leg 
. Using the Pythagorean Theorem , we know the height is 
. Now that we know the height, the area is
.
Video Solution 2
~savannahsolver
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. 
