Difference between revisions of "2015 AMC 8 Problems/Problem 6"
| Line 5: | Line 5: | ||
===Solution 1=== | ===Solution 1=== | ||
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}=\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}=\textbf{(B) }420.</math> | ||
| + | |||
| + | ===Solution 2=== | ||
| + | Splitting the isosceles triangle in half, we get a right triangle with hypotenuse 29 and leg 21. By Pythagorean Theorem the height is 20 so the area is | ||
| + | <math>\dfrac{(20)(42)}{2} = \boxed{\textbf{(B)}~420}</math>. | ||
==See Also== | ==See Also== | ||
{{AMC8 box|year=2015|num-b=5|num-a=7}} | {{AMC8 box|year=2015|num-b=5|num-a=7}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 16:35, 25 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 29 and leg 21. By Pythagorean Theorem the height is 20 so 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. 
