View source for 2015 AMC 8 Problems/Problem 6

==Problem==

In <math>\bigtriangleup ABC</math>, <math>AB=BC=29</math>, and <math>AC=42</math>. What is the area of <math>\bigtriangleup ABC</math>?

<math>\textbf{(A) }100\qquad\textbf{(B) }420\qquad\textbf{(C) }500\qquad\textbf{(D) }609\qquad \textbf{(E) }701</math>


==Video Solution==

https://www.youtube.com/watch?v=Bl3_W2i5zwc

==Solutions==
===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}=\boxed{\textbf{(B) }420}</math>.

===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
<math>\dfrac{(20)(42)}{2} = \boxed{\textbf{(B)}~420}</math>.

==Video Solution 2==
https://youtu.be/HNj4U3S827E

~savannahsolver

==See Also==

{{AMC8 box|year=2015|num-b=5|num-a=7}}
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