MHB What is the area of an isosceles triangle with side lengths 6, 6, and 4?

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The area of the isosceles triangle with side lengths 6, 6, and 4 is calculated using two methods. First, by splitting the triangle, the height is found to be 4√2, leading to an area of 8√2. The second method, Heron's formula, confirms this area calculation, yielding the same result of 8√2. Both methods validate the correctness of the area calculation. The final area of the triangle is 8√2.
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Side lengths are a=6, b=6, c=4. Find the area

A=1/2*b*h

I split the triangle in half to find the height. Since the base is 4, that divided the base into 2:

2^2+h^2=6^2

4+h^2=36

h^2=32

h=4*sqrt(2)
===========

1/2*4*4*sqrt(2)=area of 8*sqrt(2)

Did I do this correctly? I do not have an answer key to the study guide I was given.
 
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Let's check your answer using Heron's formula...the semi-perimeter \(s\) is 8, hence:

$$A=\sqrt{8(8-6)(8-6)(8-4)}=\sqrt{2^7}=8\sqrt{2}\quad\checkmark$$
 

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