Question 1108310
<br>
The triangle is isosceles, with two sides of length 6.  Then using the side with length 4 as the base, the height of the triangle will be the length of the other leg of a right triangle with one leg 2 (half of the 4) and hypotenuse 6.  So the height is
{{{sqrt(6^2-2^2) = sqrt(36-4) = sqrt(32) = 4*sqrt(2)}}}<br>
Then the area of the triangle is one-half base times height: {{{2(4*sqrt(2)) = 8*sqrt(2)}}}<br>
Then, using your diagram with 6 as the base, the area of the triangle is one-half base times the altitude you show (call it h):
{{{8*sqrt(2) = 3h}}}
{{{h = 8*sqrt(2)/3}}}<br>
And then the unknown x we are looking for is the other leg of a right triangle with one leg h and hypotenuse 6:
{{{x^2+(8*sqrt(2)/3)^2 = 6^2}}}
{{{x^2+128/9 = 36 = 324/9}}}
{{{x^2 = 196/9}}}
{{{x = 14/3}}}