Question 149171
First let's draw the pyramid (note: my drawings are not to scale).


<img src="http://i150.photobucket.com/albums/s91/jim_thompson5910/pyramid.jpg" alt="Photobucket - Video and Image Hosting">


If we just look at the base, we get something like this (Note: The hypotenuse was found through pythagorean's theorem {{{sqrt(6^2+6^2)=sqrt(36+36)=sqrt(72)=6*sqrt(2)}}}):


<img src="http://i150.photobucket.com/albums/s91/jim_thompson5910/pyramid_base.jpg" alt="Photobucket - Video and Image Hosting">





Now take half of the hypotenuse to get {{{3*sqrt(2)}}}. This figure will form the leg of the triangle. Now let's one half of a triangular cross section to get this:


<img src="http://i150.photobucket.com/albums/s91/jim_thompson5910/upright_triangle.jpg" alt="Photobucket - Video and Image Hosting">



From the drawing, we can see that legs are {{{h}}} and {{{3*sqrt(2)}}} while the hypotenuse is {{{6}}}



So to find "h", we need to use pythagorean's theorem:


{{{a^2+b^2=c^2}}}



{{{h^2+(3*sqrt(2))^2=6^2}}} Plug in {{{a=h}}}, {{{b=3*sqrt(2)}}}, and {{{c=6}}}



{{{h^2+(3*sqrt(2))^2=36}}} Square 6 to get 36.



{{{h^2+18=36}}} Square {{{3*sqrt(2)}}} to get {{{(3*sqrt(2))^2=(3*sqrt(2))(3*sqrt(2))=(3*3)*(sqrt(2)*sqrt(2))=9*2=18}}}



{{{h^2=36-18}}} Subtract {{{18}}} from both sides.



{{{h^2=18}}} Combine like terms on the right side.



{{{h=sqrt(18)}}} Take the square root of both sides. Note: only the positive square root is considered.



{{{h=3*sqrt(2)}}} Simplify.



So the height is {{{h=3*sqrt(2)}}} or {{{h=4.24264}}}



So the height of the pyramid to the nearest hundredth of an inch is 4.25 inches