Question 1205291
.
A right pyramid on a base of 4cm square has a slant edge of 6cm. 
Calculate the volume of the pyramid
~~~~~~~~~~~~~~~~~~~


<pre>
I believe that the pyramid is REGULAR, which means that its base is a square.


If so, then the side of this square is {{{sqrt(4)}}} = 2 cm,
 
and the diagonal of the square is {{{2*sqrt(2)}}} cm.



Draw the height (the altitude) of the pyramid from the upper vertex to the base of the pyramid.


You will get a right-angled triangle with one leg on the base as the half
of the diagonal of the square, so the length of this leg is  {{{sqrt(2)}}} cm long.


The hypotenuse of this right-angled triangle is 6 cm (it is the slant edge).


Hence, by Pythagoras, the altitude is  h = {{{sqrt(6^2-(sqrt(2))^2)}}} = {{{sqrt(36-2)}}} = {{{sqrt(34)}}}  cm.


Thus the volume of the pyramide is 

    {{{(1/3)*base_area*h}}} = {{{(1/3)*4*sqrt(34)}}} = {{{(4/3)*sqrt(34)}}} = 7.7746 cm^3  (approximately).    <U>ANSWER</U>
</pre>

Solved.