Question 10.

It is given that the base of the pyramid is square and each of the four sides are equilateral triangles.

Length of each side of the equilateral triangle = 20cm

Since the side of the triangle will be common to the square as well, the side of the square = 20cm

Let h be the vertical height of the pyramid ie OA

OB = 10 since it is half the side of the square 

AB is the height of the equilateral triangle i.e 10$\sqrt{\ 3}$

AOB is a right angle, so applying the Pythagorean formula, we get

$OA^2+ OB^2= AB^2$

$h^2$ + 100= 300

h=10$\sqrt{\ 2}$