When the triangle is rotated, the base forms a circle with diameter=BC=12cm, so the radius of the base is 6cm. The area of the base is πr^2=36π sq cm.

We now need the height of the cone. Going back to the triangle, the height of the cone is equal to the height of the triangle. Drop a perpendicular from A on to BC and call the base of the perpendicular where it meets BC, N. So AN=height. By Pythagoras, AN^2+BN^2=AB^2, so AN^2+36=100 and AN=√64=8 cm.

The volume of the cone is (base area) x (height) ÷ 3=36π * 8 / 3= 96π cc=301.6cc approx.

256π doesn't seem to relate to this problem. But read on. There is another way to rotate the triangle. If we fix the base BC and rotate the triangle keeping its base fixed we get a different cone, a double cone, in fact. This cone has a height of 6cm and a base radius of 8cm. The area of the base is 64π and the volume of one cone is 64π * 6 / 3=128π. The volume of the double cone is therefore 256π cc, the expected answer!