A right circular cone of radius 9 centimeter and height 12 cm is cut p

Not a GMAT question. One does not need to know the curved surface area of a frustum of a cone. Still:



We need to find he curved surface area of the lower part of the cone, so the area of the surface without the bottom and top. It's area = \(\pi*h*(R + r)\).

Since the height is cut in the ratio 1:2, then the height of the frustum is \(h_1= 8\) cm and the height of the smaller, upper cone is \(h_2= 4\) cm. Next, from similar triangles property: \(\frac{r}{R} =\frac{h_2}{H}\) (where H is the height of the big cone and h2 is the height of the smaller, upper cone). r/9 = 4/12 --> r = 3.

Therefore, the area of the frustum is \(\pi*h_1*(R + r)=\pi*8(9 + 3)=96\pi\).

P.S. Please follow the rules when posting a question: https://gmatclub.com/forum/rules-for-po ... 33935.html Thank you.