jmc

We start by drawing the frustum. Let the top and bottom circles have centers $O_1$ and $O_2$ respectively, and label points $A$ and $B$ on the circumferences as shown such that $O_1$, $O_2$, $A$, and $B$ lie in the same plane.



Because the frustum was cut from a right circular cone, $\angle A{O}_1{O}_2$ and $\angle B{O}_2{O}_1$ are both right angles. We drop a perpendicular from $A$ to $\overline{O_{2}B}$ and let the intersection point be $X$. Then $O_{1}AX{O}_2$ is a rectangle and $$XB=O_{2}B-O_{1}A=6-3=3.$$Pythagorean theorem on right $△AXB$ gives $$AB=\sqrt{A{X}^2+B{X}^2}=\sqrt{4^2+3^2}=5.$$Thus the slant height of the frustum is 5.

Extend $\overline{O_1{O}_2}$ and $\overline{AB}$ above the frustum, and let them intersect at point $C$. $C$ is the tip of the full cone that the frustum was cut from. To compute the lateral surface area of the frustum, we compute the lateral surface area of the full cone and subtract off the lateral surface area of the smaller cone that was removed.



To find the height of the whole cone, we take a vertical cross-section of the cone that includes $O_1$, $O_2$, $A$, and $B$. This cross-section is an isosceles triangle.



$△C{O}_{1}A$ and $△C{O}_{2}B$ are similar, so $$\frac{C{O}_1}{C{O}_2}=\frac{CA}{CB}=\frac{O_{1}A}{O_{2}B}=\frac{3}{6}.$$Thus $C{O}_1=4$ and $CA=5$ (and we see the small removed cone has half the height of the full cone). Also, $CB=10$.

Now we unroll the lateral surface area of the full cone. (The desired frustum lateral area is shown in blue.)



When unrolled, the full cone's lateral surface area is a sector whose arc length is the cone's base perimeter and whose radius is the cone's slant height. So, the sector has arc length $2\cdot \pi \cdot 6=12\pi$ and radius $10$. A full circle with radius 10 has arc length $2\cdot \pi \cdot 10=20\pi$, so the sector has $\frac{12\pi }{20\pi }=\frac{3}{5}$ of the circle's arc length and thus has 3/5 of the circle's area. Thus, the full cone has lateral surface area $$\frac{3}{5}\pi (10^2)=60\pi .$$Similarly, the small removed cone's lateral surface area is a sector with radius 5 and arc length $2\cdot \pi \cdot 3=6\pi$ (which is $3/5$ of the arc length of a full circle with radius 5), so its lateral surface area is $$\frac{3}{5}\pi (5^2)=15\pi .$$The lateral surface area of the frustum, in blue, is the full cone's lateral surface area minus the small removed cone's lateral surface area, which is $$60\pi -15\pi =\boxed{45\pi }.$$