We have AD = 5
Radius = 2

The cone is 3/4 circumference of the circle = 3/4 * 2 * pi * 2 = 3*pi

Now, consider semi-circle PCBQ.
The arcs PC + BQ = 1/4 of the entire circle (since cone is 3/4 and the semi-cirlce on top is 1/2. So that leaves 1/4). There are 360 degrees in a circle. So 1/4 = 90degrees. Measure of arc CB = 90 degress. Since arc CB is subtended by angle COB, angle COB = 90.

In triangle COB, CO = OB = radius = 2 and angle COB = 90
This is a 45-45-90 triangle. So, CB = 2 * sqrt2 and CD = DB = 1/2 (2 * sqrt2) = sqrt2

In triangle ADB, AD^2 + DB^2 = BA^2
5^2 + (sqrt2)^2 = BA^2
BA = sqrt27 = 3sqrt3 = AC

Perimeter = 3/4 circumference of circle + BA + AC = 3*pi + 3sqrt3 + 3sqrt3 = 3*pi + 6sqrt3

Whoo! That's a doozy. Thanks a bunch.

Can u explain this part pls..

The arcs PC + BQ = 1/4 of the entire circle (since cone is 3/4 and the semi-cirlce on top is 1/2. So that leaves 1/4). There are 360 degrees in a circle. So 1/4 = 90degrees. Measure of arc CB = 90 degress. Since arc CB is subtended by angle COB, angle COB = 90.

From given info, the part of the ice cream that sits on top of the cone is 1/4 of the circumference. Since the entire circumference of the circle has 360 degrees when measured from the center, the angle from the center measuring 1/4th of this would have (360/4)=90 degrees.

Brilliant ..You guys are amazing..I got part way through the solution and then had to make an intelligent guess..GivemeanID, you are truly my inspiration..