ice cream cone

hey, i'm kinda week in geometry. Can you explain how you got 2sqrt2 ?

Triangle ABC with heigth AH, the circle with the center O.

AB^2 = (1/2BC)^2 + AH^2

The triangle BCO is a right triangle (because chord BC has the length of 1/4 circumference of the circle)

=> BC^2 = BO^2 + CO^2 = 8
=> BC = 2sqrt2

the key for solving the above problem is knowing how to find the perimeter of a square inscribed inside a circle!

part one

finding the circal circumference

circal r = 2

circumference whole circle 2*pi*r = 4*pi

finding 3/4 of circumference = 3/4*4*pi = 3*pi

part two

draw a square inscribed inside a circle (look at the attachment)

the diagonal of the square is r=2*2=4 (diameter of circle).

the square forms two 45-45-90 triangles.

sqrt(2)*x=4

x=4/sqrt(2)

so the ratio is 1:1:sqrt(2) = 4/sqrt(2):4/sqrt(2):4

meaning the line in the attachment marked (1) equal 4/sqrt(2) and half the line is [4/sqrt(2)]/2

now:

from the triangle with height 5 we can make two smaller right triangles with base [4/sqrt(2)]/2.

using the Pythagorean theorem will yield [[4/sqrt(2)]/2]^2+5^2=x^2

2+25 = x^2

x = sqrt(27)

so we can conclude that the sign circumference is pi*3+2*sqrt(27) or in other words pi*3+3*2sqrt(3) = pi*3+6sqrt(3)

and as was said before the answer is then (B)

where did you get the idea to inscribe the square within the 3/4 circle? this is partly a rhetorical question.

The main hint here is that the circumference is 3/4 the rest 1/4 should be considered as 90 degree angle of the triangle. This will easen work a lot.

Ans: B

Anyone trying POE here...

1st part is easy... 3pi. I eliminated 4 and 5.

The second part needs to be greater than 10(2*height). eliminate A

So, u r left with B and C. I chose B before working it out the right way.