Problem 5.
Area removed from a pumpkin by carving for halloween

I have a pumpkin that is a perfect sphere with a diameter of 14 inches. If I cut out two eyes and a nose that are equilateral triangles of two inches on a side and I cut out a rhomboid mouth that is two inches high and five inches wide, what percentage of the pumpkin shell have I removed?

Answer: Approximately two and one half percent

1. Area of a sphere: A=4 pi r^2
A= 4 * pi * 7^2
A= 615.75 square inches

2. Area of cut outs
a) Mouth Area A=w*h
A=5 * 2 = 10

b) Area of 2 eyes plus nose
Area of triangle = sqrt of s(s-a)(s-b)(s-c) where s=(a+b+c)/2 and a,b,c are the sides
Area of 1 eye: A = sqrt 3(3-2)(3-2)(3-2) where s = (2=2=2)/2 = 3
= sqrt 3(1)(1)(1)
= sqrt 3 = 1.73
Area of 2 eyes + 1 nose = 1.73 * 3 = 5.19

c) Area of total cut out = 5.19 + 10 = 15.19 square inches

3. Percentage of Area removed (neglecting surface curvature of cutouts)
15.19 / 615.75 = .0247 = approx 2 1/2 percent