Question Shannon is designing a snowman for a winter mural. She wants the head of the snowman to have a radius of 5 inches (
5 in.), and the midsection of the snowman to have a radius of 7 in. How much bigger is the area of the snowman's
midsection than the area of its head? In other words, how many more square inches of material will be needed to create
the midsection of the snowman compared to the souare inches needed to create its head?

75.36

Step-by-step explanation:

Area of the top of the snowman

A=πr^2

A=3.14(5)^2

A=78.5

A=πr^2

A=3.14(7)^2

A=153.86

The difference of the areas:

153.86-78.5

=75.36

there are lots of ways.   pick's theorem tells us if the vertices are lattice points the area a is

a = b/2 + i - 1

where b is the number of boundary lattice points and i is the number of interior lattice points.

in the figure i count b=16, i=50 so

a = 8 + 50 - 1 = 57

let's try the shoelace formula.   the area is half the sum of the cross products of the sides.  

a(-2,6)   -2(2)-6(-5) = 26

b(-5,2)   -5(-3) - 2(-2) = 19

c(-2,-3) -2(-3) - (-3)(4) = 18

d(4,-3)   4(3) - (-3)(5) = 27

e(5,3)   5(3) - 3(1) = 12

f(1,3)     1(6) - 3(-2) = 12

a(-2,6)

each of the cross products is twice the area of the triangle with the associated side and one vertex the origin. so the area of abo=26/2=13.

area = (1/2) (26 + 19 +   18 + 27 + 12 + 12) = 57

that checks!

2+2 obliviously is 4.