2b) A farmer is building a pen inside a barn. The pen will be in the shape of a right triangle.

The farmer has 14 feet of barn wall to use for one side of the pen and wants another side of the pen to be 15 feet long.

a. To the nearest tenth of a foot, find all possible lengths for the third side of the triangle.

b. The farmer wants the area of the pen to be as large as possible. What length should he choose for the third side? Justify your answer.

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, 2.5)

(2.5, 3]

step-by-step explanation:

i think its these two not sure [2, 2.5)

(2.5, 3]