Afarmer wishes to fence a rectangular area behind his barn. the barn forms one end of the rectangle and the length of the rectangle is three times the width. how many linear feet of fence must he buy if the perimeter of the rectangle is 320 feet

280 feet

Step-by-step explanation:

Let width of rectangle be x

Since we are given that  the length of the rectangle is three times the width.

⇒ Length = 3x

Now, perimeter of rectangle =

Perimeter of rectangular area:

Thus the width of the rectangular area is 40 feet .

Length = 3x= 40*3=120 feet.

Since we are given that The barn forms one end of the rectangle

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Hence he must buy 280 feet of fence to form barn if the perimeter of the rectangle is 320 feet                            

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!