A rectangle is three times as long as it is widen. if it perimeter is 56cm, find the width of the rectangle

Hi there! :)

w = 7 cm.

Step-by-step explanation:

Given:

P = 56

Use the formula P = 2l + 2w to solve for the perimeter of the rectangle.

Let w = width, and

3w = length

Plug these into the equation:

56 = 2(3w) + 2(w)

56 = 6w + 2w

Combine like terms:

56 = 8w

Divide both sides by 8:

w = 7 cm.

The width of rectangle is 7 cm.

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!