Two rectangles have the same area of 70m2. It is given that the length of the first rectangle is 4m longer and the width is 2m shorter than the dimensions of the second rectangle. Find the lengths of the sides of these rectangles.

second rectangle

longer = 10

width = 7

first rectangle

longer = 14

width= 5

Step-by-step explanation:

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!

135 is your answer

step-by-step explanation:

try 180 times three then divide that number by 4 (sides) to get your answer. im glad to : )