What are the demensions of a rectangle if the perimiter is 20 and the area is 24

I simply listed the factors of 24, and tested which ones worked in this equation: P = 2a + 2b.

So the factors listed were:

24 x 1

12 x 2

8 x 3

6 x 4

After testing each one of these, 6 and 4 had worked for me.

P = 2(6) + 2(4)

P = 12 + 8

P = 20

Therefore, the dimensions of this rectangle are 6 and 4.

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!

57 732

step-by-step explanation:

i dont really.know