The area of rectangle a is twice the area of rectangle b the perimeter of rectangle a is 20 units greater than rectangle b what could the dimensions of the two rectangles be

The possible dimensions are;

If Rectangle B has a dimension of 1 unit x 2 units, then Rectangle A has a dimension of 0.315 units x 12.685 units

Step-by-step explanation:

Let;

Length of Rectangle A be a

Width of Rectangle A be b

Length of Rectangle B be c

Width of Rectangle B be d

Thus;

Area of Rectangle A = a × b

Area of Rectangle B = c × d

We are told that the area of Rectangle A is twice the area of Rectangle B:

Thus;

2cd = ab - - - - - eq. 1

perimeter of Rectangle A = 2a + 2b

perimeter of Rectangle B = 2c + 2d

We are told that the perimeter of Rectangle A is 20 units greater than the perimeter of Rectangle B. Thus, we now have;

20 + 2c + 2d = 2a + 2b - - - - eq. 2

We have 4 unknowns which are (a, b, c and d) but only 2 equations, so we need to reduce to 2 unknown variables and calculate the other ones. In this way, one of the infinite solutions is obtained.

Let's assume that c = 1 and d = 2, we obtain:

From eq 1, we have;

2 * 1 * 2 = a*b

ab = 4 or a = 4/b

From eq 2, we have;

20 + 2(1) + 2(2) = 2a + 2b

26 = 2a + 2b

Putting a = 4/b into this, we have;

26 = 2(4/b) + 2b

Multiply through by b to get;

26b = 8 + 2b²

So,we have;

2b² -26b + 8 = 0

Using quadratic formula for this,

b = 0.315 or 12.685

When, b = 12.685, a = 4/12.685 = 0.315

When, b = 0.315, a = 4/0.315 = 12.685

So, the possible dimensions are;

If Rectangle B has a dimension of 1 unit x 2 units, then Rectangle A has a dimension of 0.315 units x 12.685 units