Cousin's 3rd Grade Math

There is no solution.

Suppose the first rectangle has length x and width y,the second p and q
then 2x+2y=12 {perimeter definition)
2p+2q=18

Dividing through by 2, the equations become,
x+y=6
p+q=9

The area of a rectangle =length * breadth
So since the areas are equal,
xy=pq,

so all the information is contained in the 3 equations
x+y=6
p+q=9
xy=pq

so we have 3 equations with 4 unknowns.
Impossible to solve without a fourth equation.

I suspect that the student was learning about simultaneous equations, and one of the first things taught that is that you need the same number of equations as the number of unknowns.

PS, What age is 3rd Grade?

Nine. I have a nine year old in 3rd grade in the US (public school). She is learning basic multiplication.

The fourth equation is area equals a value such that for real solutions the value is greater than zero and less than nine.

So far I have not found a third grader who can find a solution to this problem. The two I have talked to are able to find the areas and perimeters of rectangles.

That's not an equation, only a limitation.

Assuming your cousin is only 9, then I doubt that the question would be so demanding.
I think they are required to try different values, using integers only.
Here are the 3 equations again:
So for 2 integers to add up to 6(similarly for 9), they can be (1,5),(2,4),(3,3), then by trial and error
x+y=6 , putting x=4 and y=2 satisfies the equation and the area, xy=8
p+q=9 , putting p=1 and q=8 satisfies the equation and the area, pq=8
xy=pq

So the dimensions are 2X4 and 1X8

Was this really a question to 9 year old?

I just re-read this thread and noticed you already put up my solution.

Well, call it what you will, equation, function, limitation. The fact is that for every area from zero to nine, there exists a unique set of width and length for each rectangle a total of four unknowns for which there are four equations. Or if you pick a width for one of the rectangles, there exists a unique set of a length and area for that rectangle with a corresponding a width and length for the other rectangle, which will have the same area.

My aunt insists that my cousin's third grad math text has this problem. I think you may have the right idea about what solution a third grader might find, namely the 1x8, 2x4 solution. I suspect that all other possible solutions require a knowledge of quadratic equations.

My grandfather was about my age when John Kennedy became president. He says that Kennedy quoted this poem by Robert Frost.

I shall be telling this with a sigh
Somewhere ages and ages hence:
Two roads diverged in a wood, and I—
I took the one less traveled by,

I have shown this math problem to dozens of people, So far most all of them, including me, have followed a similar road. At first there appears not to be enough information for a solution. Then, Holy Achimedes, there is at least one solution. Then, by golly, there are many solutions. Not many of us have taken the road less traveled.