Answer:
yes
Explanation:
let's say that both playgrounds have the same length (12m) and width.
and let's say that every additional corner in B happens at 1/4 of the length (every 3m).
so, we are dealing with 4 different "sub-shapes" in B :
1. top rectangle 3×width
2. upper parallelogram width×height
3. lower parallelogram width×height
4. bottom rectangle 3×width
now let's compare these "sub-areas" with playground A.
the area of A = 4 × 3×width = 12×width.
but in B, only 1. and 4. have directly the same area as in A.
the area of a parallelogram is, as mentioned above, baseline × height, so here, width×height.
in order to have the same area as the corresponding rectangle in A, h must be 3, as is the length of the rectangle counterpart in A.
and that must be the case, as the overall length of B is the same as for A.
if we would have taken just the middle rectangles of A and bent them into such parallelograms, then the side lengths would have stayed the same, but the shape and its height would have sunk into itself and decreased like a compressed accordion. and so, the overall length of B would have decreased too.
but it did not.
so, the inner heights of the bent parallelogram pieces is also 3m, the same as the length of the rectangle in A.
a parallelogram with baseline width and height = 3 has the same area as a rectangle width × length with length = 3.
therefore, yes, both have the same area.