Comparing perimeters

We get $a ^ { 2 } + b ^ { 2 } = c ^ { 2 }$ by Pythagorean theorem, so $4 ^ { 2 } + a ^ { 2 } = \sqrt { 97 } ^ { 2 }$ . Then, $16 + a ^ { 2 } = 97 \rightarrow a ^ { 2 } = 81$ . Then, we get $a = \pm 9$ , but the length cannot be negative(I would be happy if there were the negative length, age, etc.), so we get $a = 9$ . Then the perimeter of the rectangle A is $2 ( 4 + 9 ) = 2 \times 13 = 26$ .

Next, we work on rectangle B. We get $3 ^ { 2 } + a ^ { 2 } = \sqrt { 153 } ^ { 2 }$ . Then, $9 + a ^ { 2 } = 153 \Rightarrow a ^ { 2 } = 144$ . We get $a = \pm 12$ , but the length cannot be negative also, so the other length of the rectangle B is $12$ . Then we get $2 ( 3 + 12 ) = 2 \times 15 = 30$ .

Since $26 < 30$ , so rectangle B has a longer length than rectangle B.

In rectangle A, using the Pythagorean theorem, we find that the other side is 9. Therefore, the perimeter of rectangle A is 26. In rectangle B, using the Pythagorean theorem, we find that the other side is 12. Therefore, the perimeter of rectangle B is 30. In the end, Rectangle B has the larger perimeter.