Calvin likes to measure

Let $S$ be the side of the larger square, let $P$ be the perimeter, and let $A$ be the area. Then $S = 10$ , $P = 40$ , $A = 100$ . Now, the new square is $30\%$ less in perimeter than the perimeter of the larger square, or $70\%$ of it. If $p$ is the perimeter of the new square, the $p = 0.7\times40$ , or $28$ . The side of the new square, $s = \frac{28}{4}$ , and subsequently, the area $a$ must be $s^{2} = (7)^{2} = 49$ , or $a = 49$ . To find percent difference, we do $\frac{A-a}{A} = \frac{(100)-(49)}{(100)} = \frac{51}{100} = \boxed{51\%}$ .