We are given the area and perimeter of two squares that form a right triangle. We are asked to find the length of the bigger square. To do that we may use the Pythagorean theorem. Let "a" and "b" be the sides of a triangle and "c" the length of the hypothenuse, we have the following relationship:
[tex]c^2=a^2+b^2[/tex]Now, "a" and "b" are the size of the given squares. For the square which area is given, we can use the following formula for the area of a square:
[tex]A_{\text{square}}=a^2[/tex]We can solve for "a" by taking square root on both sides, like this:
[tex]a=\sqrt[]{A_{square}}[/tex]Replacing the value given for the area, we get:
[tex]\begin{gathered} a=\sqrt[]{81in^2} \\ a=9in \end{gathered}[/tex]Now, for the square which perimeter is given, we can use the fact that the perimeter of a square is the sum of all its sides, like this:
[tex]\begin{gathered} P_{\text{square}}=b+b+b+b \\ P_{\text{square}}=4b \end{gathered}[/tex]solving for "b" we get:
[tex]b=\frac{P_{\text{square}}}{4}[/tex]Replacing the known value for the perimeter, we get:
[tex]b=\frac{48in}{4}=12in[/tex]Now that we have both sides "a" and "b" we may replace this in the Pythagorean theorem, like this:
[tex]\begin{gathered} c^{2^{}}=a^2+b^2 \\ c^2=9^2+12^2 \end{gathered}[/tex]Solving the operations:
[tex]\begin{gathered} c^2=81+144 \\ c^2=225 \end{gathered}[/tex]Now we solve for "c" by taking square roots on both sides, like this:
[tex]\begin{gathered} c=\sqrt[]{225} \\ c=15 \end{gathered}[/tex]Therefore, the side length of the largest square is 15 in.