\( 1 \leftarrow \begin{array}{l}A B C D \text { is a rectangle. Find the length of each diagonal. } \\ A C=3(x-3) \text { and } B D=x+15\end{array} \) \( A C=\square, B D=\square \)

To find the length of each diagonal in rectangle ABCD, we can apply the property that both diagonals are equal in length. We have \( AC = 3(x - 3) \) and \( BD = x + 15 \). Setting these two expressions for the diagonals equal gives us: \[ 3(x - 3) = x + 15. \] Now, solve for \( x \): 1. Distribute on the left side: \( 3x - 9 = x + 15 \). 2. Subtract \( x \) from both sides: \( 3x - x - 9 = 15 \) or \( 2x - 9 = 15 \). 3. Add \( 9 \) to both sides: \( 2x = 24 \). 4. Finally, divide by \( 2 \): \( x = 12 \). Next, substitute \( x \) back into the equations for \( AC \) and \( BD \): - \( AC = 3(12 - 3) = 3(9) = 27 \). - \( BD = 12 + 15 = 27 \). So, the length of each diagonal is: \( AC = 27, BD = 27 \). Thus, \( A C = 27, B D = 27 \).