Let \( A B C D \) be a cyclic quadrilateral, such that \( A B=10, B C=9, A C=\sqrt{181} \), and \( C D=\sqrt{37} \). Find \( A D \).

To find \( AD \) in the cyclic quadrilateral \( ABCD \), we can use the properties of cyclic quadrilaterals and the Law of Cosines. Given: - \( AB = 10 \) - \( BC = 9 \) - \( AC = \sqrt{181} \) - \( CD = \sqrt{37} \) We will first find the length of \( AD \) using the Law of Cosines in triangles \( ABC \) and \( ACD \). ### Step 1: Find \( \angle ACB \) Using the Law of Cosines in triangle \( ABC \): \[ AC^2 = AB^2 + BC^2 - 2 \cdot AB \cdot BC \cdot \cos(\angle ACB) \] Substituting the known values: \[ (\sqrt{181})^2 = 10^2 + 9^2 - 2 \cdot 10 \cdot 9 \cdot \cos(\angle ACB) \] Calculating the squares: \[ 181 = 100 + 81 - 180 \cdot \cos(\angle ACB) \] Simplifying: \[ 181 = 181 - 180 \cdot \cos(\angle ACB) \] This implies: \[ 0 = -180 \cdot \cos(\angle ACB) \] Thus, we have: \[ \cos(\angle ACB) = 0 \] This means \( \angle ACB = 90^\circ \). Therefore, triangle \( ABC \) is a right triangle with \( AC \) as the hypotenuse. ### Step 2: Find \( AD \) using Ptolemy's Theorem Since \( ABCD \) is a cyclic quadrilateral, we can apply Ptolemy's theorem, which states: \[ AC \cdot BD = AB \cdot CD + AD \cdot BC \] Let \( AD = x \). We need to find \( BD \). We can find \( BD \) using the right triangle \( ABC \): Using the Pythagorean theorem in triangle \( ABC \): \[ AB^2 + BC^2 = AC^2 \] Calculating: \[ 10^2 + 9^2 = (\sqrt{181})^2 \] This confirms: \[ 100 + 81 = 181 \] Now, we need to find \( BD \). We can use the Law of Cosines in triangle \( ACD \): \[ AD^2 = AC^2 + CD^2 - 2 \cdot AC \cdot CD \cdot \cos(\angle ACD) \] Since \( \angle ACB = 90^\circ \), \( \angle ACD = 90^\circ \) as well (because \( ABCD \) is cyclic). Thus, \( \cos(\angle ACD) = 0 \): \[ AD^2 = AC^2 + CD^2 \] Substituting the known values: \[ AD^2 = (\sqrt{181})^2 + (\sqrt{37})^2 \] Calculating: \[ AD^2 = 181 + 37 = 218 \] Taking the square root: \[ AD = \sqrt{218} \] ### Final Answer Thus, the length of \( AD \) is: \[ \boxed{\sqrt{218}} \]