15 Quadrilateral ABCD is inscribed in circle \( Q . A B=12 . B C=16 \). \( C D=10 \), and \( \angle A B C \) is a right angle. Find the measare of \( \overline{A B} \) in simplified radical form.

To find the measure of \( \overline{AD} \) in the inscribed quadrilateral \( ABCD \), we first recognize that since \( ABCD \) is a cyclic quadrilateral and \( \angle ABC \) is a right angle, we can apply the Pythagorean theorem in triangle \( ABC \). Given: - \( AB = 12 \) - \( BC = 16 \) - \( CD = 10 \) - \( \angle ABC = 90^\circ \) We can find \( AC \) using the Pythagorean theorem: \[ AC = \sqrt{AB^2 + BC^2} = \sqrt{12^2 + 16^2} = \sqrt{144 + 256} = \sqrt{400} = 20 \] Now, for the cyclic quadrilateral, we use the property relating the lengths of opposite sides. According to Ptolemy's theorem for cyclic quadrilaterals, we have: \[ AC \cdot BD = AB \cdot CD + AD \cdot BC \] Thus, if we denote \( AD \) as \( x \), we have: \[ 20 \cdot BD = 12 \cdot 10 + x \cdot 16 \] Next, we also need to find \( BD \), which can be done using the inscribed angle theorem, taking into account triangle properties. However, since we do not have \( BD \) yet, let's express it considering the cyclic nature and the right angle: By properties of cyclic quadrilaterals, we also know that: \[ AD = \sqrt{AC^2 - CD^2} \] By substituting the known values: \[ AD = \sqrt{20^2 - 10^2} = \sqrt{400 - 100} = \sqrt{300} = 10\sqrt{3} \] Therefore, the measure of \( \overline{AD} \) in simplified radical form is \[ \overline{AD} = 10\sqrt{3} \]