### Problem Overview**Given:**- Point \(D\) is on circle \(C\) with diameter \(AB\) as shown.- \(AC = 6.5\) inches- \(BD = 5\) inches**Find:** Length of \(AD\).### Diagram ExplanationThe image displays a circle \(C\) with diameter \(AB\). Point \(D\) is located on the circle, forming a triangle \(ABD\) within the circle. \(AC\) (a segment from \(A\) to the midpoint of the diameter \(AB\)) is given as 6.5 inches and segment \(BD\) is given as 5 inches.### Steps to Solve1. **Determine the Lengths:** - Since \(AC\) represents the radius of the circle, and \(C\) is the midpoint of \(AB\), the length of the radius is 6.5 inches. - Since \(AB\) is the diameter, the length of the diameter of the circle is \(2 \times AC = 2 \times 6.5 = 13\) inches.2. **Apply the Pythagorean Theorem:** - Since \(D\) lies on the circle and \(ABD\) forms a right triangle with \(AB\) as its hypotenuse, we use the Pythagorean theorem. - \(AD\) and \(BD\) form the two perpendicular sides of the right triangle \(ABD\) with hypotenuse \(AB\). Using the Pythagorean theorem in the right triangle \(ABD\): \[ AD^2 + BD^2 = AB^2 \] Substituting the known values: \[ AD^2 + 5^2 = 13^2 \] \[ AD^2 + 25 = 169 \] \[ AD^2 = 144 \] \[ AD = \sqrt{144} = 12 \]### Solution- The length of \(AD\) is **12 inches**.

Name: ### Problem Overview**Given:**- Point \(D\) is on circle \(C\) with d
Uploaded: 2024-03-20T06:46:14.000Z
Channel: Bartleby
Description: ### Problem Overview**Given:**- Point \(D\) is on circle \(C\) with diameter \(AB\) as shown.- \(AC = 6.5\) inches- \(BD = 5\) inches**Find:** Length of \(AD\).### Diagram ExplanationThe image displays a circle \(C\) with diameter \(AB\). Point \(D\