Q. In the figure below AD is tangent to circle O, AD=15, and BC=16, find AB
Identify Theorem: Identify the relevant theorem or property to use for solving the problem. Since AD is tangent to circle O at point A, and we know AD and BC, we can use the tangent-secant theorem which states that the square of the length of the tangent segment (AD) equals the product of the lengths of the whole secant segment (AC) and its external part (AB).
Set Up Equation: Set up the equation based on the tangent-secant theorem: AD2=AB×AC. We know AD=15, so 152=AB×AC.
Find AC: We need to find AC. Since BC is given as 16 and it is part of the secant that extends from A through the circle to C, we can express AC as AB + BC. Therefore, AC=AB+16.
Substitute in Equation: Substitute AC in the tangent-secant theorem equation: 152=AB×(AB+16). This simplifies to 225=AB2+16AB.
Solve Quadratic Equation: Solve the quadratic equation AB2+16AB−225=0. We can use the quadratic formula, AB=2a−b±b2−4ac. Here, a=1, b=16, and c=−225.
Calculate Roots: Calculate the discriminant and the roots: Discriminant = 162−4⋅1⋅(−225)=256+900=1156. AB=[−16±1156]/2. AB=[−16±34]/2.
Find Possible Values: Find the possible values for AB: AB=(−16+34)/2=18/2=9 and AB=(−16−34)/2=−50/2=−25. Since a length cannot be negative, AB=9.
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