Question

In the adjacent figure \(PT\) is a tangent drawn from an external point \(P\) touching the circle at \(T\). \(PAB\) is a secant cutting the circle at \(A\) and \(B\). If \(PT=5\) cm and \(PA=4\) cm, then \(AB\) (in cm) is

• A. \(2.25\)
• B. \(4.25\)
• C. \(5.25\)
• D. \(6.25\)

Hint:

For tangent and secant from the same external point, always use \(PT^2=PA\cdot PB\).

Answer 1

The Correct Option is A

Concept: Use Tangent-Secant Theorem: \[ PT^2=PA\cdot PB \] Given: \[ PT=5,\quad PA=4 \] Substitute: \[ 5^2=4\cdot PB \] \[ 25=4PB \Rightarrow PB=\frac{25}{4}=6.25 \] Now: \[ AB=PB-PA \] \[ AB=6.25-4=2.25 \] Thus, \[ \boxed{2.25} \]