Question

AB is tangent to circle Q at point A, BE is a secant segment, and FG is a chord BC = 6 cm DG = 4cm DF = 3 cm and DE = 2cmWhat is AB round your answer to the nearest tenth?

Answer 1

Answer:

AB = 9.2cm (to the nearest tenth).

Explanation:

Chords CE and GF intersect at point D.

First, we apply the theorem of intersecting chords:

$\begin{gathered} CD\times DE=DG\times DF \\ CD\times2=4\times3 \\ CD=\frac{12}{2} \\ CD=6\operatorname{cm} \end{gathered}$

Next, AB is a tangent while line BE is a secant.

AB and BE intersect at B.

Using the theorem of intersecting tangent and secant:

$AB^2=BC\times BE$

First, find the length of BE.

$\begin{gathered} BE=BC+CD+DE \\ =6+6+2 \\ BE=14\operatorname{cm} \end{gathered}$

Substitute BE=14cm and BC=6 cm into the formula:

$\begin{gathered} AB^2=BC\times BE \\ AB^2=6\times14 \\ AB^2=84 \\ AB=\sqrt[]{84} \\ AB=9.2\text{ cm (to the nearest tenth)} \end{gathered}$

The length of AB is 9.2cm (to the nearest tenth).