375,699 views
44 votes
44 votes
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?

AB is tangent to circle Q at point A, BE is a secant segment, and FG is a chord BC-example-1
User Max Markov
by
2.8k points

1 Answer

26 votes
26 votes

Answer:

AB = 9.2cm (to the nearest tenth).

Step-by-step explanation:

Chords CE and GF intersect at point D.

First, we apply the theorem of intersecting chords:


\begin{gathered} CD* DE=DG* DF \\ CD*2=4*3 \\ CD=(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* 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* BE \\ AB^2=6*14 \\ 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).

AB is tangent to circle Q at point A, BE is a secant segment, and FG is a chord BC-example-1
AB is tangent to circle Q at point A, BE is a secant segment, and FG is a chord BC-example-2
AB is tangent to circle Q at point A, BE is a secant segment, and FG is a chord BC-example-3
User Marcos Marin
by
3.0k points