Question

given the circle below with secants \\( \overline { f g h } \\) and \\( \overline { j i h } \\). if \\( g h = 13 \\), \\( i h = 12 \\) and \\( j i \\) is 3 more than \\( f g \\), find the length of \\( j i \\). round to the nearest tenth if necessary.

Explanation:

Step1: Use the secant - secant formula

If two secants \(FGH\) and \(JIH\) are drawn to a circle from an external point \(H\), then \(FH\times GH=JH\times IH\). Let \(FG = x\), then \(JI=x + 3\), \(FH=x + 13\), and \(JH=(x + 3)+12=x+15\).
Substitute into the formula: \((x + 13)\times13=(x + 15)\times12\).

Step2: Expand the equation

Expand the left - hand side: \(13x+169\). Expand the right - hand side: \(12x + 180\).
So, \(13x+169=12x + 180\).

Step3: Solve for \(x\)

Subtract \(12x\) from both sides: \(13x-12x+169=12x-12x + 180\), which gives \(x=11\).

Step4: Find the length of \(JI\)

Since \(JI=x + 3\) and \(x = 11\), then \(JI=11+3=14\).

Answer:

\(14\)