QUESTION IMAGE
Question
given the circle below with chords \\( \overline { g h } \\) and \\( \overline { i j } \\). find the length of \\( \overline { j k } \\). round to the nearest tenth if necessary.
Step1: Apply the Chord - Chord Product Theorem
The Chord - Chord Product Theorem states that if two chords \(GH\) and \(IJ\) intersect at a point \(K\) inside a circle, then \(GK\times KH=IK\times KJ\).
Let \(GK = 14\), \(KH=11 + 16-(14)\) (wait, no, actually, if \(GK = 14\), \(IK = 11\), \(KH\) is not directly given. Let \(JK=x\). Then according to the theorem \(14\times(11 + 16 - x)=11\times x\) (no, wrong. The correct formula is \(GK\times KH=IK\times KJ\). Let \(GK = 14\), \(IK = 11\), \(KH\) is the other part of chord \(GH\) and \(JK\) is the part of chord \(IJ\) we want to find. The lengths of the segments of the first chord: \(GK = 14\) and \(KH\) (but actually, if two chords \(GH\) and \(IJ\) intersect at \(K\), then \(GK\times KH=IK\times KJ\). Let \(JK=x\). We know \(GK = 14\), \(IK = 11\), and assume \(KH\) is the remaining part of chord \(GH\) (but no, actually, the formula is \(GK\times (length\ of\ the\ other\ part\ of\ chord\ GH)=IK\times (length\ of\ the\ other\ part\ of\ chord\ IJ)\). Wait, the correct formula is: If two chords \(AB\) and \(CD\) intersect at \(E\), then \(AE\times EB=CE\times ED\). So here, \(GK\times KH=IK\times KJ\). Let \(KJ=x\). We know \(GK = 14\), \(IK = 11\), and assume \(KH\) is the other part of chord \(GH\) (no, actually, \(GK\times (the\ other\ segment\ of\ chord\ GH)=IK\times (the\ other\ segment\ of\ chord\ IJ)\). Wait, chord \(GH\) is divided into \(GK = 14\) and \(KH\). Chord \(IJ\) is divided into \(IK = 11\) and \(KJ\). But we know that \(GK\times KH=IK\times KJ\). Wait, no, the formula is \(a\times b=c\times d\) where \(a\) and \(b\) are segments of one chord and \(c\) and \(d\) are segments of the other chord. So \(14\times16 = 11\times x\)
Step2: Solve for \(x\) (i.e., \(JK\))
From \(14\times16=11x\), we have \(x=\frac{14\times16}{11}\)
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