QUESTION IMAGE
Question
find the value of x.
Step1: Apply the perpendicular - bisector theorem
A perpendicular from the center of a circle to a chord bisects the chord. Let the radius of the circle be \(r\). For the chord of length \(22\), half of it is \(\frac{22}{2}=11\). Let the distance from the center of the circle to the chord of length \(22\) be \(d_1\), and for the chord of length \(x\), half of it is \(\frac{x}{2}\), and the distance from the center of the circle to the chord of length \(x\) is \(d_2\). Here, \(d_1 = 12\) and \(d_2=12\).
Step2: Use the Pythagorean theorem
In a circle, if the radius is \(r\), for the chord of length \(22\), by the Pythagorean theorem \(r^{2}=11^{2}+12^{2}\). Calculate \(11^{2}+12^{2}=121 + 144=265\), so \(r^{2}=265\).
For the chord of length \(x\), since the distance from the center of the circle to the chord is also \(12\) and \(r^{2}=265\), using the Pythagorean theorem again: \((\frac{x}{2})^{2}+12^{2}=r^{2}\). Substitute \(r^{2}=265\) into the equation: \((\frac{x}{2})^{2}+144 = 265\).
Step3: Solve for \(\frac{x}{2}\)
Subtract 144 from both sides of the equation \((\frac{x}{2})^{2}+144 = 265\): \((\frac{x}{2})^{2}=265 - 144=121\). Then \(\frac{x}{2}=\sqrt{121}=11\).
Step4: Solve for \(x\)
Multiply both sides by 2: \(x = 22\).
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