A circle has centre O and radius 6 cm. PQ and PR are two equal chords of the circle, each of length 8 cm. What is the length of the chord QR (in cm)?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(\displaystyle\frac{16\sqrt{5}}{3}\)
P is at equal distances from Q and R, and so is O; hence OP is perpendicular to QR and bisects it at a point M. Pythagoras' theorem in the right triangles PMQ and OMQ gives QM.
Let OM \(=d\); then PM \(=6-d\) and QM\(^2=6^2-d^2\)
In \(\triangle PMQ\): \(8^2=(6-d)^2+(36-d^2)\)
\(64=36-12d+d^2+36-d^2=72-12d\), so \(\displaystyle d=\frac{2}{3}\) cm
QM\(^2\displaystyle=36-\frac{4}{9}=\frac{320}{9}\), so QM \(\displaystyle=\frac{8\sqrt{5}}{3}\) cm
QR \(=2\times\) QM \(\displaystyle=\frac{16\sqrt{5}}{3}\) cm
