A chord AC subtends an angle of 90° at the center O. If the radius of the circle is 8 cm, what is the length of the chord AC?
8√2 cm
Draw the two radii OA and OC to the ends of the chord. Each radius equals the circle's radius, so \(OA = OC = 8\ \text{cm}\), and the angle between them is \(\angle AOC = 90\degree\).
Triangle OAC is therefore right-angled at O, with OA and OC as the two legs and the chord AC as the hypotenuse.
By the Pythagoras theorem, \(AC = \sqrt{OA^2 + OC^2}\).
Substitute the radii: \(AC = \sqrt{8^2 + 8^2} = \sqrt{64 + 64} = \sqrt{128}\).
Simplify: \(\sqrt{128} = \sqrt{64 \times 2} = 8\sqrt{2}\) cm.
Key concept: a chord subtending \(90\degree\) at the centre, together with the two radii, forms an isosceles right triangle, so the chord is \(\sqrt{2}\) times the radius.
Hence the length of chord AC is 8√2 cm.
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