find the length of the chord which subtends 90 degrees at the centre of circle with radius 5cm?
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the two radii and the chord will form a right angled triangle.
let the chord be x.
see the diagram. using pythagoras theorem,
x² = 5² + 5²
⇒ x² = 25 + 25
⇒ x² = 50
⇒ x = √50 = 5√2 cm
Length of chord is 5√2 cm.
let the chord be x.
see the diagram. using pythagoras theorem,
x² = 5² + 5²
⇒ x² = 25 + 25
⇒ x² = 50
⇒ x = √50 = 5√2 cm
Length of chord is 5√2 cm.
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The Chord be AB. Center be O of the circle.
The radii AO and BO are perpendicular, it is given. AOB is an isosceles right angle triangle at O.
Hence according to Pythagoras theorem , we have
AB² = AO² + BO² = 5² + 5² = 50 cm²
Length of the chord = AB = √50 = 5√2 cm
The radii AO and BO are perpendicular, it is given. AOB is an isosceles right angle triangle at O.
Hence according to Pythagoras theorem , we have
AB² = AO² + BO² = 5² + 5² = 50 cm²
Length of the chord = AB = √50 = 5√2 cm
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