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Sagot :
First of all, sorry for not including a picture - I hope I can describe this well enough:
Call the centre of the circle O, the chord AB (touching the circumference at A and B) and the midpoint of AB M. From circle theorems, we know that a radius, passing through the midpoint of a chord, will be perpendicular to it. Therefore we have a right-angled triangle OMA, with the hypotenuse OA forming the radius of length 13in, the line OM being 5in as stated in the question, and the side AM being an unknown.
By Pythagoras' theorem, AM = √(13^2 - 5^2) = 12in (it's a 5,12,13 Triple)
Because M is the midpoint of AB, this value needs to be doubled to get the length of AB as 24in. One inch is 2.54cm (3sf), so the length of the chord is:
24 * 2.54 = 60.96cm, which rounds to 61.0cm (3sf)
I hope this helps
Call the centre of the circle O, the chord AB (touching the circumference at A and B) and the midpoint of AB M. From circle theorems, we know that a radius, passing through the midpoint of a chord, will be perpendicular to it. Therefore we have a right-angled triangle OMA, with the hypotenuse OA forming the radius of length 13in, the line OM being 5in as stated in the question, and the side AM being an unknown.
By Pythagoras' theorem, AM = √(13^2 - 5^2) = 12in (it's a 5,12,13 Triple)
Because M is the midpoint of AB, this value needs to be doubled to get the length of AB as 24in. One inch is 2.54cm (3sf), so the length of the chord is:
24 * 2.54 = 60.96cm, which rounds to 61.0cm (3sf)
I hope this helps
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