Question

21. Find the length of a chord which is at a distance of 12cm from the centre

of a circle of radius 13cm.​

Answer 1

Answer: Let AB be a chord of circle with centre O and radius 13cm. Draw OM perpendicular AB and join OA.

In the right triangle OMA, we have

 OA2 = OM2 + AM2

⇒  132 = 122 + AM2

⇒  AM2  = 169 - 144 = 25                                                                                                                                   ⇒  AM = 5cm.

As the perpendicular from the centre of a chord bisects the chord.Therefore,

AB = 2AM = 2 x 5 = 10cm

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Answer 2

GiveN :-

• A chord is at a distance of 12 cm from the centre .
• The radius of the circle is 13 cm .

To FinD :-

• The lenght of the chord.

SolutioN :-

Given that a chord is at a distance of 12cm from the centre of a Circle of radius 13cm . We need to find the lenght of the chord . We know that ,

• A perpendicular drawn from the centre of a circle to a chord bisects the chord . Using this Therom , we can find the length of chord . For diagram refer to attachment .

$\red{\star}\underline{\textsf{ In triangle ABO we have , }}$

$\sf:\implies \pink{ OB^2 + AB^2= AO^2}\\\\\sf:\implies (12cm)^2+AB^2 = (13cm)^2\\\\\sf:\implies 144cm^2 + AB^2=169 cm^2\\\\\sf:\implies AB^2=168cm^2-144cm^2\\\\\sf:\implies AB^2=25cm^2 \\\\\sf:\implies AB =\pm 5cm \\\\\sf:\implies\boxed{\pink{\frak{ AB = 5cm }}}$

Hence the lenght of the chord will be two times this distance = 5 cm*2 = 10 cm .