Question

Radius of a circle with centre 0is10cm . Find the length of the chord if the chord is at a distance of 6cm from the centre​

Answer 1

✬ Chord = 16 cm ✬

Explanation:

Given:

• Length of radius of circle is 10 cm.
• Distance of chord from the centre is 6 cm.

To Find:

• What is the length of the chord ?

Solution: Let in the circle with centre O and in ∆OBA we have

• OA = 10 cm , Radius {hypotenuse}

• OB = 6 cm {perpendicular}

• AB = Half chord {base}

• AC = Chord

• ∠OBA = 90°

In ∆OBA , applying Pythagoras Theorem

★ Pythagoras Theorem : H² = P² + B² ★

$\implies{\rm }$ OA² = OB² + AB²

$\implies{\rm }$ 10² = 6² + AB²

$\implies{\rm }$ 100 – 36 = AB²

$\implies{\rm }$ √64 = AB

$\implies{\rm }$ 8 = AB

So the length of AB i.e half chord is 8 cm therefore ,

➯ AC = 2 × 8

➯ AC = 16 cm

Hence, the length of the chord of the circle is 16 cm

Answer 2

Given : O is the centers of Circle , Radius of circle is of 10 cm & the chord is at a distance of 6cm from the centre [O] .

Exigency To Find : The Length of the chord .

⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀

❍ Let's say that Chord of circle be AB , The Distance from the Radius of the circle be ON & Radius of the circle be OA .

⠀⠀⠀⠀⠀⠀As , We can see that it is forming Right angled triangle [ $\triangle$ ONA ] :

⠀⠀⠀⠀Here ,

⠀⠀⠀▪︎⠀AN is the half chord of circle [ Base ]

⠀⠀⠀▪︎⠀ON is the radius of circle [ Perpendicular ]

⠀⠀⠀▪︎⠀OA is the Distance from center to chord of circle [ Hypotenuse ]

⠀⠀⠀▪︎⠀$\angle$ ONA is 90⁰ [ Right angled triangle ]

[ Kindly Refer to given attachment ( image ) ]

$\dag\:\:\sf{ As,\:We\:know\:that\::}\\\\ \qquad\bigstar\:\:\bf By\:Pythagoras\:Theorem\::$

$\qquad \dag\:\:\bigg\lgroup \sf{(Hypotenuse)^2 = (Base)^2 + (Perpendicular)^2 }\bigg\rgroup$

⠀⠀⠀Now ,

⠀⠀In $\triangle$ ONA :

$\qquad:\implies \sf \:(OA)^2 \: = \: (AN)^2 \: + \: (OA)^2 \:$

⠀⠀⠀⠀⠀⠀$\underline {\boldsymbol{\star\:Now \: By \: Substituting \: the \: known \: Values \::}}$

$\qquad:\implies \sf \:(OA)^2 \: = \: (AN)^2 \: + \: (OA)^2 \:$

$\qquad:\implies \sf \:(10)^2 \: = \: (AN)^2 \: + \: (6)^2 \:$

$\qquad:\implies \sf \:100 \: = \: (AN)^2 \: + \: (6)^2 \:$

$\qquad:\implies \sf \:100 \: = \: (AN)^2 \: + \: 36 \:$

$\qquad:\implies \sf \:(AN)^2 \: = \: 100 - \: 36 \:$

$\qquad:\implies \sf \:(AN)^2 \: = \: 64 \:$

$\qquad:\implies \sf \:AN \: = \: \sqrt {64} \:$

$\qquad:\implies \bf \:AN\: = \: 8\:$

$\qquad:\implies \frak{\underline{\purple{\:AN\: = 8\:cm\: }} }\:\:\bigstar$

Therefore,

⠀⠀⠀▪︎⠀AN is the half of AB

$\qquad:\implies \sf\dfrac{1}{2} \:of\: AB = A N \:$

$\qquad:\implies \sf\dfrac{1}{2} \:of\: AB = 8 \:$

$\qquad:\implies \sf\dfrac{1}{2} \:\times\: AB = 8 \:$

$\qquad:\implies \sf\: AB = 8\times 2 \:$

$\qquad:\implies \bf\: AB = 16 \:$

$\qquad:\implies \frak{\underline{\purple{\:AB\: = 16\:cm\: }} }\:\:\bigstar$

Therefore ,

⠀⠀⠀▪︎⠀AB is the chord of the circle is 16 cm .

Therefore,

⠀⠀⠀⠀⠀$\therefore {\underline{ \mathrm {\:Chord \:\:[\ A B\ ]\:of\:Circle \:is\:\bf{16\:cm}}}}$