Question

Two concentric circles are of raddii 5cm and 4cm.Find the length of the chord of the larger circle which touches a smaller circle

Answer 1

Given: We're provided with Two Concentric Circles, C₁ and C₂ with the center O. In Circle, C₂ AB be the larger Chord. & It is touching the Smaller Circle, C₁ at point P.

Need to find: The Length of Larger Chord, AB.

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AnSwer: We're joining OP, OA and OB together. OP = 4 cm as it's radius of smaller Circle also, OA = OB = 5 cm as radius of Larger Circle.

• As we can see, OP ⊥ AB.

$\underline{\bigstar\:{\mathcal{\pink{BY\;USING\; PYTHAGORAS\; THEOREM\;:}}}}$⠀⠀

$:\implies\sf \Big(OA\Big)^2 = \Big(OP\Big)^2 + \Big(AP\Big)^2 \\\\\\:\implies\sf \Big(5\Big)^2 = \Big( 4\Big)^2 + \Big(AP\Big)^2\\\\\\:\implies\sf 25 = 16 + AP^2\\\\\\:\implies\sf 25 - 16 = AP^2\\\\\\:\implies\sf 9 = AP^2\\\\\\:\implies\sf AP^2 = 9\\\\\\:\implies\sf AP = \sqrt{9}\\\\\\:\implies\underline{\boxed{\pmb{\frak{\pink{AP = 3\:cm}}}}}\;\bigstar$

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• Since the perpendicular from the center of a circle to a Chord bisects the Chord.

Therefore,

↠ AP = PB = 3 cm

↠ AB = AP + PB

↠ AB = 3 + 3

↠ AB = 6 cm

∴ Hence, the Length of Chord of the Larger Circle is 6 cm.

Answer 2

Given :-

Two concentric circles are of raddii 5cm and 4cm.

To Find :-

Length of the chord of the larger circle which touches a smaller circle

Solution :-

Let us assume that the point is O

The chord of larger circle be AB and it is touching the smaller circle by point K

Therefore

AK = KB

Also,

OK is perpendicular to the chord of larger circle i.e AB

By using pythagoras theorem

H² = P² + B²

OA² = OK² + AK²

AK² = OA² - OK²

AK² = (5)² - (4)²

AK² = 25 - 16

AK² = 9

√AK² = √9

AK = 3

Now

2AK = AB

2(3) = AB

6 = AB