Question

Two concentric circles are of radii 5 cm and 3 cm. Find the length of the chord of the larger circle which touches the smaller circle.​

Answer 1

Answer:

8cm

Step-by-step explanation:

Let O be the centre of the concentric circle of radii 5 cm and 3 cm respectively. Let AB be a chord of the larger circle touching the smaller circle at P.  

Then

AP=PB and OP⊥AB

Applying Pythagoras theorem in △OPA, we have

OA²  =OP²  + AP²

 

⇒25=9+AP²

⇒AP²  =16⇒AP=4 cm

∴AB=2AP=8 cm

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Answer 2

Answer:

hope it helps you!!!!!!!!

Step-by-step explanation:

Let the radius of the bigger circle be 'R' and the radius of the smaller circle be 'r'.

It is given that R=5cm=OB and r=3cm=OD and AB is the chord whose length we have to find.

AD=BD and OD⊥AB(radius is perpendicular to a chord and it divides the chord into two equal parts)

therefore, ΔODB is a right angled triangle

where, OD²+BD²=OB²

           (3)²+BD²=(5)²

            9+BD²=25

            BD²=25-9

            BD²=16

            BD=4cm

      Since AD=BD=4cm

      therefore, AB=AD+BD

                      AB=4+4

                      AB=8cm