Question

two concentric circles are radius 5cm and 3cm drawn.find the length of the chord of the larger circle which touches the smaller circle​

Answer 1

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

Answer 2

Let the two concentric circles be centered at point O. And let PQ be the chord of the larger circle which touches the smaller circle at point A. Therefore, PQ is tangent to the smaller circle.

OA ⊥ PQ (As OA is the radius of the circle)

Applying Pythagoras theorem in ΔOAP, we obtain

OA2 + AP2 = OP2

32 + AP2 = 52

9 + AP2 = 25

AP2 = 16

AP = 4

In ΔOPQ,

Since OA ⊥ PQ,

AP = AQ (Perpendicular from the center of the circle bisects the chord)

PQ = 2AP = 2 × 4 = 8

Therefore, the length of the chord of the larger circle is 8 cm.