In the given figure, O is the center of two concentric circles of radii 6 cm and 10 cm. AB is a chord of outer circle which touches the inner circle. The length of chord AB is

Given: Two concentric circles (say C_{1} and C_{2}) with common center as O and radius r_{1} = 6 cm(inner circle) and r_{2} = 10 cm (outer circle) respectively.

To Find : Length of the chord AB.

As AB is tangent to circle C_{1} and we know that "Tangent at any point on the circle is perpendicular to the radius through point of contact"

So, we have,

OP ⏊ AB

∴ OPB is a right - angled triangle at P,

By Pythagoras Theorem in △OPB

[i.e. (hypotenuse)^{2} = (perpendicular)^{2} + (base)^{2} ]

We have,

(OP)^{2} + (PA)^{2} = (OA)^{2}

r_{1}^{2} + (PA)^{2} = r_{2}^{2}

(6)^{2} + (PA)^{2}= (10)^{2}

36 + (PA)^{2} = 100

(PA)^{2} = 64

PA = 8 cm

Now, PA = PB ,

[ as perpendicular from center to chord bisects the chord and OP ⏊ AB]

So,

AB = PA + PB = PA + PA = 2PA = 2(8) = 16 cm

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