Two concentric circles are of radii 10 cm and 6 cm. Find the length of the chord of the larger circle which touches the smaller circle.

Radius of the larger circle, R = 10 cm

Radius of the smaller circle, r = 6 cm 

Concept Used:​

The length of the chord of the larger circle that touches the smaller circle can be found by forming a right triangle. Since the chord is tangent to the smaller circle, it will be perpendicular to the radius of the smaller circle at the point of tangency. 

Pythagoras theorem:

​$$\text{Hypotenuse}^2 = \text{Base}^2 + \text{Perpendicular}^2$$ 

Solution: 

In △OCA 

By the concept and Pythagoras theorem;

​$$AC^2 = OA^2-OC^2 \\ AC^2 = 10^2 - 6^2 \\ AC^2 = 100 - 36 \\ AC^2 = 64 \\ AC = \sqrt {64} \\ AC = 8 \, cm$$​​

Since, AB = AC + CB ( tangent property)  

  AC = CB ( Perpendicular bisect chord)

Thus length of chord AB = 2$$\times$$ 8 = 16 cm​
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