Two concentric circles are of diameters 30 cm and 18 cm. Find the length of the chord of the larger circle which touches the smaller circle.

Given: Two concentric circles are of diameters 30 cm and 18 cm.

To find: The length of AB.

Theorem Used:

1) Tangent to a circle at a point is perpendicular to the radius through the point of contact.

2) Pythagoras theorem:

In a right-angled triangle, the squares of the hypotenuse is equal to the sum of the squares of the other two sides.

Explanation:

In the diagram AB is the chord touching the smaller circle.

As O^’B is a tangent to radius O^’O,

By the theorem (1) stated,

O^’O ⊥ O^’B

So, we have the right-angled triangle OO'B

By Pythagoras theorem,

OB² = O^’O² + O^’B²

Now since the chord of the larger circle which touches the smaller circle is bisected at the point of contact

We have,

AB= 2 × 24= 48 cm