Two concentric circles with centre O are of radii 3 cm and 5 cm. Find the length of chord AB of the larger circle which touches the smaller circle at P.

Question :Two concentric circles with centre O are of radii 3 cm and 5 cm. Find the length of chord AB of the larger circle which touches the smaller circle at P.

The correct answer :Let Q be the center of the smaller circle, and let R be the point where the chord AB intersects the radius PO extended through O.

Since PO is perpendicular to AB, we have two right triangles, OPR and OQR. Let x be the length of RP, and let y be the length of RQ. Then, we have:

x² + 3² = 5² (by the Pythagorean theorem in triangle OPR) y² + 3² = 4² (by the Pythagorean theorem in triangle OQR)

Simplifying, we get:

x² = 16 y² = 7

Taking the square root of both sides, we get:

x = 4 y ≈ 2.646

Now, we can use the fact that OP = OQ = 3 cm to find the length of PB. Since OB is a radius of the larger circle, we have OB = 5 cm.

Therefore, PB = OB - OP = 5 - 3 = 2 cm.

Finally, we can use the fact that RP = 4 cm and PB = 2 cm to find the length of chord AB:

AB = RP + PB = 4 + 2 = 6 cm

Therefore, the length of chord AB of the larger circle which touches the smaller circle at P is 6 cm.