Question

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

Answer 1

Procedure for construction:

Draw a line segment of length AB = 10 cm. Bisect AB by constructing a perpendicular bisector of AB. Let M be the mid-point of AB.

With M as centre and AM as radius, draw a circle. Let it intersect the given circle at the points P and Q.

Join PB and QB. Thus, PB and QB are the required two tangents.

Justification: Join AP. Here ∠APB is an angle in the semi-circle. Therefore, ∠APB = 90°. Since AP is a radius of a circle, PB has to be a tangent to a circle. Similarly, QB is also a tangent to a circle.

In a Right ∆APB, AB2 = AP2 + PB2 (By using Pythagoras Theorem)

PB2 = AB2 – AP2 = 102 — 62 = 100 – 36 = 64

PB = 8 cm.

Answer 2

The length of the chord is 16 cm.

Step-by-step explanation:

See the attached diagram.

There are two concentric circles, the larger one has radius (R) = 10 cm and the smaller one has radius (r) = 6 cm.

Now, the center is O.  

Now, the large circle's chord AB touches the smaller circle at C.

In Δ OBC, ∠ BCO = 90° and OB² = OC² + BC²

⇒ R² = r² + BC²

⇒ BC² = 10² - 6² = 64

⇒ BC = 8 cm .

Now, AB = 2 × BC = 2 × 8 = 16 cm. (Answer)