Question

In two concentric circles, a chord of length 24 cm of larger circle becomes a tangent to the smaller circle whose radius is 5 cm. Then the radius of the larger circle is

• A. 8 cm
• B. 10 cm
• C. 12 cm
• D. 13 cm

Answer 1

The Correct Option is D

Given: Two concentric circles with a chord of length 24 cm in the larger circle, which acts as a tangent to the smaller circle of radius 5 cm.

Step 1: Understanding the Geometry 

Let \( O \) be the common center of both circles, and let \( AB \) be the chord of the larger circle that becomes a tangent to the smaller circle at some point \( P \).

• \( OP = 5 \) cm (radius of the smaller circle)
• \( AB = 24 \) cm (chord length of the larger circle)

Step 2: Applying the Perpendicular Bisector Theorem

The line \( OP \) is perpendicular to the chord \( AB \), which means it bisects \( AB \) at point \( M \), where:

\[ AM = MB = \frac{24}{2} = 12 \text{ cm} \]

Let \( R \) be the radius of the larger circle (\( OA = OB = R \)). Using the right-angled triangle \( OMA \):

\[ OA^2 = OM^2 + AM^2 \] \[ R^2 = 5^2 + 12^2 \] \[ R^2 = 25 + 144 = 169 \] \[ R = \sqrt{169} = 13 \text{ cm} \]

Final Answer: 13 cm

Answer 2

Let $r_1$ be the radius of the smaller circle and $r_2$ be the radius of the larger circle. We are given that $r_1 = 5$ cm.

The chord of length 24 cm of the larger circle is tangent to the smaller circle. This means that the perpendicular distance from the center O to the chord is equal to the radius of the smaller circle. Let M be the point where the perpendicular from O meets the chord. Then OM = 5 cm.

Since the perpendicular from the center to a chord bisects the chord, we have MB = $\frac{1}{2} \times 24 = 12$ cm.

In the right-angled triangle OMB, we have $OM^2 + MB^2 = OB^2$, where OB is the radius of the larger circle, $r_2$.

So, $5^2 + 12^2 = r_2^2$

$25 + 144 = r_2^2$

$169 = r_2^2$

$r_2 = \sqrt{169} = 13$ cm

Final Answer: The final answer is $\boxed{13}$