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In two concentric circles centered at $O$, a chord $AB$ of the larger circle touches the smaller circle at $C$. If $OA = 3.5$ cm, $OC = 2.1$ cm, then $AB$ is equal to

In two concentric circles centered at $O$, a chord $AB$ of the larger circle touches the smaller circle at $C$. If $OA = 3.5$ cm, $OC = 2.1$ cm, then $AB$ is equal to

(A) 5.6 cm 
(B) 2.8 cm
(C) 3.5 cm  
(D) 4.2 cm 

This Question has 1 answers.

We are given two concentric circles centered at $O$, where the chord $AB$ of the larger circle touches the smaller circle at $C$. Given:

$OA = 3.5$ cm,
$OC = 2.1$ cm.

We need to find the length of $AB$.

### Step 1: Identify Right Triangle
Since $AB$ is a chord of the larger circle and touches the smaller circle at $C$, the radius $OC$ is perpendicular to $AB$ at $C$. This means that $OC$ bisects $AB$, so:

$AC = CB = \frac{AB}{2}$.

In the right triangle $OAC$, using the Pythagorean theorem:

$ \displaystyle OA^2 = AC^2 + OC^2 $

Substituting the given values:

$ \displaystyle (3.5)^2 = AC^2 + (2.1)^2 $

$ \displaystyle 12.25 = AC^2 + 4.41 $

$ \displaystyle AC^2 = 12.25 - 4.41 = 7.84 $

$ \displaystyle AC = \sqrt{7.84} = 2.8 $

### Step 2: Find $AB$
Since $AB = 2AC$, we get:

$ \displaystyle AB = 2 \times 2.8 = 5.6 $

### Final Answer:
The length of $AB$ is $5.6$ cm.

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