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Let $\text{C}1$ and $\text{C}2$ be concentric circles such that the diameter of $\text{C}1$ is $2 \; \text{cm}$ longer than that of $\text{C}2.$ If a chord of $\text{C}1$ has length $6 \; \text{cm}$ and is a tangent to $\text{C}2,$ then the diameter, in $\text{cm},$ of $\text{C}1$ is

Given that, the diameter of $\text{C1}$ is $2 \; \text{cm}$ longer than $\text{C2}.$

So, the radius of $\text{C1}$ should be $1 \; \text{cm}$ longer than $\text{C2}.$

Let the radius of $\text{C1} = \text{R}\; \text{cm}$

So, the radius of $\text{C2} = (\text{R} -1) \; \text{cm}$

The chord of $\text{C1}$ has length $6 \; \text{cm}$ and is a tangent to $\text{C2}.$

Now, we can draw the diagram,

In $\triangle \text{OCB},$ using the Pythagoras’ theorem,

$\text{(OB)}^{2} = \text{(OC)}^{2} + \text{(CB)}^{2}$

$\Rightarrow \text{R}^{2} = \text{(R}-1)^{2} + 3^{2}$

$\Rightarrow \text{R}^{2} = \text{R}^{2} + 1 – 2 \text{R} + 9$

$\Rightarrow 2 \text{R} = 10$

$\Rightarrow \boxed { \text{R} = 5 \; \text{cm}}$

So, diameter of $\text{C1} = 2 \text{R} = 2(5) = 10 \; \text{cm}.$

$\therefore$ The diameter of $\text{C1}$ is $10 \; \text{cm}.$

Correct Answer$: 10$

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