Chapter 9: Q3. (page 337)

Find $A B .$ In Exercise $3,$ $\bar{C B}$ is tangent to $⊙ A .$

Short Answer

Expert verified

The value of, $A B = 2 \sqrt{7}$ .

Step by step solution

Step 1. Given information.

The figure here given is,

$\bar{C B}$ is a tangent to circle with centre A.

Step 2. Concept Used.

Theorem $9.2$ : tangent of the circle is perpendicular to the radius of the circle.

Theorem $8.2$ : The Square of the hypotenuse of right triangle is equal to the sum of the squares of the sides in a right triangle.

Consider the given figure:

It is given that $\bar{C B}$ is a tangent to circle with centre A.

So, by theorem $9.2$ ,

it can be said that $\bar{C B}$ is perpendicular to $\bar{A B}$ .

That is,

$m ∠ A B C = 90$

Let $A B = x$

Consider the right triangle $Δ A B C$ .

Now, $\bar{A C}$ is the hypotenuse and has length 8and the lengths of the legs are $A B = x$ and $C B = 6$ .

So, by theorem $8.2$ ,

$\begin{matrix} A C^{2} = A B^{2} + B C^{2} \\ 8^{2} = x^{2} + 6^{2} \\ 64 = x^{2} + 36 \end{matrix}$

$\begin{matrix} 64 - 36 = x^{2} \\ 28 = x^{2} \end{matrix}$

Step 3. Now, take square root of both sides.

$\begin{matrix} \sqrt{x^{2}} = \sqrt{28} \\ x = 2 \sqrt{7} \end{matrix}$

Therefore, the value of, $A B = 2 \sqrt{7}$ .

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Find $A B .$ In Exercise $3,$ $\bar{C B}$ is tangent to $⊙ A .$

Short Answer

Step by step solution

Step 1. Given information.

Step 2. Concept Used.

Step 3. Now, take square root of both sides.

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Step by step solution

Step 1. Given information.

Step 2. Concept Used.

Step 3. Now, take square root of both sides.

One App. One Place for Learning.

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Find $A B .$ In Exercise $3,$ $\bar{C B}$ is tangent to $⊙ A .$