Question

Find $AB.$In Exercise $3,$$\overline{CB}$ is tangent to $\odot A.$

Short answer

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

Step-by-step solution

Step 1. Given information.

The figure here given is,

$\overline{CB}$ 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 $\overline{CB}$ is a tangent to circle with centre A.

So, by theorem $9.2$,

it can be said that $\overline{CB}$is perpendicular to $\overline{AB}$.

That is,

$m\angle ABC=90$

Let $AB=x$

Consider the right triangle $\Delta ABC$.

Now, $\overline{AC}$ is the hypotenuse and has length 8and the lengths of the legs are $AB=x$ and $CB=6$.

So, by theorem $8.2$,

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

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

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

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

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