Chapter 9.2, Problem 3MRE

Solution

To find: $AB$

  $AB=2\sqrt{7}$

Given:

The figure:

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

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.

Explanation:

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 $△ABC$ .

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

So, by theorem 8-2,

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

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

Now, take square root of both sides,

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

So, $AB=2\sqrt{7}$