Chapter 9.4, Problem 3ST1

Solution

To find: $OP$

  $OP=15$

Given:

  $\overline{NP}$ is a tangent to circle with centre O at P .

$NO=25$ And $NP=20$

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 figure representing the given question:

It is given that $\overline{NP}$ is a tangent to circle with centre O at P .

So, by theorem 9-2, it can be said that $\overline{NP}$ is perpendicular to $\overline{OP}$ .

That is,

  $m\angle OPN=90$

Let $OP=x$

Consider the right triangle $△OPN$ .

Now, $\overline{NO}$ is the hypotenuse and has length 25 and the lengths of the legs are $OP=x$ and $NP=20$ .

So, by theorem 8-2,

  $\begin{array}{c}N{O}^2=O{P}^2+N{P}^2\\ {25}^2=x^2+{20}^2\\ 625=x^2+400\end{array}$

  $\begin{array}{c}625-400=x^2\\ 225=x^2\end{array}$

Now, take square root of both sides,

  $\begin{array}{c}\sqrt{x^2}=\sqrt{225}\\ x=15\end{array}$

So, $OP=15$