Question

A tangent PQ at point P of a circle of radius 5cm meets a line through the centre O at a point Q so that OQ =12cm. Length PQ is:

Answer 1

Hint: To solve this question, we must have some basic knowledge of geometry, its identities and also Pythagoras’s theorem. First of all, we will study the diagram of the above question which will make the solution of the question easier. From the diagram, we will see that the points O, P and Q form a right-angled triangle, so we can simply find the length PQ by applying the Pythagoras’s theorem. This will be our final answer. The Pythagoras’s theorem is stated as follows,
\[{{(hypotenuse)}^{2}}={{(base)}^{2}}+{{(perpendicular)}^{2}}\]

Complete step by step answer:
First of all, we will see the diagram of the question which is given below,

Let us assume that the length PQ is x.
As we can see in the above figure,
\[\begin{align}
  & \text{OP}=5cm \\
 & \text{OQ}=12cm \\
 & \text{PQ}=\text{x}cm \\
\end{align}\]
As we know, that the angle between the tangent and radius of a circle is right angle.
Therefore,
\[\angle \text{OPQ}={{90}^{\circ }}\]
Now, we can see that the triangle has met all the requirements for applying Pythagoras’s theorem to it.
 So, applying Pythagoras’s theorem to the triangle as stated above, we can see that,
\[\begin{align}
  & {{(hypotenuse)}^{2}}={{(base)}^{2}}+{{(perpendicular)}^{2}} \\
 & {{12}^{2}}={{5}^{2}}+{{x}^{2}} \\
 & x=\sqrt{{{12}^{2}}-{{5}^{2}}} \\
 & x=\sqrt{144-25} \\
 & x=\sqrt{119} \\
\end{align}\]
Now, taking the square root of the above equation, we get,
\[x=10.9cm\]

Hence, the length PQ is 10.9cm

Note: To solve this question, don’t forget to make the diagram before solving as this will make solving the question easier. Also, for Pythagoras’s theorem, put the values of perpendicular, base and hypotenuse carefully at the right place. Never become nervous if you don’t see a perfect square number inside the square root.