Answer:
In the above figure, the line that is drawn from the centre of the given circle to the tangent PQ. is perpendicular to PQ.
And so, $\mathrm{OP} \perp \mathrm{PQ}$
Using Pythagoras theorem in triangle $\triangle \mathrm{OPQ}$ we get,
$$
\begin{array}{l}
\mathrm{OQ}^{2}=\mathrm{OP}^{2}+\mathrm{PQ}^{2} \\
\Rightarrow(12)^{2}=5^{2}+\mathrm{PQ}^{2} \\
\Rightarrow \mathrm{PQ}^{2}=144-25 \\
\Rightarrow \mathrm{PQ}^{2}=119 \\
\Rightarrow \mathrm{PQ}=\sqrt{119} \mathrm{~cm}
\end{array}
$$
So, option D i.e. $V 119 \mathrm{~cm}$ is the length of PQ.
A tangent PQ at a point P of a circle of radius $5 \mathrm{~cm}$ meets a line through the centre $\mathrm{O}$ at a point $Q$ so that $O Q=12 \mathrm{~cm}$. Length $P Q$ is :
(A) $12 \mathrm{~cm}$
(B) $13 \mathrm{~cm}$
(C) $8.5 \mathrm{~cm}$
(D) $\sqrt{1} 19 \mathrm{~cm}$
(A) $12 \mathrm{~cm}$
(B) $13 \mathrm{~cm}$
(C) $8.5 \mathrm{~cm}$
(D) $\sqrt{1} 19 \mathrm{~cm}$
A tangent PQ at a point P of a circle of radius $5 \mathrm{~cm}$ meets a line through the centre $\mathrm{O}$ at a point $Q$ so that $O Q=12 \mathrm{~cm}$. Length $P Q$ is :
(A) $12 \mathrm{~cm}$
(B) $13 \mathrm{~cm}$
(C) $8.5 \mathrm{~cm}$
(D) $\sqrt{1} 19 \mathrm{~cm}$
(A) $12 \mathrm{~cm}$
(B) $13 \mathrm{~cm}$
(C) $8.5 \mathrm{~cm}$
(D) $\sqrt{1} 19 \mathrm{~cm}$