Answer :
[tex]\boxed{D) \sqrt{119} \: cm}[/tex]
Solution :
In the above figure, the line that is drawn from the centre of the given circle to the targent PQ is perpendicular to PQ.
And So, OP ⊥ PQ
Using Pythagoras theorem in triangle ΔOPQ we get ,
[tex]\Longrightarrow\: OQ^2 = OP^2 + PQ^2[/tex]
[tex] \: \: \: \: \: \: \: \: \: \: \: \: [/tex]
[tex]\Longrightarrow\: (12)^2 = 5^2+PQ^2[/tex]
[tex]\: \: \: \: \: \: \: \: \: \: \: \: [/tex]
[tex]\Longrightarrow\: PQ^2 = 144-25[/tex]
[tex]\: \: \: \: \: \: \: \: \: \: \: \: [/tex]
[tex]\Longrightarrow\: PQ^2 = 119[/tex]
[tex]\: \: \: \: \: \: \: \: \: \: \: \: [/tex]
[tex]\Longrightarrow\: PQ = \sqrt{119} [/tex]
[tex]\: \: \: \: \: \: \: \: \: \: \: \: [/tex]
So, option D) √119 cm is the length of PQ.
