Let TR=y.
Since OT is perpendicular bisector of PQ.
∴PR=QR=4 cm
In right triangle ORP, we have
OP2=OR2+PR2 ------Pythagoras theorem
⟹OR2=OP2−PR2=52−45=9
⟹OR=3cm
In right triangles PRT and OPT, we have
TP2=TR2+PR2 ------Pythagoras theorem
and, OT2=TP2+OP2
⟹OT2=(TR2+PR2)+OP2 [substituting the value of TP^2]
⟹(y+3)2=y2+16+25
⟹6y=32
⟹y=163
TR=163
∴TP2=(163)2+42=2569+16=4009
⟹TP=203 cm
![1031378_1009644_ans_aec6392c714a421793fbe92f64b0b716.png](https://search-static.byjusweb.com/question-images/toppr_invalid/questions/1031378_1009644_ans_aec6392c714a421793fbe92f64b0b716.png)