O is the centre of this circle. Tangent drawn from a point P, touches the circle at Q. If PQ = 24 cm and OQ = 10 cm, then what is the value of OP?

26 cm

According to the concept

\(\angle\)OQP = \({90}^\circ\)

Hence, OP is the hypotenuse of \(\Delta \)OQP which is a right angled triangle.

Now, OP =√(\( {24 }^{2 } \) + \( {10 }^{2 } \)) = 26 cm

Therefore, the value of OP is 26 cm.