If \(PQ\) and \(PR\) are tangents to the circle with centre \(O\) and radius \(4 \text{ cm}\) such that \(\angle QPR = 90^{\circ}\), then the length \(OP\) is
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Whenever tangents from an external point are perpendicular to each other, the quadrilateral formed by the center, the two points of contact, and the external point is a square. The distance from the center to the external point is always \(r\sqrt{2}\).
To solve the problem, we need to find the length of \(OP\). Given that \(PQ\) and \(PR\) are tangents to the circle from point \(P\) and \(\angle QPR = 90^\circ\), and the radius of the circle \(OQ = OR = 4 \text{ cm}\).
We can use the property of tangents that states a tangent drawn from an external point is perpendicular to the radius at the point of tangency. Thus, both \(OQ\) and \(OR\) are perpendicular to the tangents \(PQ\) and \(PR\), respectively.
Given that \(\angle QPR = 90^\circ\), the triangle \(\triangle OQR\) forms a right-angle triangle with \(OP\) as the hypotenuse.
We apply the Pythagorean Theorem in triangle \(\triangle OPQ\) or \(\triangle OPR\):
\[ OP^2 = OQ^2 + OR^2 \]
Since \(OQ = OR = 4 \text{ cm}\), substituting into the equation gives:
\[ OP^2 = 4^2 + 4^2 = 16 + 16 = 32 \]
Thus, the length of \(OP\) is:
\[ OP = \sqrt{32} = 4\sqrt{2} \text{ cm} \]
Therefore, the correct answer is \(4\sqrt{2} \text{ cm}\).
