In the given figure, from an external point P, two tangents PQ and PR are drawn to a circle ofradius 4 cm with centre O. If ∠QPR = 90°, then length of PQ i

The problem is a classic example of a right triangle problem in geometry. Here are the steps to solve it:

Step 1: Identify the right triangle. In this case, the triangle is QPR, with ∠QPR being the right angle.

Step 2: Understand that a tangent to a circle is perpendicular to the radius of the circle at the point of tangency. Therefore, OQ and OR are each 4 cm long (the radius of the circle), and ∠OQP and ∠ORP are each right angles.

Step 3: Recognize that triangle OQP and triangle ORP are congruent right triangles (they have the same length hypotenuse and the same length leg). Therefore, ∠OPQ = ∠OPR = 45°.

Step 4: Now, triangle QPR is also a right triangle, with ∠QPR = 90°. Since ∠OPQ = ∠OPR = 45°, then ∠PQR = ∠PRQ = 45°. This means triangle QPR is a 45-45-90 triangle.

Step 5: In a 45-45-90 triangle, the length of the hypotenuse is √2 times the length of each leg. Therefore, the length of PQ (and also PR) is 4√2 cm.