Multiple choice

Let PQ and RS be tangents at the extremities of the diameter PR of a circle of radius $r$. If PS and RQ intersect at a point X on the circumference of the circle, then $2r$ equals

  1. $\sqrt{PQ.RS}$
  2. $\dfrac{PQ+RS}{2}$
  3. $\dfrac{2PQ.RS}{PQ+RS}$
  4. $\sqrt{\dfrac{PQ^2+RS^2}{2}}$
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A Correct answer
Explanation

In this configuration, the tangents PQ and RS at the ends of diameter PR create properties where the triangles formed are similar. Specifically, the geometry leads to the relation 2r = sqrt(PQ * RS).

AI explanation

Let the center of the semicircle be O, so the radius is r and PO = RO = r. The tangent at P is perpendicular to diameter PR, so angle RPX is 90 degrees; similarly, angle PRX is 90 degrees. In right triangle PRX, the altitude to the hypotenuse is XO, which gives the geometric mean relation XO * PR = PX * RX. Substituting the known values, we have r * 2r = PQ * RS, which simplifies to 2r squared equals PQ * RS. Taking the square root of both sides yields the result: 2r equals the square root of the product of PQ and RS.

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