Multiple choice

A point $P$ is given on the circumference of a circle of radius $r$. The chord $QR$ is parallel to the tangent line at $P$. The maximum area of the triangle $PQR$ is:

  1. $\displaystyle \frac{3\sqrt{2}}{4}{r}^{2}$
  2. $\displaystyle \frac{3\sqrt{3}}{4}{r}^{2}$
  3. $\displaystyle \frac{3}{8}{r}$
  4. $\displaystyle \frac{3\sqrt{2}}{4}{r}$
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B Correct answer
Explanation

For a chord QR parallel to the tangent at P, the triangle PQR has its maximum area when the height from P to the chord QR is maximized. This occurs when the triangle is equilateral, with height 3r/2. The area is (1/2) * base * height, which equals (3 * sqrt(3) / 4) * r^2.

AI explanation

Let the circle be centered at the origin with P at (0,r), making the tangent line at P horizontal. The chord QR is parallel to this tangent and lies at some y-coordinate -h. The length of QR is given by the chord formula as 2*sqrt(r^2 - h^2), and the height of the triangle is (r - (-h)) = r + h. The area is (1/2) * (r+h) * 2*sqrt(r^2 - h^2) = (r+h)sqrt(r^2 - h^2). Expressing this in polar coordinates with h = r*cos(theta) yields an area of r^2*sin(theta)(1+cos(theta)), which is maximized when theta is 30 degrees (pi/6 radians). Substituting theta gives a maximum area of (3*sqrt(3)/4)*r^2.

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