Multiple choice

A square with side length $l$ is inscribed in a semi-circle such that one side of the square is on the diameter of the semicircle. The perimeter of the semicircle is given by:

  1. $l \pi \sqrt{5}$
  2. $\displaystyle \frac{\pi \sqrt{5} l}{2}$
  3. $\sqrt{5} l \left ( \dfrac{\pi}{2} + 1 \right )$
  4. $\displaystyle \frac{\sqrt{5} l }{2} (\pi + 1)$
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C Correct answer
Explanation

Let the square have side l. The vertices on the semi-circle are at (l/2, l) and (-l/2, l). The radius R satisfies R^2 = (l/2)^2 + l^2 = 5l^2/4, so R = l*sqrt(5)/2. Perimeter of semi-circle = pi*R + 2R = R(pi+2) = (l*sqrt(5)/2)(pi+2) = sqrt(5)*l(pi/2 + 1).

AI explanation

For the square placed in the semicircle, the top two vertices lie on the circular arc, creating a right triangle from the center to a top vertex with legs equal to l and l/2. By the Pythagorean theorem, the radius squared equals l squared plus l/2 squared, which simplifies to 5l squared/4, so the radius is l times the square root of 5 divided by 2. The perimeter of the semicircle includes the half-circumference and the diameter, so it is pi times R plus 2R, which simplifies to R times (pi plus 2). Substituting the radius gives the perimeter as l times the square root of 5 multiplied by (pi/2 plus 1).

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