Q.$\triangle PQR$ is an equilateral triangle with side 18 cm. A circle is drawn on the segment QR as diameter. Find the length of the arc of this circle within the triangle.

Step 1: Set up coordinates: $Q=(0,0)$, $R=(18,0)$, and since $\triangle PQR$ is equilateral with side $18$, $P=(9, 9\sqrt3)$. The circle on $QR$ as diameter has centre $M=(9,0)$ and radius $9$.

Step 2: Any point $X$ on this circle satisfies $\angle QXR = 90°$ (angle in a semicircle). Checking side $PQ$: parametrising $Q$ to $P$ and substituting into the circle equation shows the circle meets $PQ$ exactly at its midpoint (distance $9$ from $Q$); by symmetry it meets $PR$ at its midpoint too.

Step 3: These two midpoints, as seen from the circle's centre $M$, lie at angles $120°$ and $60°$ from the positive x-direction, so the arc between them on the side nearer $P$ (through the point directly above $M$) subtends a central angle of $120°-60°=60°=\frac{\pi}{3}$ radian, and lies entirely inside the triangle (its highest point, at height $9$, is well below apex $P$'s height $9\sqrt3\approx15.6$).

Step 4: Arc length $=r\theta = 9\times\frac{\pi}{3}=3\pi$ cm $\approx9.42$ cm.

Use coordinates to locate where the circle on QR as diameter meets sides PQ and PR (their midpoints, by the angle-in-a-semicircle property), find the 60° central angle between those points on the side toward P, and apply s = rθ.

• Assuming the relevant arc is the major arc (through the far side, mostly outside the triangle) rather than the 60° minor arc toward P.
• Not realising the circle intersects PQ and PR exactly at their midpoints, and instead guessing an angle.
• Forgetting the radius of this circle is half of QR (9 cm), not the full 18 cm.