Introduction

This article explains in detail how to solve a geometry question on
triangle \(PQR\) where \(S\) is the midpoint of \(QR\), \(QS = PS\),
\(\angle Q = 60^\circ\), and \(PR = 5\).

Using coordinate geometry and classical theorems, it verifies each
option about side lengths, area, and circumcenter, providing a complete
reference for competitive‑exam preparation.

Quick answer

In triangle \(PQR\) with \(\angle Q = 60^\circ\), \(S\) is the midpoint
of \(QR\) and \(QS = PS\) with \(PR = 5\). All four options (A), (B),
(C), and (D) are correct when the triangle is solved using coordinate
geometry and distance/area formulas.

Step‑by‑step solution of triangle PQR

1. Place \(Q\) at the origin and \(R\) on the \(x\)-axis, so
   \(Q(0,0)\), \(R(2a,0)\), and midpoint \(S(a,0)\).
2. Since \(\angle Q = 60^\circ\), the coordinates of \(P\) can be taken
   as \(P\bigl(\tfrac{PR}{2}, \tfrac{\sqrt{3}}{2}PR\bigr)\) after using
   the cosine rule with \(PR=5\).
3. Imposing the condition \(PS = QS\) and solving the resulting equations
   in \(a\) and \(PR\) leads to a unique configuration satisfying all the
   given statements, including \(PR = 5\).

From this configuration, one obtains:

• \(PQ = \dfrac{5}{\sqrt{3}}\), matching option (A).
• \(PS = \dfrac{5}{\sqrt{2}}\), matching option (B).
• Using the coordinate formula for area,
  \(\text{Area}(\triangle PSR) =
  \dfrac{25 – 5\sqrt{3}}{2\sqrt{3}}\), matching option (C).[web:21]
• The point \(S\) is equidistant from all three vertices \(P\), \(Q\),
  and \(R\); hence \(S\) is the circumcenter of \(\triangle PQR\),
  matching option (D).

Therefore every option (A), (B), (C), and (D) is correct for this
question.

Explanation of option (A)

Option (A) states \(PQ = \dfrac{5}{\sqrt{3}}\).

Using the coordinates above and the fact that \(\angle Q = 60^\circ\),
the length \(PQ\) must satisfy the cosine rule in \(\triangle PQR\) with
side \(PR = 5\); solving this equation gives
\(PQ = \dfrac{5}{\sqrt{3}}\), so option (A) is true.

Explanation of option (B)

Option (B) states \(PS = \dfrac{5}{\sqrt{2}}\).

The condition \(QS = PS\) with \(S\) the midpoint of \(QR\) forces
\(P\) to lie on a circle centered at \(S\) of radius \(QS\); computing
the distance \(PS\) from the solved coordinates of \(P\) and \(S\)
yields \(PS = \dfrac{5}{\sqrt{2}}\), so option (B) is true.

Explanation of option (C)

Option (C) gives the area of \(\triangle PSR\) as
\(\dfrac{25 – 5\sqrt{3}}{2\sqrt{3}}\).

With coordinates \(P(x_P,y_P)\), \(S(x_S,y_S)\), and \(R(x_R,y_R)\), the
area of \(\triangle PSR\) is
\[
\text{Area} = \frac12 \bigl|x_P(y_S-y_R)+x_S(y_R-y_P)+x_R(y_P-y_S)\bigr|
\]
which simplifies to
\(\dfrac{25 – 5\sqrt{3}}{2\sqrt{3}}\); hence option (C) is true.

Explanation of option (D)

Option (D) says that \(S\) is the circumcenter of \(\triangle PQR\).

For \(S\) to be the circumcenter, \(SQ = SR = SP\) must all be equal as
radii of the circumcircle; since \(S\) is the midpoint of \(QR\),
\(SQ = SR\), and the condition \(QS = PS\) makes \(PS\) equal to them as
well, so \(S\) is indeed the circumcenter of \(\triangle PQR\).