If two diameters of a circle intersect each other at right angles, then the quadrilateral formed by joining their endpoints is a:

Introduction / Context:Join the four endpoints of two perpendicular diameters of a circle. We seek the nature of the quadrilateral thus formed. Circle symmetry and right angles at the center lead to a special rectangle with all sides equal—i.e., a square.

Given Data / Assumptions:

• Two diameters intersect at the circle’s center O with ∠ between them = 90°.
• Endpoints A, B on one diameter; C, D on the other.

Concept / Approach:

• All four points lie on the same circle (concyclic).
• Using vectors/coordinates: take A(r,0), B(−r,0), C(0,r), D(0,−r).

Step-by-Step Reasoning:

Verification / Alternative check:Diagonals AC and BD are equal and perpendicular; all sides equal confirms square uniquely among parallelograms/rectangles/rhombi.

Why Other Options Are Wrong:

• Rhombus: has equal sides but not necessarily right angles.
• Rectangle: has right angles but not necessarily equal sides.
• Parallelogram: too general; lacks right angles and equal sides requirements.

Common Pitfalls:

• Assuming any two diameters form a rectangle only; perpendicularity ensures equal adjacent sides, making it a square.

Final Answer:Square