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:
Compute side lengths: AB = 2r, BC = √[(0−(−r))^2 + (r−0)^2] = √(r^2 + r^2) = r√2Similarly, CD = AB and DA = BC = r√2; adjoining sides are equal.Angles are right angles (due to symmetry and equal slopes), hence it is a rectangle with equal sides ⇒ square.
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
Discussion & Comments