Consider the following conditions on a parallelogram PQRS, labelled anticlockwise:
I length of PQ = length of QR
II The diagonal PR intersects the diagonal QS at right angles
III ∠ PQR = ∠ QRS
Which of these conditions is/are individually sufficient for the parallelogram PQRS to be a square?
- AI sufficient: yes, II sufficient: yes, III sufficient: yes
- BI sufficient: yes, II sufficient: yes, III sufficient: no
- CI sufficient: yes, II sufficient: no, III sufficient: yes
- DI sufficient: yes, II sufficient: no, III sufficient: no
- EI sufficient: no, II sufficient: yes, III sufficient: yes
- FI sufficient: no, II sufficient: yes, III sufficient: no
- GI sufficient: no, II sufficient: no, III sufficient: yes
- HI sufficient: no, II sufficient: no, III sufficient: no
Show the answer and worked solution
answer · H
- AI sufficient: yes, II sufficient: yes, III sufficient: yes
- BI sufficient: yes, II sufficient: yes, III sufficient: no
- CI sufficient: yes, II sufficient: no, III sufficient: yes
- DI sufficient: yes, II sufficient: no, III sufficient: no
- EI sufficient: no, II sufficient: yes, III sufficient: yes
- FI sufficient: no, II sufficient: yes, III sufficient: no
- GI sufficient: no, II sufficient: no, III sufficient: yes
- HI sufficient: no, II sufficient: no, III sufficient: no
Each condition turns the parallelogram into something, but a square needs both equal sides and right angles. Condition I says two adjacent sides are equal, which makes it a rhombus — a non-square rhombus satisfies it. Condition II, diagonals meeting at right angles, is exactly the rhombus condition again, so the same counterexample works. Condition III says two consecutive interior angles are equal, and since consecutive angles of a parallelogram add to 180∘ each must be 90∘ — that gives a rectangle, and a non-square rectangle satisfies it. So no condition on its own is sufficient.