- A40 cm
- B56 cm
- C28 cm
- D20 cm
Step-by-Step Solution
The diagonal of the rectangle is cm, and both diagonals are equal. By the mid-point theorem each side of PQRS is half a diagonal, cm, so PQRS is a rhombus with perimeter cm.
Answer: A) 40 cm
Rekha draws a rectangle ABCD of 16 cm by 12 cm and joins the mid-points P, Q, R, S of its sides. Find the perimeter of the orange shape PQRS.
Class 9QuadrilateralsGeometry — Quadrilaterals
Key idea: Mid-point theorem applied to quadrilaterals
**Result**: The figure formed by joining the mid-points of the sides of any quadrilateral (in order) is a **parallelogram**.
**Proof sketch**: Let ABCD be any quadrilateral. Let P, Q, R, S be midpoints of AB, BC, CD, DA respectively. Draw diagonal AC.
- In ΔABC: PQ ∥ AC and PQ = (1/2)AC (mid-point theorem).
- In ΔACD: SR ∥ AC and SR = (1/2)AC (mid-point theorem).
- So PQ ∥ SR and PQ = SR. One pair equal AND parallel → PQRS is a parallelogram. ∎
If ABCD is a rectangle, PQRS becomes a rhombus. If ABCD is a rhombus, PQRS becomes a rectangle. Surprising and elegant.
**Proof sketch**: Let ABCD be any quadrilateral. Let P, Q, R, S be midpoints of AB, BC, CD, DA respectively. Draw diagonal AC.
- In ΔABC: PQ ∥ AC and PQ = (1/2)AC (mid-point theorem).
- In ΔACD: SR ∥ AC and SR = (1/2)AC (mid-point theorem).
- So PQ ∥ SR and PQ = SR. One pair equal AND parallel → PQRS is a parallelogram. ∎
If ABCD is a rectangle, PQRS becomes a rhombus. If ABCD is a rhombus, PQRS becomes a rectangle. Surprising and elegant.
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