Question

(c) In the given figure ABCD is a rectangle P, Q, R and S are the mid points of the sides AB,BC,CD and AD
respectively.
(i) Prove that PQRS is a rhombus.
(ii) If the length of AC is 6 cm, find the perimeter of PQRS.​

Answer 1

Perimeter of PQRS = 12 cm.

Step-by-step explanation:

(i) Join AC.

In Δ ABC, as per Triangle Mid segment theorem

PQ = AC/2 and PQ ║ AC                                         (1)

similarly, in Δ ADC

SR = AC/2 and SR ║ AC                                          (2)

From (1) and (2), it can be said that

PQ = SR

since PQ ║ AC and SR ║ AC , PQ ║ SR

PQRS is a parallelogram.                                        (3)                    

BC = AD (Opposite sides of a rectangle)

BC/2 = AD/2

QC = SR                                                                    (4)

In ΔSDR and ΔQCR

from (3) QC = SR

∠QCR = ∠SDR = 90°

Since R is the midpoint of CD, CR = DR

Hence, ΔQCR ≅ ΔSDR by Side Angle Side(SAS) congruence theorem

It implies that, QR = SR                                            (5)

From (3) and (5) it can be proved that PQRS is a rhombus.

ii) Given AC = 6 cm.

From PQ = SR = AC/2

PQ = SR = 3 cm.

Perimeter of rhombus = 4a

                                    = 4*3 = 12cm

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