Question

Show that the quadrilateral, formed by joining the
mid-points of the consecutive sides of a rectangle, is a rhombus.​

Answer 1

Answer:

Let ABCD be a rhombus and P, Q, R and S be the mid-points of sides AB, BC, CD and DA, respectively

Join AC and BD.

From triangle ABD,

we have SP = 1/2 BD and SP || BD (Because S and P are mid-points)

Similarly, RQ = 1/2 BD and RQ || BD

Therefore, SP = RQ and SP || RQ

So, PQRS is a parallelogram. (1) Also,AC ⊥ BD (Diagonals of a rhombus are perpendicular)

Further PQ || AC (From ∆BAC)

As SP || BD, PQ || AC and AC ⊥ BD,

therefore, we have SP ⊥ PQ, i.e. ∠SPQ = 90º. (2)

Therefore, PQRS is a rectangle[From (1) and (2)]

Answer 2

Answer:

Hey mate,here's your answer

Step-by-step explanation:

A rectangle ABCD in which P,Q,R and S are the mid-points of sides AB,BC,CD and DA respectively. 

PQ,QR,RS and SP are joined. 

Join AC. In ΔABC,P and Q are the mid-points of sides AB and BC respectively.

∴PQ∣∣AC and PQ=21AC   .....(i) (Mid pt. Theorem)

In ΔADC,R and S are the mid-points of sides CD and AD respectively.

∴SR∣∣AC and SR=21AC   ....(ii)  (Mid. pt. Theorem)

From (i) and (ii), we have

PQ∣∣SR and PQ=SR

⇒PQRS is a parallelogram.

Now ABCD is a rectangle. 

∴AD=BC

⇒21AD=21BC

⇒AS=BQ   .....(iii)

In ΔS,APS and BPQ, we have

AP=BP   (P is the mid-point of AB)

∠PAS=∠PBQ  (Each is equal to 90∘)

and AS=BQ  (From (iii))

∴ΔAPQ≅ΔBPQ (SAS)

⇒PS=PQ

∴PQRS is a parallelogram whose adjacent sides are equal.

⇒PQRS is a rhombus.

Hope it's helpful.

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