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in Parallelograms by (26.9k points)
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The figure formed by joining the mid-points of the adjacent sides of a rectangle is a :

A. Square 

B. Rhombus 

C. Trapezium 

D. None of these

1 Answer

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by (27.3k points)
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Best answer

Option : (B)

Given : 

ABCD is a rectangle and P, Q, R, S are their midpoints 

To Prove : 

PQRS is a rhombus 

Proof : 

In ΔABC, 

P and Q are the mid points 

So, 

PQ is parallel AC 

And, 

PQ = \(\frac{1}{2}\)AC 

(The line segment joining the mid points of 2 sides of the triangle is parallel to the third side and half of the third side) 

Similarly, 

RS is parallel AC 

And, 

RS = \(\frac{1}{2}\)AC 

Hence, 

Both PQ and RS are parallel to AC and equal to \(\frac{1}{2}\)AC.

Hence, 

PQRS is a parallelogram 

In triangles APS & BPQ, 

AP = BP 

(P is the mid-point of side AB) 

∠PAS = ∠PBQ (90° each) 

AS = BQ 

(S and Q are the mid points of AD and BC respectively and since opposite sides of a rectangle are equal, so their halves will also be equal) 

ΔAPS ≅  ΔBPQ 

(By SAS congruence rule) 

PS = PQ (By c.p.c.t.) 

PQRS is a parallelogram in which adjacent sides are equal. 

Hence, 

PQRS is a rhombus.

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