Question

show that the quadrilateral formed by joining the mid-points of
the pairs of adjacent sides of a rectangle is a rhombus.
[hint . Join AC. Then PQ || AC and PQ-
Also SR || AC and SR - AC
PQRS is a Il gm.
Now, triangle ASP congurent to​ triangle BQP. So, PS = PQ




please send me the correct answer i will mark u as brainlist ​

Answer 1

Solution:-

Given that S, R, p and q are mid points.

So,

•In ∆ADC

SR = 1/2 AC ( mid point theorem)

SR || AC ( " " " )

•In ∆ABC

PQ = 1/2 AC ( mid point theorem)

PQ || AC (------------- " -----------)

SR = PQ AND SR || PQ (because, SR = 1/2 AC AND PQ = 1/2 AC) and (because SR || AC and PQ || AC)

One pair of opposite angle is equal and parallel, So PQRS is a parallelogram.

• ∆ASP congruent ∆BQP [using SAS congruency]
• angle SAP = angle QBP ( each angle of rectangle is 90 )
• AP = BP ( p is the mid point of AB)
• AS = BQ ( S and Q are the mid points )
• So, PS = PQ {By CPCT}

NOW IN A PARALLELOGRAM, ADJACENT SIDES ARE EQUAL.

• Hence, it is a rhombus (because the adjacent sides are equal and it is a parallelogram)

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