Math, asked by jjjjrsv7, 1 year ago

ABCD is a rhombus and pqrs are the midpoints of the side ab bc cd and da. pllz show the figure, draw the figure and send

Answers

Answered by sonabrainly
1

Answer:


Step-by-step explanation:


 Given-  ABCD is a rhombus and P, Q, R and S are the mid-points of the sides AB, BC, CD and DA respectively.


To Prove-PQRS is a rectangle


Construction,

AC and BD are joined.


Proof,

In ΔDRS and ΔBPQ,

DS = BQ (Halves of the opposite sides of the rhombus)

∠SDR = ∠QBP (Opposite angles of the rhombus)

DR = BP (Halves of the opposite sides of the rhombus)

Thus, ΔDRS ≅ ΔBPQ by SAS congruence condition.

RS = PQ by CPCT --- (i)

In ΔQCR and ΔSAP,

RC = PA (Halves of the opposite sides of the rhombus)

∠RCQ = ∠PAS (Opposite angles of the rhombus)

CQ = AS (Halves of the opposite sides of the rhombus)

Thus, ΔQCR ≅ ΔSAP by SAS congruence condition.

RQ = SP by CPCT --- (ii)

Now,

In ΔCDB,

R and Q are the mid points of CD and BC respectively.

⇒ QR || BD  

also,

P and S are the mid points of AD and AB respectively.

⇒ PS || BD

⇒ QR || PS

Thus, PQRS is a parallelogram.

also, ∠PQR = 90° 

Now,

In PQRS,

RS = PQ and RQ = SP from (i) and (ii)

∠Q = 90°

Thus, PQRS is a rectangle.





jjjjrsv7: draw the figure plzz
sonabrainly: ok
sonabrainly: but mark it the brainliest
Answered by AspiringLearner
0

Answer: The figure has been attached.

When the midpoints of a rhombus are joined, the resulting figure will be a rectangle.



Hope this helps...


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