Q. 124.3( 3 Votes )

The figure formed A. Square

B. Rectangle

C. Trapezium

D. None of these

Answer :

To prove: That the quadrilateral formed by joining the mid points of sides of a rhombus is a rectangle.

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


Construction: Join AC


Proof: In ΔABC, P and Q are the mid points of AB and BC respectively


Therefore,


PQ || AC and PQ = AC (i) (Mid-point theorem)


Similarly,


RS || AC and RS = AC (ii) (Mid-point theorem)


From (i) and (ii), we get


PQ RS and PQ = RS


Thus, PQRS is a parallelogram (A quadrilateral is a parallelogram, if one pair of opposite sides is parallel and equal)


AB = BC (Given)


Therefore,


AB = BC


PB = BQ (P and Q are mid points of AB and BC respectively)


In ΔPBQ,


PB = BQ


Therefore,


BQP = BPQ (iii) (Equal sides have equal angles opposite to them)


In ΔAPS and ΔCQR,


AP = CQ (AB = BC = AB = BC = AP = CQ)


AS = CR (AD = CD = AD = CD = AS = CR)


PS = RQ (Opposite sides of parallelogram are equal)


Therefore,


APS CQR (By SSS congruence rule)


APS = CQR (iv) (By c.p.c.t)


Now,


BPQ + SPQ + APS = 180°


BQP + PQR + CQR = 180°


Therefore,


BPQ + SPQ + APS = BQP + PQR + CQR


SPQ = PQR (v) [From (iii) and (iv)]


PS || QR and PQ is the transversal,


Therefore,


SPQ + PQR = 180o (Sum of adjacent interior an angles is 180o)


SPQ + SPQ = 180° [From (v)]


2 SPQ = 180°


SPQ = 90°


Thus, PQRS is a parallelogram such that SPQ = 90o


Hence, PQRS is a rectangle.

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