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 = 180^{o} (Sum of adjacent interior an angles is 180^{o})

∠SPQ + ∠SPQ = 180° [From (v)]

2 ∠SPQ = 180°

∠SPQ = 90°

Thus, PQRS is a parallelogram such that ∠SPQ = 90^{o}

Hence, PQRS is a rectangle.

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