Question

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ABCD is a rectangle and P, Q, R and S are mid - points of all the sides AB, BC, CD and DA respectively. Show that the quadrilateral PQRS is a rhombus.


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Answer 1

Step-by-step explanation:

Data: ABCD is a rectangle and P, Q, R and S are mid-points of the sides AB, BC, CD and DA respectively. To Prove: PQRS is a rhombus. Construction : Diagonals AC and BD are drawn. Proof: In ∆ABC, P and Q are the mid-points of AD and BC. ∴ PQ || AC (Mid-point theorem) PQ = 1 2 12AC ………….. (i) Similarly, in ∆ADC, S and R are the mid-points of AD and CD. ∴ SR || AC SR = 1 2 12AC …………… (ii) Similarly, in ∆ABD, SP || BD SP = 1 2 12BD ……………….. (iii) Similarly, in ∆BCD, QR || BD QR = 1 2 12BD ……………… (iv) From (i), (ii), (iii) and (iv), PQ = QR = SR = PS and Opposite sides are parallel. ∴ PQRS is a rhombus

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Answer 2

Answer:

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ABCD is a rectangle and P, Q, R and S are mid - points of all the sides AB, BC, CD and DA respectively. Show that the quadrilateral PQRS is a rhombus.

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Here, we are joining A and C.

In ΔABC

P is the mid point of AB

Q is the mid point of BC

PQ∣∣AC [Line segments joining the mid points of two sides of a triangle is parallel to AC(third side) and also is half of it]

$PQ= \frac{1}{2} AC$

In ΔADC

R is mid point of CD

S is mid point of AD

RS∣∣AC [Line segments joining the mid points of two sides of a triangle is parallel to third side and also is half of it]

$RS = \frac{1}{2} AC$

So, PQ∣∣RS and PQ=RS [one pair of opposite side is parallel and equal]

In ΔAPS & ΔBPQ

AP=BP [P is the mid point of AB)

∠PAS=∠PBQ(All the angles of rectangle are 90° )

AS=BQ

∴ΔAPS≅ΔBPQ(SAS congruency)

∴PS=PQ

BS=PQ & PQ=RS (opposite sides of parallelogram is equal)

∴ PQ=RS=PS=RQ[All sides are equal]

∴ PQRS is a parallelogram with all sides equal

∴ So PQRS is a rhombus.

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