Question

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Show that the quadrilateral formed by joining the midpoints of the pairs of adjacent sides of rhombus is a rectangle ​

Answer 1

Question :-

Show that the quadrilateral formed by joining the midpoints of the pairs of adjacent sides of rhombus is a rectangle .

Answer :-

Consider △ ABC

We know that P and Q are the mid points of AB and BC

By using the midpoint theorem

We know that PQ || AC and PQ = ½

AC Consider △ ADC

We know that RS || AC and RS = ½

AC It can be written as PQ || RS and PR = RS = ½ AC ……. (1) Consider △ BAD

We know that P and S are the mid points of AB and AD

Based on the midpoint theorem

We know that PS || BD and PS = ½

DB Consider △ BCD

We know that RQ || BD and RQ = ½

DB It can be written as PS || RQ and PS = RQ = ½ DB …….

(2) By considering equations (1) and (2)

The diagonals intersects at right angles in a rhombus

So we get ∠ EQF = 90o

We know that RQ || DB

So we get RE || FO

In the same way SR || AC

So we get FR || OE

So we know that OERF is a parallelogram.

We know that the opposite angles are equal in a parallelogram.

So we get ∠ FRE = ∠ EOF = 90o

So we know that PQRS is a parallelogram having ∠ R = 90o

⏩ Therefore, it is proved that the quadrilateral formed by joining the midpoints of the pairs of adjacent sides of a rhombus is a rectangle.⏭️

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Answer 2

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Let ABCD be the rectangle and P,Q,R,S be the midpoints of AB,BC,CD,DA respectively.

Join AC, a diagonal of the rectangle.

In ΔABC, we have:

PQ∣∣AC and PQ=21AC [By midpoint theorem]

Again, in ΔDAC, the points S and R are the mid points of AD and DC, respectively.

SR∣∣AC and SR=21AC [By midpoint theorem]

Now, PQ∣∣AC and SR∣∣AC and PQ∣∣SR

Also, PQ=SR [Each equal to 21AC] . . . . . . . (i)

So, PQRS is a parallelogram.

Now, in ΔSAP and ΔQBP, we have:

AS=BQ,∠A=∠B=90∘andAP=BP

i.e.,ΔSAP∼ΔQBP

PS=PQ . . . . . . . . . (ii)

Similarly, ΔSDR∼ΔQCR

SR=RQ . . . . . . . . (iii)

From (i), (ii) and (iii), we have:

PQ=PS=SR=RQ

Hence, PQRS is a rhombus.