Question

QUESTION

➡️Show that quadrilateral formed by joining the mid points of the pair of adjacent sides of a rectangle is a rhombus.

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Answer 1

Answer:

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.

i hope you understand bye

Answer 2

Step-by-step explanation:

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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= 1/2

AC [By midpoint theorem]

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

SR∣∣AC and SR= 1/2

AC [By midpoint theorem]

Now, PQ∣∣AC and SR∣∣AC and PQ∣∣SR

Also, PQ=SR [Each equal to

1/2 AC] . . . . . . . (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.

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