Question

Prove that the figure obtained by joining the mid-point of the adjacent sides of a rectangle is a rhombus.

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Answer 1

Answer:

The figure is shown below:

Let ABCD be a rectangle where P,Q,R,S are the midpoint of AB,BC,CD,DA. We need to show that PQRS is a rhombus

For help we draw two diagonal BD and AC as shown in figure Where BD=AC (Since diagonal of rectangle are equal)

Proof:

From △ABD and △BCD

PS= 1/2 BD=QR and PS∥BD∥QR

2PS=2QR=BD and PS∥QR .......(1)

Similarly 2PQ=2SR=AC and PQ∥SR ........(2)

From (1) and (2) we get

PQ=QR=RS=PS

Therefore PQRS is a rhomus.

Hence proved

Step-by-step explanation:

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Answer 2

The figure is shown below:

Let ABCD be a rectangle where P,Q,R,S are the midpoint of AB,BC,CD,DA. We need to show that PQRS is a rhombus

For help we draw two diagonal BD and AC as shown in figure Where BD=AC (Since diagonal of rectangle are equal)

Proof:

From △ABD and △BCD

PS=

2

1

BD=QR and PS∥BD∥QR

2PS=2QR=BD and PS∥QR .......(1)

Similarly 2PQ=2SR=AC and PQ∥SR ........(2)

From (1) and (2) we get

PQ=QR=RS=PS

Therefore PQRS is a rhomus.

Hence proved