Question

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

Answer 1

Answer:

Hi mate

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=

2

1

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=

2

1

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

o

)

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.

Answer 2

Given :

• ABCD is a rectangle.
• P,Q,R and S are the mid-points of AB,BC,CD and DA respectively.
• PQ,QR,RS and SP are joined.

To prove :

• Quadrilateral PQRS is a rhombus.

Construction :

• Join AC

Proof :

In ∆ABC,

∵ P and Q are the mid points of AB and DC respectively.

∴ PQ||AC and PQ = 1/2 AC ...(1)

In ∆ADC,

∵ S and R are the mid points of AD and DC respectively.

∴ SR||AC and SR = 1/2 AC ...(2)

From(1) and (2)

PQ||SR and PQ = SR ....(3)

∴Quadrilateral PQRS is a parallelogram.

In rectangle ABCD,

AD = BC (Opposite sides of a rectangle are equal)

:$\implies{\sf{\frac{1}{2}AC =\frac{1}{2}BC}}$

(Halves of equals are equal.)

:$\implies{\sf{ AS = BQ}}$

In ∆APS and ∆BPQ,

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

AS = BQ ( Proved above )

∠PAS = ∠PBQ ( Each 90°)

∴ ∆APS ≅ ∆BPQ. (SAS congruency Axiom)

∴ PS = PQ (C.P.C.T) .....(4)

In view of (3) and (4), PQRS is a rhombus.