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 the quadrilateral PQRS is a rhombus.
​

Answer 1

Answer:

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.

pls mark me brainiest

Answer 2

$\LARGE{\underline{\frak{\purple{Answєr :}}}}$

Answєr :

Data : ABCD is a rhombus and P, Q, R and S are the mid-points of the sides AB, BC, CD and DA respectively.

To prove : PQRS is a rectangle.

Construction : Diagonals AC and BD are drawn.

Proof : To prove PQRS is a rectnagle, one of its angle should be right angle.

In ∆ADC, S and R are the mid points of AD and DC.

∴ SR || AC

SR = 1212AC (mid-point formula)

In ∆ABC, P and Q are the mid points AB and BC.

∴ PQ || AC PQ = ½AC.

g ∴ SR || PQ and SR = PQ

$\small{\boxed{\sf{{∴ PQRS is a parallelogram}}}}$

But diagonals of a rhombus bisect at right angles. 90° angle is formed at ‘O’.

∴ ∠P = 90°

∴ PQRS is a parallelogram, each of its angle is right angle.

This is the property of rectangle.

$\small{\boxed{\sf{{∴ PQRS is a rectangle}}}}$

Hᴏᴘᴇ Tʜɪs Hᴇʟᴘs Yᴏᴜ