Question

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$\rightarrow$ ABCD is a rectangle 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

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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.

Step-by-step explanation:

Hey Dear! here is your answer

Answer 2

Answer:

PQ = QR = SR = PS

Step-by-step explanation:

In ∆ ABC

P in mid - point of AB & Q is mid - point of BC

•: PQ || AC and PQ = 1/2 AC........(1)

In ∆ ADC

Ris mid - point of CD & S is mid - point of AD.

•: RS || AC and RS = 1/2 AC......(2)

from equation (1) & (2)

PQ|| RS and PQ = RS

since in quadrilateral PQRS, one pair of opposite sides is equal and parallel to each other, it is a parallelogram.

•: PS||QR and PS = QR

( opposite sides of parallelogram).....(3)

In ∆ BCD, Q and R are the mid - point of sides BC and CD respectively point theorem) ..... (4)

•: QR|| BD and QR = BD ( mid - point theorem) ...(4)

however, the diagonal of a rectangle are equal..

•: AC = BD.....(5)

by using equation (1),(2),(3),(4) ,(5) , we obtain.

PQ = QR = SR = PS..

I hope it will help you