Prove that the quadrilateral formed by joining the mid points of the consecutive sides of
the rectangle is a rhombus.
Answers
Answer:
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Step-by-step explanation:
A rectangle ABCD in which P,Q,R and S are the mid-points of sides AB,BC,CD and DA respectively.
PQ,QR,RS and SP are joined.
Join AC. In ΔABC,P and Q are the mid-points of sides AB and BC respectively.
∴PQ∣∣AC and PQ=
2
1
AC .....(i) (Mid pt. Theorem)
In ΔADC,R and S are the mid-points of sides CD and AD respectively.
∴SR∣∣AC and SR=
2
1
AC ....(ii) (Mid. pt. Theorem)
From (i) and (ii), we have
PQ∣∣SR and PQ=SR
⇒PQRS is a parallelogram.
Now ABCD is a rectangle.
∴AD=BC
⇒
2
1
AD=
2
1
BC
⇒AS=BQ .....(iii)
In ΔS,APS and BPQ, we have
AP=BP (P is the mid-point of AB)
∠PAS=∠PBQ (Each is equal to 90
∘
)
and AS=BQ (From (iii))
∴ΔAPQ≅ΔBPQ (SAS)
⇒PS=PQ
∴PQRS is a parallelogram whose adjacent sides are equal.
⇒PQRS is a rhombus.
Answer:
Length=l
Breadth=b
Let the diagonal intersect each other at O
As it can be seen from image
Side EH=EF=FG=GH=
2
l
2
+b
2Hence are all sides are congruent
Also, EO=OG=
2
b
HO=OF=
2
l
Hence the diagonal bisect each other.
And as it can be seen from the figure that they intersect at 90
∘
Hence, due to all the reasons above it is a proved that the quadilateral formed is rhombus
