show that the quadrilateral formed by joining the mid points of the pairs of the adjacent sides of a rectangle is a rhombus.
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Answers
Answer:
The quadrilateral formed by joining the midpoint of the consecutive side of a rectangle is a rhombus is proved
Solution:
The figure is attached below
A rectangle ABCD in which P, Q, R, S are the midpoints of the sides AB, BC, CD and DA respectively
PQ, QR, RS and SP are joined
To prove: PQRS is a rhombus
Construction: Join AC
Proof:
In Triangle ABC, P and Q are the mid-points of sides AB and BC respectively
So, PQ is parallel to AC and also we get:
\mathrm{PQ}=\frac{1}{2} \times A CPQ=
2
1
×AC ---- eqn 1
In Triangle ADC, R and S are the mid-points of sides CD and AD respectively
So, SR is parallel to AC and we get,
\mathrm{SR}=\frac{1}{2} \times A CSR=
2
1
×AC ----- eqn 2
From (1) and (2) , we can say PQRS is a parallelogram since opposites sides are parallel to each other and are also Equal
That is, PQ ll SR and PQ = SR ----- eqn 3
\begin{gathered}\begin{array}{l}{\text { Therefore } A D=B C} \\\\ {\frac{1}{2} \times A D=\frac{1}{2} \times B C} \\\\ {A S=B Q}\end{array}\end{gathered}
Therefore AD=BC
2
1
×AD=
2
1
×BC
AS=BQ
AS = BQ ---- eqn 4
In triangle APS and BPQ, we have
AP = BP [ P is the midpoint of AB]
Angle PAS = angle PBQ [ each equal to 90 degre]
and AS = BQ [ from eqn 4]
The side angle side theorem states that two triangles are equal if two sides and the angle between those two sides are equal.
So by SAS criterion of congruence, we have
angle APS = angle BPQ
By CPCT theorem,
“Corresponding Parts of Congruent Triangles” states that if we take two or more triangles which are congruent to each other then the corresponding angles and the sides of the triangles are also congruent to each other i.e., their corresponding parts are equal to each other.
PS = PT ---- eqn 5
From eqn 3 and eqn 5 we obtain that PQRS is a parallelogram such that PS = PQ .i.e two adjacent sides are equal
Hence PQRS is a rhombus
Learn more about quadrilateral
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