Question

show that the quadrilateral formed by joining the midpoint of the consecutive side of a rectangle is a rhombus.

Answer 1

example no. 13 is your answer

Answer 2

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:

$\mathrm{PQ}=\frac{1}{2} \times A C$  ---- 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,

$\mathrm{SR}=\frac{1}{2} \times A C$  ----- 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

$\begin{array}{l}{\text { Therefore } A D=B C} \\\\ {\frac{1}{2} \times A D=\frac{1}{2} \times B C} \\\\ {A S=B Q}\end{array}$

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

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