Question

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ABCD is a rectangle and P,Q,R,S are mid-points of the side AB, BC,CD and DA respectively. Show that the quadrilateral PQRS is a rhombus.

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Answer 1

Answer:

Let the point P join with R and Q join with S

by joining the midpoints of the opposite sides of a rectangle it forms four rectangle which are equal both in length and breadth

Now, join the points P and Q it will be the diagonal of the small rectangle as with the other three rectangles to they also form the diagonal of the respective rectangles. Since the rectangles are equal their diagonals are also equal

therefore four diagonals are equal and from that we can say the quadrilateral formed will be a rhombus

Answer 2

Step-by-step explanation:

Given:-

• ABCD is a rectangle.
• P,Q,R,S are mid-points of the side AB, BC,CD and DA respectively.

To show :-

• PQRS is a rhombus.

Construction:-

• Join AC.

Proof :-

In ∆ABC

P and Q are mid points of sides AB and BC.

We know, The line segment joining the mid points of two sides of a triangle is parallel to the third side of triangle. [Mid point theorem]

PQ || AC

PQ = $\sf \cfrac{1}{2}$ AC ------(i)

In ∆ADC

By mid point theorem

SR || AC and,

SR = $\sf \cfrac{1}{2}$ AC -------(ii)

From equation (i) and (ii)

PQ || SR and PQ = SR -------(iii)

In Quadrilateral PQRS, it's one opposite sides PQ and SR is parallel and equal. By equation (iii).

By this, PQRS is a parallelogram

AD = BC ------(iv) [Beacuse Opposite sides of rectangle are equal]

$\sf \cfrac{1}{2}$ AC = $\sf \cfrac{1}{2}$ BC

AS = BQ

In ∆APS and ∆BPQ

AD = BP [P is mid point of side AB]

AS = BQ [As we proved it above]

∠PAS = ∠PBQ [All angles of rectangle are of 90°]

By using Congruency SAS,

∆APS is congruent to ∆BPQ.

PS = PQ ------(v) [By CPCT]

By equation (iii) and (v)

• Opposite sides of rhombus are equal and parallel.

Therefore,

PQRS is a rhombus.

Hence, Proved!!