Get Answers to all your Questions

header-bg qa

Show that the quadrilateral formed by joining the mid-points of the consecutive sides of a square is also a square.
Hint:  Prove all the sides are equal and diagonals are equal 

Answers (1)

best_answer

Solution.
Given: In a square ABCD; P, Q, R and S are the mid-points of AB, BC, CD and DA.
To Prove: PQRS is a square
Construction: Join AC and BD

Proof : Here ABCD is a square
\ AB = BC = CD = AD
P, Q, R, S are the mid-points of AB, BC, CD and DA.
In
\triangle ADC,
SR\parallel AC

SR=\frac{1}{2}AC  {using mid-point theorem} …..(1)
In
\triangle ABC,
PQ\parallel AC

PQ=\frac{1}{2}AC {using mid-point theorem} …..(2)
From equation 1 and 2
SR\parallel PQ  and SR= PQ=\frac{1}{2}AC   
 …..(3)
Similarly, SP\parallel BD  and BD\parallel RQ

\ SP=\frac{1}{2}BD and  RQ=\frac{1}{2}BD   {using mid-point theorem}
 SP=RQ=\frac{1}{2}BD
Since diagonals of a square bisect each other at right angles.
 AC = BD
\Rightarrow SP=RQ=\frac{1}{2}AC …..(4)
From equation 3 and 4
SP = PQ = SP = RQ
{All sides are equal}
In quadrilateral OERF
OE\parallel FR and OF\parallel ER

\angle EOF=\angle ERF=90^{\circ}
In quadrilateral RSPQ
\angle SRQ=\angle ERF=\angle SPQ=90^{\circ} 
 {Opposite angles in a parallelogram are equal}

\angle RSP + \angle SRQ + \angle RQP + \angle QRS = 360^{\circ}
90^{\circ} +90^{\circ}+ \angle RSP + \angle RQP = 360^{\circ}
\angle RSP + \angle RQP = 180^{\circ}
\angle RSP = \angle RQP = 90^{\circ}
Since all the sides are equal and angles are also equal, so PQRS is a square.
Hence Proved

Posted by

infoexpert26

View full answer