Question

The figure formed by joining the mid-points of the adjacent sides of a square is a

Options

• rhombus

• square

• rectangle

• parallelogram

Answer 1

We get a square by joining the mid-points of the sides of a square.

It is given a square ABCD in which P, Q, R and S are the mid-points AB, BC, CD and DArespectively.

PQ, QR, RS and SP are joined.

Join AC and BD.

In , ΔABC P and Q are the mid-points AB and BC respectively.

Therefore,

PQ || AC and ${\displaystyle PQ=\frac{1}{2}AC}$……(i)

Similarly, In ΔADC, R and S are the mid-points CD and AD respectively.

Therefore,

 SR || AC and${\displaystyle SR=\frac{1}{2}AC}$……(ii)

From (i) and (ii), we get

PQ || SR and PQ = SR

Therefore, PQRS is a parallelogram. …… (iii)

Now ABCD is a square.

Therefore,

       AB = CD

${\displaystyle \frac{1}{2}AB=\frac{1}{2}CD}$

PB = RC …… (iv)

Similarly,

BQ = CQ …… (v)

In ΔPBQ and ΔRCQ, we have:

PB = RC (From equation (iv))

∠PBQ =∠RCQ (Each is a right angle)

BQ = CQ (From equation (v))

So, by SAS congruence criteria, we get:

ΔPBQ ≅ ΔRCQ

By Corresponding parts of congruent triangles property we have:

PQ = QR…… (vi)

From (iii) and (vi) we obtain that PQRS is a parallelogram such that PS = PQ.

But, PQRS is a parallelogram.

QR = PS

So,PQ =QR

and  RS = PS…… (vii)

Now, PQ || AC (From equation (i))

Therefore,

 PM || NO…… (viii)

SinceP and S are the mid-points AB and AD respectively

PS || BD 

Therefore,

 PN || MO …… (ix)

Thus, in quadrilateral PMON, we have:

 PM || NO

and PN || MO (From equation (viii) and (ix))

Therefore, quadrilateral PMON is a parallelogram.

Also,

∠MPN = ∠MON

∠MPN = ∠BOA (Because ∠MON = ∠BOA)

∠MPN = 90° (Because diagonals of a square are perpendicular)

∠QPS = 90°

Therefore,PQRS is a quadrilateral such that PQ = QR ,

RS = QR and RS = PS .Also, ∠QPS = 90°.

Hence, PQRS is a square.

Hence the correct choice is (b).