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

A. Rhombus 

B. Square 

C. Rectangle 

D. Parallelogram

1 Answer

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Best answer

Option : (B)

Let ABCD is a square such that AB = BC = CD = DA, AC = BD and P, Q, R and S are the mid points of the sides AB, BC, CD and DA respectively.

In ΔABC,

P and Q are the mid-points of AB and BC respectively.

Therefore,

PQ ‖ AC and PQ = \(\frac{1}{2}\)AC (Mid-point theorem) ..(i)

Similarly,

In ΔADC,

SR ‖ AC and SR = \(\frac{1}{2}\)AC (Mid-point theorem) ..(ii)

Clearly, 

PQ ‖ SR and PQ = SR 

Since, 

In quadrilateral PQRS, 

One pair of opposite sides is equal and parallel to each other. 

Hence, 

It is a parallelogram. 

Therefore, 

PS ‖ QR and PS = QR 

(Opposite sides of a parallelogram) ...(iii)

In ΔBCD,

Q and R are the mid-points of sides BC and CD respectively 

Therefore, 

QR ‖ BD and QR = \(\frac{1}{2}\)BD (Mid-point theorem) ...(iv)

However, 

The diagonals of a square are equal.

Therefore,

AC = BD ...(v) 

By using equation (i), (ii), (iii), (iv) and (v), we obtain 

PQ = QR = SR = PS 

We know that, 

Diagonals of a square are perpendicular bisector of each other. 

Therefore, 

∠AOD = ∠AOB = ∠COD = ∠BOC = 90°

Now, 

In quadrilateral EHOS, we have 

SE || OH 

Therefore, 

∠AOD + ∠AES = 180°

(Corresponding angle) 

∠AES = 180°- 90° =  90° 

Again, 

∠AES + ∠SEO = 180°(Linear pair) 

∠SEO = 180° - 90° = 90° 

Similarly, 

SH || EO 

Therefore, 

∠AOD + ∠DHS = 180°

(Corresponding angle) 

∠DHS = 180°- 90° = 90° 

Again, 

∠DHS + ∠SHO = 180° 

(Linear pair) 

∠SHO = 180° - 90° = 90° 

Again, 

In quadrilateral EHOS, we have 

∠SEO = ∠SHO = ∠EOH = 90° 

Therefore, 

By angle sum property of quadrilateral in EHOS, we get 

∠SEO + ∠SHO + ∠EOH + ∠ESH = 360° 

90° + 90° + 90° + ∠ESH = 360° 

∠ESH = 90° 

In the same manner, 

In quadrilateral EFOP, FGOQ, GHOR, we get 

∠HRG = ∠FQG = ∠EPF = 90° 

Therefore, 

In quadrilateral PQRS, we have 

PQ = QR = SR = PS and ∠ESH = ∠HRG = ∠FQG = ∠EPF = 90° 

Hence, 

PQRS is a square.

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