Question

ABCD is a square. P, Q, R, S are the mid-points of AB, BC, CD and DA respectively. By joining AR, BS, CP, DQ, we get a quadrilateral which is a 

(a) trapezium 

(b) rectangle 

(c) square 

(d) rhombus

Answer 1

(c) Square.

The quadrilateral formed by joining AR, BS, CP and DQ is EFGH. 

Now, DDCQ ≅ DPBC as, 

CD = BC                  (sides of a square) 

∠DCQ = ∠PBC = 90° 

CQ = PB             (Half of sides of a square) 

∴ ∠CDQ = ∠BCP = a (say) 

Then, ∠DQC = ∠EQC = 90° – a 

∴ In ΔEQC, ∠CEQ = 180° – (∠QCE + ∠CQE) 

= 180° – (a + 90° – a) = 90° 

⇒ ∠FEH = 90° 

Similarly, ∠EFG = ∠FGH = ∠GHE = 90° 

⇒ Vertex angles of quadrilateral EFGH are 90° each 

⇒ EFGH is a rectangle. 

Now, ΔDRF ≅ ΔCQE ⇒ RF = EQ, DF = EC 

Following the same concept, 

BH = CE = AG = DF and 

RF = EQ = PH = GS 

∴ EF = FG          (Adjacent sides of EFGH are equal) 

∴ A rectangle whose adjacent sides are equal is a square 

⇒ EFGH is a square.