Question

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

Answer 1

Step-by-step explanation:

Some of the properties of a rhombus :

1) all sides are congruent, ⇒ A B = B C = C D = D A ,

2) opposite angles are congruent, ⇒ ∠ A D C = ∠ A B C = y , and ∠ B A D = ∠ B C D = x ,

3) adjacent angles are supplementary, ⇒ x + y = 180 ∘

4) opposite sides are parallel, ⇒ A D // B C , and A B // D C ,

given that P , Q , R and S are midpoints of A B , B C , C D , and D A , respectively,

⇒ A P = P B = B Q = Q C = C R = R D = D S = S A Consider Δ A P S , as A P = A S , ⇒ Δ A P S is isosceles, ⇒ ∠ A S P = ∠ A P S = w , ⇒ x + 2 w = 180 ∘ − − − − − E q ( 1 )

Consider Δ B P Q , as B P = B Q , ⇒ Δ B P Q is isosceles, ⇒ ∠ B P Q = ∠ B Q S = z ⇒ y + 2 z = 180 ∘ − − − − − E q ( 2 )

E q ( 1 ) + E q ( 2 ) = ( x + y ) + 2 ( w + z ) = 360 ∘ ⇒ 180 + 2 ( w + z ) = 360 ∘

⇒ w + z = 90 ∘

⇒ ∠ S P Q = 180 − ( w + z ) = 180 − 90 = 90 ∘ Similarly, ∠ P Q R = ∠ Q R S = ∠ R S P = 180 − ( w + z ) = 90 ∘ Hence, P Q R S is a rectangle.

Answer 2

Given:- AB=BC=CD=AD

TO PROVE:- PQRS is a rectangle

SOLVE:-

In ∆ABC, P and Q are the midpoints of the sides AB and BC.

- PQ//AC and PQ=AC (line segments joining the midpoints of two sides of a ∆ is // to the third side)-(1)

In ∆ADC, R & S are the midpoints of the sides CD & AD.

- RS//AC & RS=AC (line segments joining the midpoints of two sides of a ∆ is // to the third side) - (2)

From (1) & (2), PQ//RS & PQ=RS.

In ∆DAB, P & S are the midpoints of AB and AD.

- PS//BD & PS=BD (line segments joining the midpoints of the sides of a ∆ is // to the third side) - (3)

In ∆BCD, Q & R are the midpoints of the sides BC & CD.

- QR//BD & QR=BD (line segments joining the midpoints of the sides of a ∆ is // to the third side) - (4)

From (3) and (4), PS//QR & PS=QR.

Hence, PQRS is a //gm, has both pairs of opposite sides are equal.

Now, angle Q= 90°

THUS, PQRS is a rectangle.