Question

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

★ Given :-

• ABCD is a rhombus.
• P,Q,R,S are the mid-poinTS of AB, BC, CD,DA respectively.
• PQ,PR,RS and SP are joined.

★To Prove :-

• PQRS is a rectangle.

★Construction :-

• Join AC and BD.

★Proof :-

To proof PQRS is Rectangle . we need to show that the all angles of PQRS should be 90° .

Now, In ∆ABC , P and Q are the mid points AB and BC.

∴ PQ || AC

PQ = 1/2 AC ・・・(Midpoint formula)(1)

Now, In ∆ADC , S and R are the mid points AD and DC.

∴ SR || AC

SR = 1/2 AC ・・・(Midpoint formula)(2)

From, 1 and 2

PQ = SR

∴ PQ || SR

So, PQRS is a parallelogram.

In figure, The diagonals of a rhombus bisect at right angles.

∴ ∠P = 90°

So, we know that all angles of parallelogram are equal.

So, All angle are right angles.

∴PQRS is a rectangle.

-By the property.

Hence proved.

Answer 2

$\\⠀⠀\bigstar \:⠀⠀⠀\underbrace\bold{\large\underline{Given: - }}$

• ABCD is a rhombus
• P, Q, R and S are the mid-points
• AB, BC, CD and DA are the sides

$\\⠀⠀\bigstar \:⠀⠀⠀\underbrace\bold{\large\underline{R.T.P: - }}$

• PQRS is a rectangle

$\\⠀⠀\bigstar \:⠀⠀⠀\underbrace\bold{\large\underline{Construction: - }}$

• Join AC & BD

$\\⠀⠀\bigstar \:⠀⠀⠀\underbrace\bold{\large\underline{Proof: - }}$

• A rectangle is a parallelogram with one angle 90°

First we will prove PQRS is a parallelogram and prove one angle is 90°

To prove PQRS is a rectnagle, one of its angle should be right angle.

In ∆ADC,

• S and R are the mid points of AD and DC.

∴ SR || AC SR = 1212AC (mid-point formula)

In ∆ABC, P and Q are the mid points AB and BC.

∴ PQ || AC PQ = 1212AC.

∴ SR || PQ and SR = PQ

∴ PQRS is a parallelogram.

But diagonals of a rhombus bisect at right angles. 90° angle is formed at ’O’.

∴ ∠P = 90°

∴ PQRS is a parallelogram, each of its angle is right angle.

This is the property of rectangle.

∴ PQRS is a rectangle.

$\large{ \pink{\mathfrak{ \underline{\overline{\mid \:\: ❥Proved\:\: \mid}}}}}$