Question

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Answer 1

Answer:

in your face

Step-by-step explanation:

Answer 2

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ABCD is rhombus and P, Q, R and S are the midpoint of the sides AB, BC, CD and DA respectively. show that the quadrilateral PQRS is a rectangle.

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Given :-

• ABCD is a rhombus.
• P, Q, R and S are the midpoint of the sides AB, BC, CD and DA respectively.

To show :-

• The PQRS quadrilateral is a rectangle.

Construction:-

• Join A & C

First we will prove PQRS is parallelogram, since parallelogram with one angle 90° is rectangle.

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In ΔABC, P is mid point of AB, Q is mid point of BC.

By mid point theorem,

$\mathtt{ \therefore \: PQ{ \parallel}AC \: \: \: and \: \: \: PQ = \frac{1}{2} AC} \: \: ........(1)$

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In ΔADC, R is the mid-point of CD, S is mid point AD.

By mid point theorem,

$\mathtt{ \therefore \: RS \parallel AC \: \: and \: \: RS = \frac{1}{2} AC}$

From (1) and (2)

In PQRS, one pair of opposite side is parallel and equal.

Hence, PQRS is a parallelogram.

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Now we have to prove PQRS is a rectangle,

$\mathtt{Since \: \: AB = BC \implies \frac{1}{2} AB = \frac{1}{2} BC}$

So PB=BQ

Now, In ΔBPQ, PB=BQ

∴∠2 = ∠1 ………(3)

In ΔAPS & ΔCQR

⇒ AP=CQ

⇒ AS=CR

⇒ PS=QR

∴ ΔAPS≅ΔCQR

⇒ ∠3=∠4 by CPCT …………(4)

Now

AB is a line

So, ∠3+∠SPQ1=180° ……….(5)

Similarly for line Bc

∠2+∠PQR+∠4=180°

∠1+∠PQR+∠3=180° ……….(6)

From 5 & 6

∠1+∠SPQ+∠3=∠1+∠PQR+∠3

∴∠SPQ=∠PQR ……….(7)

Now, PS=QR, and PQ is a transversal

So, ∠ SPQ+∠PQR=180°

∠SPQ+∠SPQ=180. 《 from 7》

2∠SPQ=180°

${ \bold{\color{green} \boxed{∠SPQ=90°}}}$

So, PQRS is a parallelogram with one angle 90°

∴ PQRS is a rectangle.

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