Question

•another question from chapter (quadrilaterals)...
pls solve the above question....

Answer 1

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The solution of your question is in attached photo. ☺

Thanks ♥♥

Answer 2

Here is your solution

In ΔABC, P and Q are the mid-points of sides AB and BC respectively.

∴ PQ || AC and PQ =AC (Using mid-point theorem) ... (1)

In ΔADC,

R and S are the mid-points of CD and AD respectively.

∴ RS || AC and RS =AC (Using mid-point theorem) ... (2)

From equations (1) and (2), we obtain

PQ || RS and PQ = RS

Since in quadrilateral PQRS, one pair of opposite sides is equal and parallel to each other, it is a parallelogram.

Let

the diagonals of rhombus ABCD intersect each other at point O.

In quadrilateral OMQN,

MQ || ON (PQ || AC)

QN || OM (QR || BD)

Therefore, OMQN is a parallelogram.

⇒ ∠MQN = ∠NOM

⇒ ∠PQR = ∠NOM

However, ∠NOM = 90° (Diagonals of a rhombus are perpendicular to each other)

∴ ∠PQR = 90°

Clearly, PQRS is a parallelogram having one of its interior angles as 90º.

Hence, PQRS is a rectangle.