Question

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

Answer 1

Given:-

• ABCD is rhombus where PQRS are formed by joining mid points.

To Prove:-

• PQRS is a rectangle.

To Construct:-

• Join AC and BD.

Proof:-

•Since RQ are formed by joining mid points so, QR || BD and QR = ½ BD _(i)

•Since PS are formed by joining mid points so, PS || BD and PS = ½ BD _(ii)

From (i) and (ii) we get,

QR || PS and QR = PS

Hence a pair if side is equal and parallel so PQRS is ||gm.

Now, since QR || BD so MR || ON __(iii)

Similarly,

NR || AC __(iv) [ Because SR || AC, since SR is formed by joining mid points ]

From (iii) and (iv) we get, MONR is a ||gm.

•As we know that diagonals of rhombus bisect at 90° so < MON = 90°

•Now, since MONR is ||gm so its opposite angles will be equal. So,

< MRN = 90°

Or < QRS = 90°

Hence PQRS is such ||gm whose one if the angle ( < QRS ) is 90° so PQRS is a rectangle.

ʜᴇɴᴄᴇ ᴘʀᴏᴠᴇᴅ.

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