Question

prove that the figure obtained by joining the midpoints of rectangle is rhombus​

Answer 1

Qᴜᴇsᴛɪᴏɴ -

Prove that the figure obtained by joining the midpoints of the adjacent sides of a rectangle is a rhombus.

Gɪᴠᴇɴ -

ABCD is a rectangle in which the mid points PQRS of the adjacent sides bisect AB, BC, CD, DA respectively.

Tᴏ ᴘʀᴏᴠᴇ -

PQRS is a rhombus.

Cᴏɴsᴛʀᴜᴄᴛɪᴏɴ -

Join AC.

Pʀᴏᴏғ -

In ∆ ABC,

PQ || AC

∴ PQ = 1/2 AC. (mid point theorem)

=> AC/2 _____(i)

Similarily,

SR || AC

∴ SR = 1/2 AC. (mid point theorem)

=> AC/2 _____(ii)

∴ PQRS is a parallelogram.

From eq. (i) and (ii)

PQ = SR. (AC/2 is common).

Now,

In ∆ SAP and ∆ QBP,

SA = QB. (midpoint of AD and BC)

AP = PB. ( midpoint of AB)

∠SAP = ∠QBP. (90°)

∴ ∆ SAP ≅ ∆ QBP. (S.A.S)

∴ SP = QP. (C.P.C.T)

Now,

SP=RQ (oppo. sides of parallelogram are equal)

PQ = SR (proved above)

PS = PQ (proved above)

PS = RQ (prove above)

∴ PQRS is a rhombus.

____________________proved____

@MissSolitary ✌️