Question

ABCD is a rhombus.AC=12cm ,BD=16cm.Find the area of the quad by joining the midpoints of sides of rhombus ABCD

Answer 1

Solution :-

Given -

ABCD is a rhombus with AC = 12 cm and BD =16 cm. 

P, Q, R and S are the mid-points of the sides AB, BC, CD and AD respectively. 

We have to find the area of the quadrilateral PQRS formed by joining the mid-points of the rhombus ABCD

In Δ ABD, S is the mid-point of AD and P is the mid-point of AB.

⇒ SP || BD

And, SP = 1/2 of BD (∴ Mid-point theorem)

⇒ 1/2*16

= 8 cm

So, SP = 8 cm

In Δ BCD, R is the mid-point of CD and Q is the mid-point of BC.

⇒ QR || BD and QR = 1/2 of BD

⇒ 1/2*16 

= 8 cm

So, QR is 8 cm

SP || BD and QR || BD ⇒ SP || QR

Also, SP = QR = 8 cm

⇒ SPQR is a parallelogram.

Similarly,  PQ = SR = 1/2 of AC

⇒ 1/2*12

= 6 cm

So, PQ = SR = 6 cm

Now,

BD || SP ⇒ SN || OM

AC || RS ⇒ ON || MS

⇒ NOMS is a parallelogram.

∠ AOD = 90° (∴ Diagonals of a parallelogram bisect at right angle)

⇒ ∠ MSN = 90° (∴ NOMS is a parallelogram)

⇒ PQRS is a rectangle.

Area of rectangle PQRS = SP × PQ

⇒ 8 × 6 

= 48 sq cm

So, area of the quadrilateral formed by joining the mid-points of the sides of

rhombus ABCD is 48 sq cm.

Answer.