Question

P, Q, R, S are respectively the midpoints of the sides AB, BC, CD and DA of || gm ABCD. Show that PQRS is a parallelogram and also show that
$\mathrm{ar}\left(\left|\right|\mathrm{gm}PQRS\right)=\frac{1}{2}\times \mathrm{ar}\left(\right||\mathrm{gm}ABCD)$.

Answer 1

Given: ABCD is a parallelogram and P, Q, R and S are the midpoints of sides AB, BC, CD and DA, respectively.
To prove: ar(parallelogram PQRS ) = $\frac{1}{2}$ × ar(parallelogram ABCD )
Proof:
In ∆ABC, PQ || AC and PQ = $\frac{1}{2}$ × AC [ By midpoint theorem]
Again, in ∆DAC, the points S and R are the mid points of AD and DC, respectively.
∴ SR || AC and SR = $\frac{1}{2}$ × AC [ By midpoint theorem]
Now, PQ || AC and SR || AC
⇒ ​PQ || SR
Also, PQ = SR = $\frac{1}{2}$ × AC
∴ PQ || SR and PQ = SR
Hence, PQRS is a parallelogram.

Now, ar(parallelogram​ PQRS) = ar(∆PSQ) + ar(∆SRQ) ...(i)
also, ar(parallelogram ABCD) = ar(parallelogram ABQS) + ar(parallelogram SQCD) ...(ii)

∆PSQ and parallelogram ABQS are on the same base and between the same parallel lines.
So, ar(∆PSQ ) =$\frac{1}{2}$ × ar(parallelogram ABQS) ...(iii)
Similarly, ∆SRQ and parallelogram SQCD are on the same base and between the same parallel lines.
So, ar(∆SRQ ) = $\frac{1}{2}$ × ar(parallelogram SQCD) ...(iv)
Putting the values from (iii) and (iv) in (i), we get:
ar(parallelogram​ PQRS) = $\frac{1}{2}$ × ar(parallelogram ABQS)​ + $\frac{1}{2}$ × ar(parallelogram SQCD)
From (ii), we get:
ar(parallelogram​ PQRS) = $\frac{1}{2}$ × ar(parallelogram ABCD)