The area of square is 16cm sq. Find the area of the square joining the mid point of the sides.

Given below shows a figure ABCD in which P, Q, R and S are the midpoints of sides AB, BC, CD and AD.

Area of square ABCD = $16 {\mathrm{cm}}^2$
So each side = AB = BC = CD = DA = $\sqrt{16}$ = 4 cm
Now AP = $\frac{1}{2}\times \mathrm{AB} = \frac{1}{2}\times 4$ = 2 cm
Similarly, AS = 2 cm
Using Pythagoras theorem in triangle SAP, we have;
$\mathrm{SP} = \sqrt{{\left(\mathrm{AP}\right)}^2+{\left(\mathrm{AS}\right)}^2} = \sqrt{2^2+2^2} = \sqrt{8} = 2\sqrt{2}$ cm
Since PQRS is a square, so we have;
each side = SP = SR = RQ = PQ = $2\sqrt{2}$ cm
So area of square PQRS = ${\left(2\sqrt{2}\right)}^2$ = $8 {\mathrm{cm}}^2$
Therefore area of the square obtained is 8 sq. cm.