Question

Find the area of a square $PQRS$ formed by joining the mid-points of sides of another square $ABCD$ of side $9\sqrt{2}cm$.

Answer 1

$\Rightarrow$ $AB=BC=CD=AD=9\sqrt{2}cm$ [ Given ]

$\Rightarrow$ $AP=\frac{1}{2}AB$ [ $P$ is midpoint of $AB$ ]

$\Rightarrow$ $AP=\frac{1}{2}\times 9\sqrt{2}=\frac{9\sqrt{2}}{2}cm$

$\therefore$ $AP=AS=\frac{9\sqrt{2}}{2}$

$\Rightarrow$ $\angle PAS={90}^o$ [ Angle of square is ${90}^o$ ]

In right angled $△PAS$

$\Rightarrow$ $(PS{)}^2=(AP{)}^2+(AS{)}^2$ [ By Pythagoras theorem ]

$\Rightarrow$ $(PS{)}^2={\left(\frac{9\sqrt{2}}{2}\right)}^2+{\left(\frac{9\sqrt{2}}{2}\right)}^2$

$\Rightarrow$ $(PS{)}^2=\frac{162}{4}+\frac{162}{4}$

$\Rightarrow$ $(PS{)}^2=\frac{324}{4}$

$\Rightarrow$ $(PS{)}^2=81$

$\therefore$ $PS=9cm$

$\Rightarrow$ Area of square $PQRS=(9{)}^2=81c{m}^2$