The square

d1

The area of the square is: $$A=l^2$$

Its perimeter is: $$P=4 \cdot l$$

And, applying the Pythagorean theorem, its diagonal is: $$d=\sqrt{2 \cdot l^2}= l\cdot \sqrt{2}$$

Calculate the area of a square which diagonal is $3$ cm.

$$ 3^2=l^2+l^2=2 \cdot l^2 $$ $$l^2=\frac{9}{2}$$ $$l=\frac{3}{\sqrt{2}} \ cm$$

$$A=\Big( \frac{3}{\sqrt{2}}\Big)^2=\frac{9}{2} \ cm^2$$

Practice exercises