The Diagonal of a Square

Relevant wiki: Pythagorean Theorem

$A=s^2=18$

$x=\sqrt{s^2+s^2}=\sqrt{2s^2}$

but: $s^2=18$

now we substitute

$x=\sqrt{2(18)}=\sqrt{36}=6$

Relevant wiki: Area of Figures

$Area \space of \space square = d^2/2 (where \space diagonal "d" is \space given) \\ d^2/2=18 => d^2=36 \\ therefore \space d=6 \space which \space is \space the \space required \space result.$

$d^2 = 18$

$d= 3 * 2^.5$

$x= 2^.5 * d = 3 *2 =6$

$A = s^2$

$s = \sqrt{A}$

$x = s\sqrt{2} = \sqrt{2A} = \sqrt{2 \times 18} = \sqrt{36} = 6$

We know that the are of a square is a^2 Hence , a=18^1/2 So a^2+a^2=x^2 18+18=x^2 36=x^2 6=x

Area of triangle=1/2 [diagonal]^2

Direct way: consider the four right triangles the interesecting diagonals make. $\frac{4}{2} \cdot ({\frac{x}{2}})^2 =18$

$x^2= 36 \Rightarrow x=6$