Ar. of a square

Let $\displaystyle L$ be the length of a side of the square.Then length of its diagonal is $\displaystyle L\sqrt { 2 }$ and its[square's] area = $\displaystyle { L}^{ 2 }$

Now, given $\displaystyle L\sqrt { 2 } =a+b$

$\Rightarrow { 2L }^{ 2 }={ \left( a+b \right) }^{ 2 }\\ \Rightarrow { L }^{ 2 }=area=\frac { { \left( a+b \right) }^{ 2 } }{ 2 }$ Tada!

Side of square is $a+b$ \times\sin 45 So, the area is side\times side. $a+b$ ^{2}\times $\sin 45$ ^{2}

Suppose a square with size $a$ and $b$ ; the diagonal $d$ equals to $\sqrt{a^{2}+b^{2}}$ . The area $A$ of the square is $a^2$ or $b^2$ because $a=b$

$d^2 = a^2 + b^2 = 2A \rightarrow \frac{d^2}{2} = A$

Now in this problem:

$d=a+b$ , $\frac{d^2}{2} = A \rightarrow \frac{(a+b)^2}{2}=A$

By using pythagorus theorm, let us consider x be the side of square. given that the length of a diagonal is a+b. hence by pythagorus theorem. side^{2} + side^{2} =diagonal^{2} . x^{2} + x^{2} = (a+b)^{2}. 2x^{2} =(a+b)^{2} area of the square is x^{2}. hence x^{2}= (a+b)^{2} /2