since the area of a square is equal to the square of one of its side's length, then the area should be equivalent to
![x^(2)](https://img.qammunity.org/2019/formulas/mathematics/high-school/x18wowxes8du7ezs6fltwoqbv6i8lovbjg.png)
.
![A = x^(2)](https://img.qammunity.org/2019/formulas/mathematics/high-school/uarsy4wdvptdhoetb0id9lxq7zbekggcku.png)
---> equation (1)
By using pythagoras rule which states that the
![x^(2) = hyp^2 - opposite^2](https://img.qammunity.org/2019/formulas/mathematics/high-school/tyybq3n4h9ao5itqv1npirhlmvps80ha2u.png)
---> equation (2)
where the opposite side's length is 8 and the hypotenuse side's length is 10
by substituting by the values in equation (2) therefore,
![x^(2) = 10^(2) - 8^(2)](https://img.qammunity.org/2019/formulas/mathematics/high-school/jayi5wua9wznkn44uv6wx018323kz3zibq.png)
substitute this value in equation (1) then
![A = x^(2) = 10^(2) -8^(2)](https://img.qammunity.org/2019/formulas/mathematics/high-school/h9c6mvkbfc71tolso3lcvks17igcskptj3.png)
where A is the area of the square whose side is x