Here in the question we are given the Diagonal of a square as 25√2 cm. We are asked to find the area of the square. Finding area is simple using [tex]\sf\:(side)^{2}[/tex] formula. But here we are not given the length of the square. So what can be done !?
Yeah you are correct we will equate the formula for diagonal of a square and the given value of the diagonal. Doing so we will get the measure of side of the square and using it we will get our final answer that is area of the square . So let’s proceed !
Answer:
area of square is side square so it is the first use Pythagoras theeorm to find side then side square
[tex]\large\sf\underline{Given\::}[/tex]
[tex]\large\sf\underline{To\:find\::}[/tex]
[tex]\large\sf\underline{Creating\:a\:road\:map\:for\:the\:Solution\::}[/tex]
Here in the question we are given the Diagonal of a square as 25√2 cm. We are asked to find the area of the square. Finding area is simple using [tex]\sf\:(side)^{2}[/tex] formula. But here we are not given the length of the square. So what can be done !?
Yeah you are correct we will equate the formula for diagonal of a square and the given value of the diagonal. Doing so we will get the measure of side of the square and using it we will get our final answer that is area of the square . So let’s proceed !
[tex]\large\sf\underline{Formula\:to\:be\:used\::}[/tex]
[tex]\large\sf\underline{Solution\::}[/tex]
Diagonal of the square = [tex]\sf\:\sqrt{2} \times side[/tex]
[tex]\sf\to\:25\sqrt{2}=\sqrt{2} \times side[/tex]
[tex]\sf\to\:\frac{25\sqrt{2}}{\sqrt{2}}=side[/tex]
[tex]\sf\to\:\frac{25\cancel{\sqrt{2}}}{\cancel{\sqrt{2}}}=side[/tex]
[tex]\small\fbox\red{★\:side\:=\:25\:cm}[/tex]
Now Side of a square = 25 cm.
Let’s find the area :
[tex]\sf\:Area\:=\:(side)^{2}[/tex]
[tex]\sf\leadsto\:Area\:=\:(25)^{2}[/tex]
[tex]\small\fbox\red{★\:Area\:=\:625\:sq\:cm}[/tex]
!! Hope it helps !!