Question

What is the area of a square whose diagonals are $12\,cm$?

Answer 1

Hint: In the question we have to find the area of the square whose diagonal is of $12\,cm$. We know that if $x$ is the length of the side of the square, then the length of the diagonal is equal to $\sqrt 2 x$. First, we will find the length of the side of the square then we will calculate the area of the square. As area of square$ = $${(side)^2}$.

Complete step by step answer:
A square is a two-dimensional figure with four equal sides and all four angles are adequate to $90^\circ $. The diagonals of the square are equal and also they bisect each other at $90^\circ $.

Now, let us assume a square ABCD whose length of diagonal is equal to $12\,cm$.
We know that if $x$ is the length of the side of the square, then the length of the diagonal is equal to $\sqrt 2 x$.
Let us assume the length of the square is equal to $x$
It is given that length of the diagonal is equal to $12\,cm$.
So, $\sqrt 2 x = 12$
By using cross multiplication, we get
$ \Rightarrow x = \dfrac{{12}}{{\sqrt 2 }} \ldots \ldots (1)$
We know that area of square$ = {(side)^2} = {x^2}$
Substitute the value of $x$ from equation $(1)$. We get,
Area of square $ = $${\left( {\dfrac{{12}}{{\sqrt 2 }}} \right)^2}$
Area of square $ = $$\dfrac{{144}}{2}$
Therefore, area of square $ = 72\,c{m^2}$
Hence, the area of the given square is equal to $72\,c{m^2}$.

Note: One should always remember that if the length of the side of the square is $x\,cm$ then the length of the diagonal is equal to $\sqrt 2 x\,cm$. Note that the length of the diagonal is greater than the sides of the square. The diagonal for the square when divided, comes out as two similar isosceles triangles.