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If diagonal of a square is 13cm, then find its side.

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Answer
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Hint:
We will use the Pythagoras theorem in the right triangle formed by the sides and diagonal of the square as: ${\text{hypotenus}}{{\text{e}}^2}{\text{ = perpendicula}}{{\text{r}}^2}{\text{ + bas}}{{\text{e}}^2}$ since all angles (vertex) of a square are of ${90^ \circ }$.

Complete step by step solution:
We need to find the length of the side of the square.
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Since all sides of the square are equal, let the sides be of length x cm each. And AC be the diagonal of the square whose length is 13cm.
We know that all the corner angles of a square are of ${90^ \circ }$, therefore, we can use Pythagoras theorem given by: ${\text{hypotenus}}{{\text{e}}^2}{\text{ = perpendicular}}{{\text{r}}^2}{\text{ + bas}}{{\text{e}}^2}$
Using Pythagoras theorem in $\vartriangle ABC$, we get
$
   \Rightarrow {\text{hypotenus}}{{\text{e}}^2}{\text{ = perpendicula}}{{\text{r}}^2}{\text{ + bas}}{{\text{e}}^2} \\
   \Rightarrow {\left( {AC} \right)^2} = {\left( {AB} \right)^2} + {\left( {BC} \right)^2} \\
 $
Substituting the values of AC, AB and BC, we get
$
   \Rightarrow {\left( {13} \right)^2} = {\left( x \right)^2} + {\left( x \right)^2} \\
   \Rightarrow 169 = 2{x^2} \\
   \Rightarrow {x^2} = \dfrac{{169}}{2} \\
   \Rightarrow x = \sqrt {\dfrac{{169}}{2}} = \dfrac{{13}}{{\sqrt 2 }} = \dfrac{{6.5 \times 2}}{{\sqrt 2 }} = 6.5\sqrt 2 {\text{cm}} \\
 $

Therefore, the side of the square is of length $6.5\sqrt 2 {\text{cm}}$.


Note:
In this question, you can get confused in the calculation of x from the equation obtained after using the Pythagoras theorem. You can calculate it directly by taking the square root of $\dfrac{{169}}{2} = 84.5$for the value of x instead of converting it into the form of $6.5\sqrt 2 $. You can also solve this question by directly applying the formula of relation between the diagonal of the square and its sides given as: Diagonal = $\sqrt 2 a$ , where a is the length of the side of the square, instead of using the Pythagoras theorem. This method will be helpful in MCQs to conserve time and get solutions quickly.