Question

The ratio of side and a diagonal of a square is:
(a) $1:\sqrt{2}$
(b) $3:\sqrt{2}$
(c) $\sqrt{2}:1$
(d) $\sqrt{2}:3$

Answer 1

Hint: Start by considering a square ABCD with the length of its sides equal to a unit. Then draw the diagonal AC and let its length be l units. As all the angles of a square are right angles, triangle ABC is a right angled triangle with base and perpendicular equal to a unit and hypotenuse l. Now use the Pythagoras theorem to get the relation between l and a, hence, the required ratio.

Complete step by step solution:
Let us start by taking a square ABCD with each of its sides measuring a unit. Also, we let the length of the diagonals to be l units.
Now let us draw the diagram of the square ABCD with all the assumed lengths for better visualisation.

As we know that all the sides of a square are equal and each of the interior angle measures $90{}^\circ $ , we can say that $\Delta ABC$ is a right isosceles triangle with measure of the base and perpendicular equal to a units and AC being the hypotenuse, i.e., hypotenuse is equal to l units.
Now applying the Pythagoras theorem for right angled triangles, we get
${{\left( Base \right)}^{2}}+{{\left( Perpendicular \right)}^{2}}={{\left( Hypotenuse \right)}^{2}}$
$\Rightarrow A{{B}^{2}}+B{{C}^{2}}=A{{C}^{2}}$
$\Rightarrow {{a}^{2}}+{{a}^{2}}={{l}^{2}}$
$\Rightarrow 2{{a}^{2}}={{l}^{2}}$
Taking square root of both the sides of the equation will give us:
$l=\sqrt{2}a\text{ }$
Now, we have got l in terms of a. Now, we have to find the ratio of side and diagonal of the square, i.e., ratio of a and l. So we will rearrange the terms in the above equation such that we get $\dfrac{a}{l}$ . So, now we will get
$\Rightarrow \dfrac{a}{l}\text{=}\dfrac{1}{\sqrt{2}}\text{ }$
Therefore, the ratio of side and a diagonal of a square is $1:\sqrt{2}$ . Hence, the answer to the above question is option (a).

Note: You need to remember the properties like the square is a quadrilateral with all sides equal and each interior angle equal to $90{}^\circ $ . Also, remember that $2{{a}^{2}}={{l}^{2}}$ actually implies $l=\pm \sqrt{2}a$ but we have considered only the positive case as the length of a line segment is always positive. Also, don’t get confused and report the ratio of length of the diagonal to the length of the side making option (c) your answer, which is completely wrong.