### Theory:

A square can be constructed using a few known measurements.

**The measurements are**:

**1.**When the side is known.

**2.**When the diagonal is known

Method \(1\): When the side is known

Let us construct a square with \(5\) \(cm\) as its side. Let us also find the area of the square obtained.Step \(1\): Draw a rough diagram for the measurements given.

**\(2\): Draw a line segment \(AB\) of length \(5\) \(cm\).**

**Step**Step \(3\): With \(A\) as centre, draw a perpendicular line.

Step \(4\): With \(A\) as centre and with \(5\) \(cm\) as radius, draw an arc on the perpendicular line. Mark the intersection as \(D\).

Step \(5\): With \(D\) as centre and with \(5\) \(cm\) as radius, draw an arc. Similarly, with \(B\) as centre and with \(5\) \(cm\) as radius, cut the existing arc. Mark the intersection as \(C\).

Step \(6\): Join \(BC\) and \(CD\) to form the desired square.

**To find the area of the square**:

\(\text{Area of a square} = \text{Side} \times \text{Side}\)

\(\text{Area of a square} = \text{Side}^2\)

Here, \(\text{Side} = 5\) \(cm\).

Therefore, \(\text{Area of a square} = 5^2\)

\(= 25\) \(cm^2\)

Method \(2\): When the diagonal is known

Let us construct a square with one of its diagonals as \(10\) \(cm\). Let us also find the area of the square obtained.

Step \(1\): Draw a rough diagram with the measurements known.

Step \(2\): Draw a line segment \(AC\) of length \(10\) \(cm\).

Step \(3\): Draw a perpendicular bisector to \(AC\) such that the bisector intersects \(AC\) at \(O\).

Step \(4\): With \(O\) as centre and with \(5\) \(cm\) as radius, draw arcs on both sides of the perpendicular bisector. Mark the intersections as \(B\) and \(D\).

Step \(5\): Join \(AB\), \(BC\), \(CD\), and \(AD\) to form the desired square.

**To find the area of the square**:

\(\text{Area of a square} = \text{Side} \times \text{Side}\)

\(\text{Area of a square} = \text{Side}^2\)

Here, side is unknown.

Therefore, we should measure the length of the side manually.

On measuring, we found that, \(\text{Side}\) \(=\) \(7.1\) \(cm\)

Now, \(\text{Area of a square} = 7.1^2\)

\(= 50.41\) \(cm^2\)