Theory:

A rectangle can be constructed using two set of measurements.
 
The two instances are:
1. When the length and breadth is known.
 
2. When a side and a diagonal is known.
Method \(1\): When the length and breadth is known
Let us construct a rectangle with length as \(10\) \(cm\) and breadth as \(4\) \(cm\). Let us also calculate the area of the rectangle thus constructed.
 
Step \(1\): Draw a rough diagram using the measurements known.
 
1324_31.svg
 
Step \(2\): Draw a line segment \(AB\) of length \(10\) \(cm\).
 
1324_30.svg
 
Step \(3\): With \(A\) as centre, draw a perpendicular line.
 
1324_32.svg
 
Step \(4\): With \(A\) as centre and with \(4\) \(cm\) as radius, draw an arc on the perpendicular line and mark the intersection as \(D\).
 
1324_33.svg
 
Step \(5\): With \(D\) as centre and with \(10\) \(cm\) as radius, draw an arc. Similarly, with \(B\) as centre and with \(4\) \(cm\) as radius, cut the existing arc and mark the intersection as \(C\).
 
1324_34.svg
 
Step \(6\): Now join \(CD\) and \(BC\) to obtain the desired rectangle.
 
1324_35.svg
 
To find the area of the rectangle:
 
\(\text{Area of the rectangle} = \text{Length} \times \text{Breadth}\)
 
\(\text{Area of the rectangle} = l \times b\)
 
We know that, \(l = 10\) \(cm\), and \(b = 4\) \(cm\).
 
Now, \(\text{Area of the rectangle} = 10 \times 4\)
 
\(= 40\) \(cm^2\)
Method \(2\): When a side and a diagonal is known
Let us construct rectangle with a side as \(8\) \(cm\) and the diagonal as \(10\) \(cm\). Let us also calculate the area of the obtained rectangle.
 
Step \(1\): Draw a rough diagram with the known measurements.
 
1324_36.svg
 
Step \(2\): Draw a line segment \(AB\) of length \(8\) \(cm\).
 
1324_37.svg
 
Step \(3\): With \(A\) as centre and with \(10\) \(cm\) as radius, draw an arc.
 
1324_38.svg
 
Step \(4\): With \(B\) as centre, draw a perpendicular line until it meets the arc already drawn. Mark the intersection as \(C\).
 
1324_39.svg
 
Step \(5\): Measure \(BC\). In this case, \(BC = 6\) \(cm\), With \(A\) as centre and with \(6\) \(cm\) as radius, draw an arc. Similarly, with \(C\) as centre and with \(8\) \(cm\) as radius, cut the arc. Mark the intersection as \(D\).
 
1324_40.svg
 
Step \(6\): Join \(AD\) and \(CD\) to form the rectangle.
 
1324_41.svg
 
To find the area of the rectangle:
 
\(\text{Area of the rectangle} = \text{Length} \times \text{Breadth}\)
 
\(\text{Area of the rectangle} = l \times b\)
 
We know that, \(l = 8\) \(cm\), and \(b = 6\) \(cm\).
 
Now, \(\text{Area of the rectangle} = 8 \times 6\)
 
\(= 48\) \(cm^2\)