We shall explain this construction through an example .
Example : Construct a quadrilateral PQRS where PQ = 4 cm,QR = 6 cm, RS = 5 cm, PS = 5.5 cm and PR = 7 cm.
A rough sketch will help us in visualising the quadrilateral.
Step 1 - From the rough sketch, it is easy to see that ∆PQR can be constructed using SSS construction condition. Draw ∆PQR in fig.
Step 2 - Now, we have to locate the fourth point S. This ‘S’ would be on the side opposite to Q with reference to PR. For that, we have two measurements.
S is 5.5 cm away from P. So, with P as centre, draw an arc of radius 5.5 cm. (The point S is somewhere on this arc)fig.
Step 3: S is 5 cm away from R. So with R as centre, draw an arc of radius 5 cm (The point S is somewhere on this arc also)in following fig.
Step 4 : S should lie on both the arcs drawn. So it is the point of intersection of the two arcs. Mark S and complete PQRS. PQRS is the required quadrilateral in following fig .
Shaalaa.com | Quadrilateral with 4 sides & 1 diagonal
Construct the following quadrilaterals.
JU = 3.5 cm
UM = 4 cm
MP = 5 cm
PJ = 4.5 cm
PU = 6.5 cm