draw a rectangle ABCD, bc=5.2cm and angle dbc = 30 degree

Dear Student,

Please find below the solution to the asked query:

Given : $\angle$ DBC  =  30$°$ , $\angle$ ACB  =  30$°$  as we know angle between same base and diagonals are same . 
We follow these steps  :

Step 1 :  Draw a line BC  =   5.2 cm

Step 2 :  Take any radius ( Less than half of BC )  and center " B "  draw a semicircle that intersect our line BC at " P " . Now with same radius and center " P " draw an arc that intersect our semicircle at " Q " . With same radius and center " Q " draw an arc that intersect our semicircle at " R " . Now with same radius and center " Q " and " R " draw an arcs these arcs intersect at " S " .

Step 3 :  Join BS , we get $\angle$ CBS  = 90$°$

Step 4 :  Take any radius ( Less than half of BC )  and center " C "  draw a semicircle that intersect our line BC at " E " . Now with same radius and center " E " draw an arc that intersect our semicircle at " F " . With same radius and center " F " draw an arc that intersect our semicircle at " G " . Now with same radius and center " F " and " G " draw an arcs these arcs intersect at " H " .

Step 5 :  Join BJ  , we get $\angle$ ABH  = 90$°$ .

Step 6 : With same radius ( Used in step 2 ) and center " P "  and " Q "  draw two arcs and these arcs intersect at " I " . Join BI and extend it so that line intersect line CH at " D " .

Step 7 : With same radius ( Used in step 4 ) and center " E "  and " F "  draw two arcs and these arcs intersect at " J " . Join CJ and extend it so that line intersect line BS at " A " .

Step 8 : Join Ad and we get our required rectangle ABCD , As :



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