Draw quadrilateral ABCD with the following measurements. Find also its area.
AB = 6 cm, BC = 6 cm, ∠BAC = 50°, ∠ACD = 30° ,∠CAD = 100°
Given. ABCD is a quadrilateral.
AB = 6 cm, BC = 6 cm, ∠BAC = 50°, ∠ACD = 30° and ∠CAD = 100°
Formula used. Area of triangle = 
× Base × Height
Steps for construction.
Step 1 : Draw a line AB of 6cm
Step 2 : Draw an angle of 50° on point A and extend ray AC’
Step 3 : Make an arc of 6cm from point B and intersect it on AC’
Mark intersection point of arc and ray AC’ as point C
Step 4 : Draw an angle of 30° on point C on AC and extend
ray to CD’
Step 5 : Make an angle of 100° on point A on AC and extend ray to
intersect ray CD’ mark the intersection point as D
Step 6 : Join BD
Step 6 : Draw perpendicular from A and C on BD and name the intersecting point as E and F respectively
Area of quadrilateral = Area of ΔACB + Area of ΔACD
= 
× AE × BD + 
× CF × BD
= 
× BD × (AE + CF)
= 
× 10.5cm × (1.4cm + 5.5cm)
= 36.225cm2
Conclusion. Hence quadrilateral ABCD is constructed and its Area is
36.225cm2