Draw quadrilateral ABCD with the following measurements. Find also its area.
AB = 5.5 cm, BC = 4.5cm, AC = 6.5 cm, ∠ CAD = 80o and ∠ ACD = 40o .
Draw line AB=5.5 cm
Draw an arc of radius 6.5 cm from point A and an other arc of radius 4.5 cm from point B both meet at point C
Join C to A and B
Draw an angle of 400 at point A and an other angle of 400 at point C. Both cut at point D
Join D to C and A
Hence ABCD is the required quadrilateral.
In triangle ABC,
s=6.5+5.5+4.52=8.25
Then area of ABC=√8.25(8.25−6.5)(8.25−5.5)(8.25−4.5)=√8.25×1.75×2.75×3.75=√148.89=12.20
In triangle ACD
ADsin400=6.5sin1000⇒AD=6.5×0.640.98=4.25cm
Then area triangle ACD =12absinθ=6.5×4.25sin400=27.63×0.64=17.68cm2
So area of ABCD quadrilateral =12.20+17.68=29.88 sq cm