In parallelogram ABCD, AB=10cm, Angle A = 70 gradus, diagonal DB is perpendicular to AB. Find CB, DB, area of ABCD.

Answer 1

angle BDA = 90-70 = 20
law of sines
sin 20 /10 =sin 70/BD
so
BD = 10 sin 70 / sin 20 = 27.5 cm
so area of triangle ABD which is half the area of the parallelogram
= (1/2)(10)(27.5)
so the whole area = 275 cm^2
CB=AB = sqrt(100+27.5^2) because hypotenuse