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

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

To find the values of CB, DB, and the area of parallelogram ABCD, we can use the given information and apply the properties of parallelograms.

1. Find CB:
Since opposite sides of a parallelogram are equal in length, we know that CB is equal to AB. Therefore, CB = AB = 10 cm.

2. Find DB:
Since diagonal DB is perpendicular to side AB, we have a right triangle formed by sides AB, DB, and diagonal AD. From the given information, we know that AB = 10 cm, and angle A = 70°.

To find DB, we can use trigonometric ratios. Let's use the sine ratio:
sin(A) = opposite/hypotenuse
sin(70°) = DB/AB
sin(70°) = DB/10

Rearranging the equation to solve for DB:
DB = sin(70°) * 10
DB ≈ 9.40 cm (rounded to two decimal places)

3. Find the area of ABCD:
To find the area of parallelogram ABCD, we can use the formula: Area = base * height.

The base of the parallelogram is AB = 10 cm, and the height is the length of the perpendicular from side CB (or AD) to DB.

Since we know that angle A is 70°, the height is DB. Therefore, the area of ABCD is:
Area = base * height
Area = AB * DB
Area = 10 cm * 9.40 cm
Area ≈ 94 cm² (rounded to two decimal places)

So, CB = 10 cm, DB ≈ 9.40 cm, and the area of ABCD ≈ 94 cm².