The required Area of the parallelogram is 43.868units.
Given that,
ABCD is a parallelogram,
AC = 9cm, DC = 11cm and DAC = 100degree
We have to find,
The area of a parallelogram.
According to the question,
By the law of sin angle,
[tex]\frac{SinA}{a} = \frac{SinD}{d} \\\frac{Sin100}{11} = \frac{SinD}{9} \\\frac{SinD}{9} = 0.0895\\SinD = 53.69[/tex]
D = 53.69
In the triangle sum of all angles in 180degree.
∠A + ∠ACD + ∠D = 180
100 + ∠ACD + 53.69 = 180
∠ACD + 153.69 = 180
∠ACD = 180-153.69
∠ACD = 26.31
The base which is 11. for height make right triangle and angle ACQ is right angle.
[tex]SinACQ = \frac{h}{AC} \\SinACQ = \frac{h}{9} \\sin26.31 = \frac{h}{9} \\0.443 = \frac{h}{9} \\h = 3.98[/tex]
Area of the parallelogram = Base × Height
Area of the parallelogram = (11)(3.98)
Area of the parallelogram = 43.868
Hence, The required Area of the parallelogram is 43.868units.
For more information about Parallelogram click the link given below.
https://brainly.com/question/9680084