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ABCD is a parallelogram. AC = 9cm, DC = 11cm, angle DAC = 100'. Calculate the area of the parallelogram

Sagot :

I was wondering why my previous post was deleted (I attached the same picture below).

I spent a long time drawing and labeling a diagram as well as writing out a detailed formulaic explanation.
View image rsl1121

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