Answer:
The area of parallelogram ABCD is[tex]78.42 \mathrm{in}^{2}[/tex]
Explanation:
Given:
AD = 12 in
[tex]m \angle C=46^{\circ}[/tex]
[tex]m \angle D B A=72^{\circ}[/tex]
To Find:
The area of parallelogram ABCD=?
Solution:
When we construct the parallelogram with the given data, we get a parallelogram formed by 12 cm as one side and an angle with 46 degrees.
The area of the parallelogram can be calculated by [tex]a b * \sin (a n g l e)[/tex]
Substituting the value of a=12 we have
[tex]\text { Area of parallelogram }=12 * \text { bsin } 46[/tex]
To find the value of b,
We know that area of a triangle can be expressed as,
[tex]\text { Area of triangle }=(A b / 2) \sin (\text {angle})[/tex]
So,
[tex](12 * B D / 2) * \sin 46=(A B * B D / 2) * \sin 72[/tex]
Cancelling BD and 2 on both sides we get,
[tex]12 * \sin 46=A B * \sin 72[/tex]
[tex]A B=12 * \frac{\sin 46}{\sin 72}[/tex]
Therefore,
[tex]b=\frac{12 \sin 46}{\sin 72}[/tex]
Substituting the value of b,
[tex]=12 *\left(\frac{12 \sin 46}{\sin 72}\right) * \sin 46[/tex]
=78.42
So the area of the parallelogram is[tex]78.42 \mathrm{in}^{2}[/tex]
