Answer:
96.16 square in
Step-by-step explanation:
Given information: In parallelogram ABCD, AD = 12 in, m∠C = 46º, m∠DBA = 72º.
Opposite angles of a parallelogram are congruent.
[tex]m\angle A=m\angle C=72^{\circ}[/tex]
According to the angle sum property, the sum of interior angles of a triangle is 180º.
[tex]\angle ADB=180.00^{\circ}-A-B=180.00^{\circ}-46.00^{\circ}-72.00^{\circ}=62.00^{\circ}[/tex]
Law of Sine:
[tex]\dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}[/tex]
Using Law of Sine we get
[tex]AB=\frac{\sin(\angle ADB)\cdot b}{\sin(\angle ABD)}=\frac{\sin(62.00^{\circ})\cdot 12.00}{\sin(72.00^{\circ})}=11.14[/tex]
It means the base of parallelogram is 11.14 in.
Draw an altitude on AB from D.
In a right angle
[tex]\sin \theta = \frac{opposite}{hypotenuse}[/tex]
In triangle ADE,
[tex]\sin A= \frac{h}{12}[/tex]
[tex]\sin (46)= \frac{h}{12}[/tex]
[tex]\sin (46)\times 12=h[/tex]
[tex]h\approx 8.632[/tex]
The height of the parallelogram is 8.632.
The area of parallelogram is
[tex]Area=base\times height[/tex]
[tex]Area=11.14\times 8.632[/tex]
[tex]Area=96.16[/tex]
Therefore, the area of parallelogram is 96.16 square in.
