Question

In parallelogram ABCD, AD = 12 in, m∠C = 46º, m∠DBA = 72º. Find the area of ABCD

Answer 1

Answer:

The area of   parallelogram ABCD  is$78.42 \mathrm{in}^{2}$

Explanation:

Given:

AD = 12 in

$m \angle C=46^{\circ}$

$m \angle D B A=72^{\circ}$

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 $a b * \sin (a n g l e)$

Substituting the value of a=12  we have

$\text { Area of parallelogram }=12 * \text { bsin } 46$

To find  the value of b,

We know that area of a triangle can be expressed as,

$\text { Area of triangle }=(A b / 2) \sin (\text {angle})$

So,

$(12 * B D / 2) * \sin 46=(A B * B D / 2) * \sin 72$

Cancelling BD and 2 on both sides we get,  

$12 * \sin 46=A B * \sin 72$

$A B=12 * \frac{\sin 46}{\sin 72}$

Therefore,

$b=\frac{12 \sin 46}{\sin 72}$

Substituting the value of b,

$=12 *\left(\frac{12 \sin 46}{\sin 72}\right) * \sin 46$

=78.42  

So the area of the parallelogram is$78.42 \mathrm{in}^{2}$

Answer 2

Answer:I need to see the problem

Step-by-step explanation: