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skad [1K]
3 years ago
10

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

Mathematics
2 answers:
dmitriy555 [2]3 years ago
6 0

Answer:

The area of   parallelogram ABCD  is78.42 \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

<u>To find  the value of b, </u>

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 is78.42 \mathrm{in}^{2}

Ratling [72]3 years ago
3 0

Answer:I need to see the problem

Step-by-step explanation:

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