Question

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

Answer 1

96.1666

Explanation:

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Answer 2

96.16 square in

Explanation:

Given information: In parallelogram ABCD, AD = 12 in, m∠C = 46º, m∠DBA = 72º.

Opposite angles of a parallelogram are congruent.

`m\angle A=m\angle C=72^(\circ)`

According to the angle sum property, the sum of interior angles of a triangle is 180º.

`\angle ADB=180.00^(\circ)-A-B=180.00^(\circ)-46.00^(\circ)-72.00^(\circ)=62.00^(\circ)`

Law of Sine:

`(a)/(\sin A)=(b)/(\sin B)=(c)/(\sin C)`

Using Law of Sine we get

`AB=(\sin(\angle ADB)\cdot b)/(\sin(\angle ABD))=(\sin(62.00^(\circ))\cdot 12.00)/(\sin(72.00^(\circ)))=11.14`

It means the base of parallelogram is 11.14 in.

Draw an altitude on AB from D.

In a right angle

`\sin \theta = (opposite)/(hypotenuse)`

In triangle ADE,

`\sin A= (h)/(12)`

`\sin (46)= (h)/(12)`

`\sin (46)* 12=h`

`h\approx 8.632`

The height of the parallelogram is 8.632.

The area of parallelogram is

`Area=base* height`

`Area=11.14* 8.632`

`Area=96.16`

Therefore, the area of parallelogram is 96.16 square in.