In parallelogram ABCD AB = BD AD = 12 sinA = 0.8, Find the area of the parallelogram.

univerkov | June 11, 2021 | education

Since, by condition, AB = BD, the triangle ABD is isosceles.

Let’s build the height BH of an equilateral triangle ABD, which will also be the median then AH = DH = AD / 2 = 12/2 = 6 cm.

In the right-angled triangle ABН, we define the cosine of the angle BAН.

Cos2BAH = 1 – Sin2BAH = 1 – 0.64 = 0.36.

CosBAH = 0.6 cm.

Then CosBAH = AH / AB.

AB = AH / CosBAH = 6 / 0.6 = 10 cm.

Then BH ^ 2 = AB ^ 2 – AH ^ 2 = 100 – 36 = 64.

BH = 8 cm.

Then the area of the parallelogram is equal to: Savsd = AD * ВН = 12 * 8 = 96 cm2.

Answer: The area of the parallelogram is 96 cm2.

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