In rhombus ABCD, angle C is 2 times less than angle B. In rhombus ABCD, angle C is 2 times less than angle B

In rhombus ABCD, angle C is 2 times less than angle B. In rhombus ABCD, angle C is 2 times less than angle B. Point M and O are the midpoints of sides AD and DC, respectively. Calculate the area of a rhombus if the area of the MBO triangle is 3 cm2.

Let’s define the angles of the rhombus. Let the size of the acute angle of the rob be equal to X0, then, by condition, the obtuse angle is 2 * X.

The sum of adjacent angles is 180, then X + 2 * X = 180.3 * X = 180.

X = 180/3 = 60.

Angle BAD = BCD = 60, angle ABC = ADc = 120.

Then the triangles ABD and BCD are equilateral, and the medians BM and BO are also the heights and bisectors of the triangles.

Then the angle MbO = 120 – AbM – СВO – 120 – 30 – 30 = 60.

Since ВM = VO, as heights, the PTO triangle is equilateral. The area of ​​an equilateral triangle is equal to: Swam = a2 * √3 / 6, where a is the side of the triangle.

Then 3 * √3 = a * √3 / 4.

a = BM = ВO = MO = 12 = 2 * √3 cm.

Let the side of the rhombus be X cm.

Then, in a right-angled triangle ABM, AM = AD / 2 = X / 2.

On the Pythagorean theorem, AB ^ 2 = AM ^ 2 + BM ^ 2.

X ^ 2 = X ^ 2/4 + (2 * √3) ^ 2.

3 * X ^ 2/4 = 12.

X ^ 2 = 12 * 4/3 = 16.

X = AB = BC = CD = AD = 4.

Determine the area of ​​the rhombus.

Savsd = AD * ВO = 4 * 2 * √3 = 8 * √3 cm2.

Answer: The area of ​​the rhombus is 8 * √3 cm2.