Find the perimeter of the rhombus ABCD if the angle is B = 120 degrees, BD = 8 cm

| December 3, 2020 | education

1) BD is a diagonal (like AC), then, by the rhombus property that the diagonals are bisected by the intersection point, BO = OD = 4 cm.
2) Now consider a triangle ABO-rectangular. We know that the angle B = 120 °, however, by the property of the rhombus that the diagonals are also bisectors for angles, the angle ABO = 60 °. Then, knowing that the sum of acute angles in a rectangular triangle is 90 °, we can find the angle BAO = 90 ° -60 ° = 30 °. Opposite the angle of 30 ° lies the leg (BO), equal to half of the hypotenuse (AB). Then AB = 2BO = 2 × 4 = 8 ( cm).
3) Consider triangles ABO and BOC-rectangular:
• side VO-common
• AO side = OS side (according to the diagonals in the rhombus)
Consequently, these triangles are equal in two legs. Then their similar elements are also equal, i.e. AB = BC.
Likewise with triangles DOC and AOD.
4) The perimeter is the sum of the sides of the polygon.
P = a + b + c + d
P (ABCD) = AB + BC + CD + AD = 8 + 8 + 8 + 8 = 32 (cm).
Answer: 32cm.



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