In parallelogram ABCD, angle B is equal to 120 degrees and the bisector of this angle divides side AD
In parallelogram ABCD, angle B is equal to 120 degrees and the bisector of this angle divides side AD into segments AE = 6 and BC = 2. Find the perimeter of the parallelogram. Determine the view of the quadrilateral ABCD and find its perimeter.
Quadrangle ABCD is a parallelogram.
By the property of parallelogram angles, angle A = angle B (internal one-sided at the intersection of parallel straight lines of the third)
Let us find the angle A, if the angle B = 120 °, according to the condition.
Angle A + angle B = 180 °.
Angle A = 180 ° – Angle B = 180 ° – 120 ° = 60 °.
BE – bisector of angle B, halves it.
Angle ABE = 120 °: 2 = 60 °.
Consider triangle ABE, the sum of the angles is 180 °.
Angle A + angle ABE + angle BEA = 180 °.
60о + 60о + angle BEA = 180о.
angle BEA = 180 ° – 120 ° = 60 °.
All angles are equal, triangle ABE is regular, equilateral.
AB = BE = AE = 6 (by condition)
AD = AE + ED = 6 + 2 = 8.
P ABCD = 2 * (AB + AD).
P = 2 * (6 + 8) = 2 * 14 = 28.
Answer: P ABCD = 28, ABCD – – parallelogram.
