The bisector EH of an obtuse angle E is drawn in the parallelogram FEOP. Point H lies on the segment FP.

| September 9, 2021 | education

The bisector EH of an obtuse angle E is drawn in the parallelogram FEOP. Point H lies on the segment FP. Find the sides and angles of the parallelogram if FH is 3 dm less than HP, and the parallelogram perimeter is 66 dm and the FEH angle is 70 °.

Determine the value of the angle FEO. Since EH is the bisector of the angle, then the angle FEO = 2 * FEH = 2 * 70 = 140. Since the sum of the adjacent angles of the parallelogram is 180, the angle EFH = 180 – FEO = 180 – 140 = 40, then the angle FHE = 180 – 70 – 40 = 70.

In a triangle FEH, the angle FEH = FHE, then the triangle FEH is equilateral FE = FH.

Let the segment НР = X cm, then, by condition, FH = X – 3 dm, then FE = X – 3 dm.

Determine the perimeter of the parallelogram P = 2 * (EF + FP) = 2 * (X – 3 + X – 3 + X) = 66 dm.

3 * X – 6 = 33.

3 * X = 39.

X = 39/3 = 13 dm.

Then FE = OP = 13 – 3 = 10 dm, FP = EO = 13 + 13 – 3 = 23 dm.

Answer: The angles of the parallelogram are 140 and 40, the sides are 10 inches and 23 inches.



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