The height lowered from the top of the obtuse angle of the parallelogram, equal to 135 degrees, is 4 cm, and divides the side

The height lowered from the top of the obtuse angle of the parallelogram, equal to 135 degrees, is 4 cm, and divides the side to which it is lowered into two equal parts. Find the perimeter and area of the parallelogram.

Given a parallelogram ABCD: ∠B = 135 °, BH = 4 cm – height to side AD, AH = DH.
1. By the theorem on the sum of the angles of a quadrangle:
∠A + ∠B + ∠C + ∠D = 360 °.
Since ABCD is a parallelogram, then ∠A = ∠C = x, ∠B = ∠D = 135 °, then:
x + 135 ° + x + 135 ° = 360 °;
2 * x = 360 ° – 270 °;
2 * x = 90 °;
x = 90 ° / 2;
x = 45 °.
Thus, ∠A = ∠C = x = 45 °.
2. △ AHB – rectangular: ∠AHB = 90 °, ∠HAB (aka ∠A) = 45 °. By the theorem on the sum of the angles of a triangle:
∠AHB + ∠HAB + ∠ABH = 180 °;
90 ° + 45 ° + ∠ABH = 180 °;
∠ABH = 180 ° – 135 °;
∠ABH = 45 °.
△ AHB – isosceles, then AH = BH = 4 cm.
3. Since AH = DH, then AH = DH = 4 cm. Then:
AD = AH + DH = 4 + 4 = 8 cm.
4. By the Pythagorean theorem:
AB = √ (AH² + BH²) = √ (16 + 16) = √32 = 4 √2 (cm).
5. Since ABCD is a parallelogram, then AB = CD = 4 √2 cm and AD = BC = 8 cm.
Perimeter ABCD:
P = 4 √2 + 8 + 4 √2 + 8 = 16 + 8 √2 (cm).
6. Area ABCD:
S = BH * AD = 4 * 8 = 32 (cm²).
Answer: P = 16 + 8 √2 cm, S = 32 cm².