Answer:
The area of a parallelogram is 360 in.²
Step-by-step explanation:
Where DG = GH
GP = 12 in.
AB = 39 in.
∠DAB + ∠ABC = 180° (Adjacent angles of a parallelogram)
Whereby ∠DAB is bisected by AG and ∠ABC is bisected by BH
Therefore, ∠GAB + ∠HBA = 90°
Hence, ∠BPA = 90° (Sum of interior angles of a triangle)
![cos(\angle GAB) = \dfrac{AP}{AB} = \dfrac{AP}{39} = \dfrac{GP}{GH} =\dfrac{12}{GH}](https://tex.z-dn.net/?f=cos%28%5Cangle%20GAB%29%20%3D%20%5Cdfrac%7BAP%7D%7BAB%7D%20%3D%20%5Cdfrac%7BAP%7D%7B39%7D%20%3D%20%5Cdfrac%7BGP%7D%7BGH%7D%20%3D%5Cdfrac%7B12%7D%7BGH%7D)
We note that ∠AGD = ∠GAB (Alternate angles of parallel lines)
∴ ∠AGD = ∠AGD since ∠AGD = ∠GAB (Bisected angle)
Hence AD = DG (Side length of isosceles triangle)
The bisector of ∠ADG is parallel to BH and will bisect AG at point Q
Hence ΔDAQ ≅ ΔDGQ ≅ ΔGPH and AQ = QG = GP
Hence, AP = 3 × GP = 3 × 12 = 36
![cos(\angle GAB) = \dfrac{AP}{AB} = \dfrac{36}{39}](https://tex.z-dn.net/?f=cos%28%5Cangle%20GAB%29%20%3D%20%5Cdfrac%7BAP%7D%7BAB%7D%20%3D%20%5Cdfrac%7B36%7D%7B39%7D)
![\angle GAB = cos^{-1} \left (\dfrac{36}{39} \right )](https://tex.z-dn.net/?f=%5Cangle%20GAB%20%3D%20cos%5E%7B-1%7D%20%5Cleft%20%28%5Cdfrac%7B36%7D%7B39%7D%20%20%5Cright%20%29)
∠GAB = 22.62°
![cos(\angle GAB) = \dfrac{36}{39} = \dfrac{12}{GH}](https://tex.z-dn.net/?f=cos%28%5Cangle%20GAB%29%20%3D%20%20%5Cdfrac%7B36%7D%7B39%7D%20%3D%20%5Cdfrac%7B12%7D%7BGH%7D)
![GH = \dfrac{39}{36} \times {12}](https://tex.z-dn.net/?f=GH%20%3D%20%20%5Cdfrac%7B39%7D%7B36%7D%20%5Ctimes%20%7B12%7D)
GH = 13 in.
∴ AD 13 in.
BP = 39 × sin(22.62°) = 15 in.
GH = √(GP² + HP²)
∠DAB = 2 × 22.62° = 45.24°
The height of the parallelogram = AD × sin(∠DAB) = 13 × sin(45.24°)
The height of the parallelogram = 120/13 = 9.23 in.
The area of a parallelogram = Base × Height = (120/13) × 39 = 360 in.²