Can someone plz help me solved this problem I need help plz help me! Will mark you as brainiest!
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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)
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
∠GAB = 22.62°
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
