In parallelogram ABCD, the diagonal is AC = 10. Find the area of the parallelogram ABCD
In parallelogram ABCD, the diagonal is AC = 10. Find the area of the parallelogram ABCD if the angle BAC = 30 °, angle DAC = 45 °.
Determine the values of the angles of the parallelogram.
Angle BAD = BAC + DAC = 30 + 45 = 750.
Angle ABC = 180 – BAD = 180 – 75 = 1050.
In triangle ABC we apply the theorem of sines.
ВС / Sin30 = AC / Sin105.
BC = AC * Sin30 / Sin105 = 10 * (1/2) / ((√3 + 1) / 2 * √2) = 10 * √2 / (√3 + 1) cm.
Angle BAC = CAD = 45, as cross-lying angles at the intersection of parallel straight lines BC and AD secant SA.
Determine the area of the triangle ABC.
Sас = ВС * АС * SinBCA / 2 = ((10 * √2 / (√3 + 1)) * 10 * √2 / 2) / 2 = 50 / (√3 + 1) cm2.
Since Savs = Sasd, then Savs = 2 8 Savs = 2 * 50 / (√3 + 1) = 100 / (√3 + 1) cm2.
Answer: The area of the parallelogram is 100 / (√3 + 1) cm2.