The longer diagonal of a parallelogram measures 62 cm and makes an angle of 30 degrees with the base. Find the area of the parallelogram if the diagonals intersect at angle of 70 degrees.

Answer 1

1405

Answer 2

Since the angle of intersection of the diagonals is 70 degrees then the other angle of intersection of the diagonals is 110 degrees.

180 = 70 + θ
where θ = the other angle of intersection
θ = 110 degrees

A parallelogram makes 4 oblique triangles. One triangle has angles : 30, 40, and 110 degrees.
Since the diagonals intersect at their midpoints, the length of a side of one oblique triangle will measure 31 degrees (which is from 62/2).
By the given values you can solve for half the length of the shorter diagonal using the law of sines.
31/sin40 = x/sin30
where x = the length of one side of the oblique triangle and also half the length of the shorter diagonal.
x = 24.11371932 cm
2(24.11371932) = 48.23 --> length of shorter diagonal

use the formula:
Area of parallelogram = (lenth of longer diagonal x lenth of shorter diagonal x sinθ)/2
where θ = angle of intersection of the two diagonals (either 70 degrees or 110 degrees)

A = ((62 cm)(48.23)sin70)/2
A = 1404.962628 or 1405 cm^2