geometry
posted by Katreena .
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.

1405

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