Finding the Area of a Parallelogram

Question

I usually know how to find it, but I can't seem to find an equation or anything that works for this one. The question is: "Find the area of a parallelogram with sides of length 6" and 8" and the diagonals intersect at an angle of 114 degrees."

I can't find a way to get the area given the angle that the diagonals hit. I've done it with diagonal lengths and with an angle at the corner, but not like this. Any help would be appreciated.

Answer 1

Check out the video and message me with any additional questions.

Answer 2

There is a way to find the area of a parallelogram using vector dot products, but I used the Law of Cosines to find the the length of the diagonals.

Since the diagonals intersect at the midpoint of each, labeling the triangles and their intersecting angle of 114 and 66 degrees, I arrived at the following 2 equations:

6^2 = c^2 + d^2 - 2cdCos66

8^2 = c^2 + d^2 - 2cdCos66

This is quite messy so I verified my answers using DESMOS which shows the answers are 2 intersecting circles.

c = 6.495 and d= 3.533 which are half the diagonal lengths.

Therefore 2c and 2d = 12.99 rounded to 13, and 7.066 rounded 7.

The area of a parallelogram is 4 times the area of one of the 4 triangles created by the intersecting diagonals. Since we know all three sides, 6,6.5 and 3.5, Herons formula can be used to calculate A1, then multiply by 4 . A1 by Heron = 3.67 (you can work the details) so

Area of the Parallelogram = 4A1 = 14.7 sq in.

Whew!