The diagonals bisect each other, and also meet at an angle of 125°. So, the long side of the parallelogram can be found using the law of cosines.
Let the parallelogram be ABCD and the center be E.
AB^2 = 3^2+4^2 - 2*3*4*cos125°
AB = 6.2262
Now, the altitude of triangle AEB is found using the law of sines:
sin(EAB)/3 = sin(125°)/6.2262
EAB = 23.25°
h/4 = sin 23.25°
h = 1.578
So, the area of the parallelogram is
1/2 * 6.2262 * (2*1.578) = 9.825
Or, using vectors, let
u = 4i
v = 3cos55° i + 3sin55° j
Then the area is
1/2 |u×v| = 9.829
close enough
The diagonal of a parallelogram 6cm and 8cm long and
they intersect at an angle of 55 degrees. calculate the area of
the parallelogram
1 answer
1 answer