Asked by Riliwan
The diagonal of a parallelogram 6cm and 8cm long and
they intersect at an angle of 55 degrees. calculate the area of
the parallelogram
they intersect at an angle of 55 degrees. calculate the area of
the parallelogram
Answers
Answered by
Steve
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
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