+41 The diagonals of a parallelogram intersect at a 45 degree angle. If the diagonal lengths are 8 cm and 10 cm, what is the area of the parallelogram. Would appreciate an answer without trig.
Thanks
+1713 Are you DurgneyDabs? I can't really tell- try logging back in.
+129852 Well.....if we can just use a "slight" bit of trig.....
The diagonals of a parallelogram bisect each other.....so.........these diagonals form four cogruent triangles
Each of these triangles will have two sides with lengths of 4 and 5 and an included angle between these sides of 45°
And the area of one of these triangles = (1/2) (product of these sides) * sine of the included angle
So.....the area of one of the triangles = (1/2) (4 * 5) * sin (45°)
[ sin (45° = √2 / 2 ]....so we have
(1/2) ( 4 * 5) ( √2/2)
(1/4) (20) * √2
And since we havefour of these triangles....the total area is
4 (1/4) * (20) * √2 =
20√2 units^2
