HallsofIvy said:The area of a parallelogram is "base times height". "Base" is the length of one side, b. The height is the length of perpendicular from a vertex to the base. Given angle [itex]\theta[/itex] the height is the "opposite side" while the length of that side, a, is the hypotenuse, h, so [itex]h= a sin(\theta)[/itex] The area is [itex]ab sin(\theta)[/itex].
insightful said:I got a solution using the Law of Cosines with the angle of intersection of the diagonals to get the lengths of the sides.
No, you can use some simple trig in one quadrant of the parallelogram to get the angle (in terms of A and diagonals d1 and d2). Remember, each quadrant has 1/4 the total area A.tiggertime said:So you were given the angle of intersection?
A parallelogram is a quadrilateral with two pairs of parallel sides. This means that the opposite sides are equal in length and parallel to each other.
To find the angles of a parallelogram, you can use the formula A = 180 - 2B, where A represents the measure of one angle and B represents the measure of the adjacent angle.
Yes, you can find the angles of a parallelogram if you know the length of one side. You can use the formula A = 180 - 2B, where A represents the measure of one angle and B represents the measure of the adjacent angle. You will need to know the length of at least one side and one angle in order to solve for the other angle.
No, it is not possible to have a parallelogram with all equal angles. A parallelogram must have two pairs of opposite angles that are equal, but the other two angles will always be different.
No, the Pythagorean Theorem cannot be used to find the angles of a parallelogram. This theorem only applies to right triangles, which a parallelogram is not.