Re: Parallelogram ABCD lies in the xy plane
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04 May 2020, 01:58
There is an easier solution, where we need to connect the diagonal AC and find it's length, which is relatively easy with the elaborate coordinate system given. We can use the distance formula between A,C points, or use an easier technique, forming a small right triangle with A,C by extending lines from the points parallel to both X and Y axes. Let's assume the point of intersection is X. So AX = 1, CX = 1 from the graph, so AC is the hypotenuse which is root(2).
We can calculate the height of the 2 triangles BDC and BDA of the parallelogram by taking half of AC, which is root(2) / 2.
The base BD of the 2 triangles of the parallelogram can also be found out either by the distance formula, or the right angle triangle method. It's 7root(2).
Area of triangle BDC = 1/2 * base * height = 1/2 * BD * root(2)/2 = 1/2 * 7root(2) * root(2)/2 which is 7/2. So the area of the parallelogram is twice the area of the triangle, so 7 * (7/2) = 7.