Practice Parallelograms

Practice Parallelograms A parallelogram is a quadrilateral whose opposite sides are parallel and which has twopairs of congruent adjacent sides. The opposite sides are called the bases of theparallelogram. The line that connects the midpoints of the pairs of opposite sides, calledthe diagonal, is also a base of the parallelogram. The area and perimeter of a rectangularparallelogram can be found using formulas similar to those used for triangles. The determinant of a matrix is a number that can be computed from the elements of thematrix. It is used in various calculations and is useful to know when solving systems oflinear equations, among other things. Determinants are used in many areas ofmathematics. They are used to find the volume of a parallelepiped, solve systems of linearequations and find eigenvalues. Determinants can also be used to find the area under acurve and find the centroid (center of mass) of a triangle or quadrilateral by taking theaverage of all its vertices. Let's begin working on this problem. The first thing we need to be careful about is knowingthat we want to take a determinant, but we need to be careful. Determinants of pairs ofvectors make sense, but those of points do not. In this example, we have four points that form the basis for a parallelogram. To calculatethe area of the parallelogram, we need to compute vectors from these points. So we have taken the vectors that connect the endpoints of the parallelogram. You will seethat vector 6, 1 is coming from point 1, 1 in the original parallelogram and 7, 2. So, vector 6, 1 is equivalent to the difference between point 7, 2 and point 1, 1. The sameway, vector 5, 2 is equivalent to the difference of the original point 6, 3 and the base point1, 1. Now that we have these two vectors, the area of our parallelogram is going to be equal tothe determinant of the two vectors. We should be careful because the area will be determined by the value of the determinant(positive or negative). It is time to compute this determinant. Now we can find six times two minus five. This gives us twelve minus five, which is seven.It is a positive one. This plus or minus, we take to be positive. If we had computed our determinant by transposing the rows, then we might have found anegative 7. To area be positive, we would choose 7. The tricky part of this problem is that the original endpoints of our parallelogram are notwhat is important for figuring out its area. What we need to find are the vectors connectingthose original endpoints together. We computed the values of 6, 1 and 5, 2 and then taking their determinant resulted in thearea of the parallelogram. That’s it.